# Abstract Nonsense

## Munkres Chapter two Section 12 & 13: Topological Spaces and Bases

Point of post: This is the solutions to Munkres Chapter two Section 12 as the heading indicates.

1.

Problem: Let $X$ be a topological space, let $A\subseteq X$. Suppose that for each $x\in A$ there is an open set $U$ containing $x$ such that $U\subseteq A$. Show that $A$ is open.

September 30, 2010

## Munkres Chapter one Section three: Relations

1.

Problem: Define two points $(x_0,y_0)$ and $(x_1,y_1)$ of the pane to be equivalent if $y_0-x_0^2=y_1-x_1^2$. Check that this is an equivalence relation and describe the equivalence classes

Proof:

Reflexivity: Clearly $y_0-x_0^2=y_0-x_0^2$

Symmetry: If $(x_0,y_0)\sim (x_1,y_1)$ then

$y_0-x_0^2=y_1-x_1^2\implies y_1-x_1^2=y_0-x_0^2$

and thus $(x_1,y_1)\sim(x_0,y_0)$

Transitivity: If $(x_0,y_0)\sim(x_1,y_1)$ then $y_0-x_0^2=y_1-x_1^2$ and $(x_1,y_1)\sim(x_2,y_2)$ implies that $y_1-x_1^2=y_2-x_2^2$. But, concatenating these inequalities and cutting out the middle gives $y_0-x_0^2=y_2-x_2^2$ and so $(x_0,y_0)\sim(x_2,y_2)$.

2.

Problem: Let $C$ be a relation on a set $A$. If $A_0\subseteq A$, define the restriction of $C$ to $A_0$ to be the relation $C\cap\left(A_0\times A_0\right)$. Show that the restriction of an equivalence relation is an equivalence relation

Proof:

Reflexivity: Note that for each $x\in A_0$ it’s true that $x\in A$ and so $(x,x)\in C$ (by $C$‘s reflexivity) and so $(x,x)\in C\cap\left(A_0\times A_0\right)$

Symmetry: If $(x,y)\in C\cap\left(A_0\times A_0\right)$ then $(x,y)\in C\text{ and }(x,y)\in A_0\times A_0$ which says that $(y,x)\in C\text{ and }(y,x)\in A_0\times A_0$ and thus $(y,x)\in C\cap\left(A_0\times A_0\right)$

Transitivity: If $(x,y),(y,z)\in C\cap\left(A_0\times A_0\right)$ then we first note that $x,y,z\in A_0$. Furthermore, since $(x,y),(y,z)\in C$ we see that $C$‘s transitivity implies that $(x,z)\in C$ and so by previous comment $(x,z)\in C\cap\left(A_0\times A_0\right)$

3.

Problem: Here is a proof that every relation $C$ that is both symmetric and transitive is also relfexive

“Since $C$ is symmetric, $x\sim y\implies y\sim x$. Since $C$ is transitive $x\sim y\text{ and }y\sim x\implies x\sim x$

What’s the problem?

Proof: He assumed there is a $y$ such that $x\sim y$

4.

Problem: Let $f:A\to B$ be surjective function. Let us define a relation on $A$ by setting

$a_0\sim a_1$ if $f(a_0)=f(a_1)$

a) Show that this is an equivalence relation

b) Let $A^*=\left\{E(x):x\in A\right\}$ (set of all equivalence classes) Show there is a bijective correspondence between $A^*$ and $B$

Proof:

a)

Reflexivity: Clearly, since $f$ is a function, $a_0=a_1\implies f(a_0)=f(a_1)$ so that $a_0\sim a_1$

Symmetry: If $f(a_0)=f(a_1)$ then evidently $f(a_1)=f(a_0)$ from where symmetry immediately foll0ws.

Transitivity: Equally easy, if $f(a_0)=f(a_1),f(a_1)=f(a_2)$ then $f(a_0)=f(a_2)$

b) Define

$\eta:A^*\to B:E(x)\mapsto f(x)$

First, it is not clear that this is actually a function (the rule of assignment implicitly stated above). To see this suppose that

$E(x)=E(y)$

then

$\eta\left(E(x)\right)=f(x)=f(y)=\eta\left(E(y)\right)$

thus $\eta$ is a well-defined function. To see that this function is bijective we first note that if

$f(x)\ne f(y)\implies x\not\sim y\implies E(x)\ne E(y)$

and then by the surjectivity of $f$ for every $y\in B$ there is some $x\in A$ such that $f(x)=y$. Thus, $\eta\left(E(x)\right)=y$ and so $\eta$ is surjective.

Remark: All we’ve done is identified the fibers together

5.

Problem: LEt 4latex S$and $S'$ be the following subsets of $\mathbb{R}^2$ $S=\left\{(x,y)\in\mathbb{R}^2:y=x+1\text{ and }0 and $S'=\left\{(x,y)\in\mathbb{R}^2:y-x\in\mathbb{R}\right\}$ a) Show that $S'$ is an equivalence relation on $\mathbb{R}$ and $S'\supseteq S$. Describe the equivalence classes of $S'$ b) Show that given any collection of equivalence relations on a set $A$, their intersection is an equivalence relation Proof: a) Reflexivity: Clearly $x-x=0$ for every $x\in\mathbb{R}$ Symmetry: This follows since $z\in\mathbb{Z}\implies -z\in\mathbb{Z}$ Transitivity: Clearly, if $x\sim y,y\sim z$ then $z-y=(z-x)-(x-y)$ and thus $z-y$ is the difference of two integers and so itself and integer. To see that $S\subseteq S'$ we merely note that $y=x+1\implies y-x=1\in\mathbb{Z}$ b) Let $\left\{R_\alpha\right\}_{\alpha\in\mathcal{A}}$ be a nonempty class of equivalence relations and let $\displaystyle \Omega=\bigcap_{\alpha\in\mathcal{A}}R_{\alpha}$ To see that $\Omega$ is an equivalence relation on $\mathbb{R}$ we first note that since $(x,x)\in R_\alpha,\text{ }\forall \alpha\in\mathcal{A}$ that $(x,x)\in\Omega$. Next, we note that if $(x,y)\in\Omega$ then $(x,y)\in R_\alpha,\text{ }\forall \alpha\in\mathcal{A}$ and so $(y,x)\in R_\alpha,\text{ }\forall\alpha\in\mathcal{A}$ and so $(y,x)\in\Omega$. Lastly, if $(x,y),(y,z)\in\Omega$ then $(x,y)(y,z)\in R_{\alpha},\text{ }\forall\alpha\in\mathcal{A}$ and so $(x,z)\in R_{\alpha},\text{ }\forall \alpha\in\mathcal{A}$ and so $(x,z)\in\Omega$ 6. Problem: Define a relation on $\mathbb{R}^2$ by setting $(x_0,y_0)<(x_1,y_1)$ if either $y_0-x_0^2, or $y_0-x_0^2=y_1-x_1^2$ and $x_0. Show that this an order relation on the plane. Proof: Comparable: Since the definition of the relation took arbitrary points in $\mathbb{R}^2$ and compared them, it’s clearly a linear ordering. Non-reflexive: This is clear since $x_0\not Transitive: Suppose that $(x_0,y_0)<(x_1,y_1)$ and $(x_1,y_1)<(x_2,y_2)$. There are several cases all of which are trivial. 7. Problem: Show that the restriction of an order relation is an order relation. Proof: Let $R$ be an order relation on $A$ and $A_0\subseteq A$. Comparability: Let $x,y\in A_0$ then $x,y\in A$ and so $(x,y)\in C$ or $(y,x)\in C$, and so $(x,y)\in C\cap\left(A_0\times A_0\right)$ or $(y,x)\in C\cap\left(A_0\times A_0\right)$ Non-reflexivity: If $x then $(x,x)\in C\cap\left(A_0\times A_0\right)\subseteq C$ which is a contradiction. Transitivity: This is identical to the transitivity proof for restrictions of equivalence relations. 8. Problem: Check that the relation $R\subseteq\mathbb{R}^2$ given by $x<_R y$ iff $x^2 or if $x^2=y^2$ then $x. Proof: Comparability: Let $x,y\in\mathbb{R}$. If $y=-x,\text{ }x\geqslant 0$ then $y<_R x$ and if $y\ne -x$ then we have by the comparability of the usual ordering on $\mathbb{R}$ that either $y^2 or $x^2 Non-reflexivity: Since $x\not< x$ and $x^2\not non-reflexivity is immediate. Transitivity: This is just several simple cases again. 9. Problem: Check that if $\left(A,<_A\right)$ and $\left(B,<_B\right)$ are two ordered sets, then the dictionary order $<_D$ is an order relation on $A\times B$ Proof: This is also relatively menial, but so important that it needs to be done: Comparability: Let $(a,b),(c,d)\in A\times B$ be arbitrary. If $a\ne c$ then the comparability of $<_A$ implies that either $a<_A c$ or $c<_A a$, and so either $(a,b)<_D (c,d)$ or $(c,d)<_D (a,b)$. If $x=y$ the comparability of $<_B$ implies that either $b<_B d$ or $d<_B b$ and so either $(a,b)<_D (c,d)$ or $(c,d)<_D (a,b)$ Non-reflexivity: Note that by the non-reflexivity of $<_B$ we see that $b\not <_B b$ and so $(a,b)\not<_D(c,d)$ Transitivity: Suppose that $(a,b)<_D (c,d)$ and $(c,d)<_D (e,f)$. Now, if $a=c$ then $(a,b)<_D (c,d)\implies b<_B c$. So, if $c=e$ we see that $(c,d)<_D (e,f)\implies d<_B f$ and so $b<_B f$ and so $(a,b)<_D (e,f)$. If $c\ne e$ then $(c,d)<_D (e,f)\implies a=c<_A e$ and so $(a,b)<_D (e,f)$. Now, if $a\ne c$ then $(a,b)<_D (c,d)\implies a<_A c$. If $c=e$ then $a<_A c=e$ and so $(a,b)<_D (e,f)$. If $c\ne e$ then $(c,d)<_D (e,f)\implies c<_A e$ and th same conclusion follows. 10. Problem: a) Show that the map $f:(-1,1)\to\mathbb{R}$:x\mapsto\frac{x}{1-x^2}$  is order preserving.

b) Show that the equation $g(y)=\frac{2y}{1+\sqrt{1+4y^2}}$ defines an inverse for $f$

Proof:

a) Suppose that $0 then $-x^2>-y^2$ and so $1-x^2>1-y^2$ and so $\frac{1}{1-x^2}<\frac{1}{1-y^2}$ and so finally $\frac{x}{1-x^2}<\frac{y}{1-y^2}$. Next, suppose that $y then $-y>-x>0$ and so the above shows that $\frac{-x}{1-x^2}<\frac{-y}{1-y^2}$ and so $\frac{x}{1-x^2}>\frac{y}{1-y^2}. If$latex x>0,y<0$this obvious. b) This is just function composition and is simple. 11. Problem: Show that an element of an ordered set has at most one immediate successor and at most one immediate predecessor. Show that a subset of an ordered set has at most one smallest element and at most one largest element. Proof: Let $a,b\in A$ and suppose that $a$ and $b$ are both immediate predecessor to $c$. Suppose that $a, then $b\in(a,c)$ contradicting $a$‘s status as an immediate predecessor. The proof for immediate successor is similar. Let $\alpha,\beta$ both be least elements of a set $A$. Then, since $\alpha\in A$ we have that $\beta\leqslant\alpha$. But, since $\beta\in A$ we have that $\alpha\leqslant \beta$. Since $\alpha<\beta$ and $\beta<\alpha$ is impossible, it follows that $\alpha=\beta$. The proof for largest elements is similar. 12. I’ve done this, but I don’t feel like writing it up. I will at request though. 13. Problem: Prove the following theorem: ” If an ordered set $A$ has the l.u.b. property, then it has the g.l.b. property.” Proof: Let $A$ have the l.u.b. property and let $\varnothing\subsetneq E\subseteq A$ be bounded below. Consider the set $L=\left\{x\in A:x\leqslant e,\text{ }\forall e\in E\right\}$. Then, $L\ne\varnothing$ (since $E$ is bounded below) and it’s bounded above (by any element of $E$). Thus, by assumption $\sup L=\alpha$ exists. We claim that $\alpha=\inf E$. To see this, first note that $\sup L\in L$, since otherwise there is some $e\in E$ such that $e and so clearly $e$ is an upper bound for $L$ which is smaller than $\alpha$, contradicting it’s definition as $\sup L$. Now, to see that $x\leqslant \alpha$ for every $x\in L$ we suppose not, then there is some $x_0\in L$ such that $\alpha, but this contradicts that $\alpha$ is an upper bound for $L$. It follows that $\alpha\geqslant x,\text{ }\forall x\in L$ and that $\alpha\in L$. Thus, $\alpha=\min L$ and so $\alpha=\inf E$. Remark: That may have been poorly written (I’m in a rush to finish these), if you have any questions just ask. 14. Problem: If $C$ is a relation on a set $A$, define a new relation on $A$ by letting $(b,a)\in D$ if $(a,b)\in C$ a) Show that $C$ is symmetric iff $C=D$ b) Show that if $C$ is an order relation, then $D$ is an order relation. c) Prove the converse of the theorem in exercise thirteen. Proof: a) This is absolutely clear. If $(a,b)\in C$ then $(b,a)\in D=C$. b) Comparability: Let $x,y\in A$ be arbitrary. We may assume WLOG that $(x,y)\in C$ and so $(y,x)\in D$ Non-reflexivity: Suppose that $(x,x)\in D$ then $(x,x)\in C$, which is a contradiction. Transitivity: Let $(x,y),(y,z)\in D$ be arbitrary, then $(z,y),(y,x)\in C$ and so $(z,x)\in C$ and therefore $(x,z)\in D$ c) It’s pretty much the same procedure. For a non-empty bounded above set $E$ we have that $U=\left\{x\in A:x\geqslant e,\text{ }\forall e\in E\right\}$ is non-empty and bounded below. Thus, $\inf U$ exists and using the same logic as before one can show that $\inf U=\sup E$. 15. Problem: Assume that the real line has the l.u.b. property. a) Show that the sets $[0,1]$ and $[0,1)$ have the l.u.b. b) Does $[0,1]^2$ in the dictionary order have the l.u.b? What about $[0,1]\times[0,1]$? What about $[0,1)\times[0,1]$ Proof: a) Let $E\subseteq[0,1]$ be bounded above by $x\leqslant 1$. By the l.u.b property of $\mathbb{R}$ we have that$$\sup_{\mathbb{R}}[0,1]=\alpha$ exists. Clearly, we have that $0\alpha$ and $\alpha\leqslant 1$, thus $\alpha\in [0,1]$

Let $E\subseteq [0,1)$ be nonempty and bounded above by $x$. We have that $\frac{x+1}{2}<1$ and certainly $\sup_{\mathbb{R}}(E)=\alpha<\frac{x+1}{2}$.

b) Ask me for these answers if there’s a strong desire too.

September 25, 2010

## Munkres Chapter one Section two: Functions

1.

Problem: Let $f:A\to B$. Let $A_0\subseteq A$ and $B_0\subseteq B$

a) Show that $A_0\subseteq f^{-1}\left(f\left(A_0\right)\right)$ and that equality holds iff $f$

b) Show that $f\left(f^{-1}\left(B_0\right)\right)\subseteq B_0$ and that equality holds iff $f$ is surjective

Proof:

a) Let $x\in A_0$, then$latex f(x)\in f\left(A_0\right)$ and so $x\in f^{-1}\left(f\left(A_0\right)\right)$. Now, suppose that $f$ is injective and let $x\in f^{-1}\left(f\left(A_0\right)\right)$ clearly then $f(x)\in f\left(A_0\right)$. It follows by injectivity that $x\in A_0$ (if this isn’t apparent, note that by definition $f(x)\in f\left(A_0\right)$ means that $f(x)=f(y)$ for some $y\in A_0$. Now, by injectivity we see that $x=y$ and so the result becomes clear). Conversely, suppose that $A_0=f^{-1}\left(f\left(A_0\right)\right)$ for every $A_0\subseteq A$. Then, in particular we see that $\{x\}=f^{-1}\left(f(\{x\})\right)$ and so

$x\ne y\implies y\notin \{x\}=f^{-1}\left(f(\{x\})\right)\implies f(y)\ne f(x)$

from where injectivity follows.

b) Let $f(x)\in f\left(f^{-1}\left(B_0\right)\right)$, then $f(x)=f(y)$ for some $y\in f^{-1}\left(B_0\right)$. It follows then that $f(y)=f(x)\in B_0$.

Now, suppose that $f$ is surjective and let $f(x)\in B_0$. Then, $x\in f^{-1}\left(B_0\right)$ and so $f(x)\in f\left(f^{-1}\left(B_0\right)\right)$. Conversely, suppose that $f\left(f^{-1}\left(B_0\right)\right)=B_0$ for every $B_0\subseteq B$. Then, in particular we see that $f\left(A\right)=f\left(f^{-1}\left(B\right)\right)=B$ and surjectivity follows.

2.

Problem: Let $f:A\to B$ and let $A_i\subseteq A$ and $B_i\subseteq B$ for $i=0,1$. Show that $f^{-1}$ preserves inclusions, unions, intersections, and differences of sets:

a) $B_0\subseteq B_1\implies f^{-1}\left(B_0\right)\subseteq f^{-1}\left(B_1\right)$

b) $f^{-1}\left(B_0\cup B-1\right)=f^{-1}\left(B_0\right)\cup f^{-1}\left(B_1\right)$

c) $f^{-1}\left(B_0\cap B_1\right)=f^{-1}\left(B_0\right)\cap f^{-1}\left(B_1\right)$

d) $f^{-1}\left(B_0-B_1\right)=f^{-1}\left(B_0\right)-f^{-1}\left(B_1\right)$

Show that $f$ preserves inclusions and unions only:

e) $A_0\subseteq A_1\implies f\left(A_0\right)\subseteq f\left(A_1\right)$

f) $f\left(A_0\cup A_1\right)=f\left(A_0\right)\cup f\left(A_1\right)$

g) $f\left(A_0\cap A_1\right)\subseteq f\left(A_0\right)\cap f\left(A_1\right)$; show that equality holds iff $f$ is injective

h) $f\left(A_0-A_1\right)\supseteq f\left(A_0\right)-f\left(A_1\right)$; show that equality holds iff $f$ is injective

Proof:

a) Let $x\in f^{-1}\left(B_0\right)$, then $f(x)\in B_0$ and so $f(x)\in B_1$ and thus $x\in f^{-1}\left(B_1\right)$

b) See number three

c) See number three

d) Let $x\in f^{-1}\left(B_0-B_1\right)$ then $f(x)\in B_0-B_1$, or $f(x)\in B-0\text{ and }f(x)\notin B_1$ it follows that $x\in f^{-1}\left(B_0\right)$ and $x\notin f^{-1}\left(B_1\right)$ (otherwise $f(x)\in f\left(f^{-1}\left(B_1\right)\right)\subseteq B_1$). This is equivalent to saying that $x\in f^{-1}\left(B_0\right)-f^{-1}\left(B_1\right)$. Conversely, let $x\in f^{-1}\left(B_0\right)-f^{-1}\left(B_1\right)$, then $x\in f^{-1}\left(B_0\right)$ and $x\notin f^{-1}\left(B_1\right)$. Thus, $f(x)\in f\left(f^{-1}\left(B_0\right)\right)\subseteq B_0$ and $f(x)\notin B_1$. It follows that $f(x)\in B_0-B_1$ or $x\in f^{-1}\left(B_0-B_1\right)$

e) Let $f(x)\in f\left(A_0\right)$ then $f(x)=f(y)$ for some $y\in A_0$ and thus (since $y\in A_1$) we see that $f(x)=f(y)$ for some $y\in A_1$ which is equivalent to saying that $f(x)\in f\left(A_1\right)$.

f) See number three

g) See number three for the first part. Clearly, it suffices to prove the second part for two sets. So, suppose that $f$ is injective and let $f(x)\in\left(A_0\cap A_1\right)$ then $x\in A_0\cap A_1$ and so $x\in A_0$ and $x\in A_1$ and so $f(x)\in f\left(A_0\right)$ and $f(x)\in f\left(A_1\right)$, or $f(x)\in f\left(A_0\right)\cap f\left(A_1\right)$. The proposed equality follows.

Conversely, suppose that $f\left(A_0\cap A_1\right)=f\left(A_0\right)\cap f\left(A_1\right)$ for every $A_0,A_1\subseteq A$. We see then in particular the following line of reasoning

$x\ne y\implies \{x\}\cap\{y\}=\varnothing$

and thus

$\varnothing=f\left(\varnothing\right)=f\left(\{x\}\cap\{y\}\right)=f\left(\{x\}\right)\cap f\left(\{y\}\right)=\{f(x)\}\cap\{f(y)\}$

and clearly singletons are disjoint iff they’re single elements are not equal. Thus, condensing

$x\ne y\implies f(x)\ne f(y)$

and injectivity follows.

h) Let $f(x)\in f\left(A_0\right)-f\left(A_1\right)$, then $f(x)\in f\left(A_0\right)$ and $f(x)\notin f\left(A_1\right)$. This, in particular says that $f(x)=f(y)$ for some $y\in A_0$ and $y\notin A_1$ (otherwise $f(y)=f(x)\in f\left(A_1\right)$. Thus, $f(x)=f(y)$ for some $y\in A_0-A_1$, or $f(x)\in f\left(A_0-A_1\right)$.

Now, suppose that $f$ is injective and let $f(x)\in f\left(A_0-A_1\right)$, then by injectivity $x\in A_0-A_1$ and so $x\in A_0$ and $x\notin A_1$. Thus, $f(x)\in f\left(A_0\right)$ and $f(x)\notin f\left(A_1\right)$. So, $f(x)\in f\left(A_1\right)-f\left(A_0\right)$. Conversely, if $f\left(A_0-A_1\right)=f\left(A_0\right)-f\left(A_1\right)$ then we see that

$f(x)=f(y)\implies \varnothing=\{f(x)\}-\{f(y)\}=f(\{x\}-f(\{y\})=f\left(\{x\}-\{y\}\right)$

but $f(A)=\varnothing$ iff $A=\varnothing$. It follows that $\{x\}-\{y\}=\varnothing$ and so $x=y$.

3.

Problem: Show that b), c), f), and g) of the last exercise hold for arbitrary unions and intersections

Proof:

a) Let $\displaystyle x\in f^{-1}\left(\bigcup_{\alpha\in\mathcal{A}}U_\alpha\right)$ then $\displaystyle f(x)\in\bigcup_{\alpha\in\mathcal{A}}U_\alpha$. Thus, there exists some $\alpha_0\in\mathcal{A}$ such that $f(x)\in U_{\alpha_0}$ and so $x\in f^{-1}\left(U_{\alpha_0}\right)$. Therefore, $\displaystyle x\in\bigcup_{\alpha\in\mathcal{A}}f^{-1}\left(U_{\alpha}\right)$.

Conversely, if let $x\in\bigcup_{\alpha\in\mathcal{A}}f^{-1}\left(U_{\alpha}\right)$, then $x\in f^{-1}\left(U_{\alpha_0}\right)$ for some $\alpha_0\in\mathcal{A}$. Therefore, $f(x)\in U_{\alpha_0}$ so that $\displaystyle f(x)\in\bigcup_{\alpha\in\mathcal{A}}U_{\alpha}$. So finally, we may conclude that $\displaystyle x\in f^{-1}\left(\bigcup_{\alpha\in\mathcal{A}}U_\alpha\right)$

c) We merely need note that

$\displaystyle A-f^{-1}\left(\bigcap_{\alpha\in\mathcal{A}}U_{\alpha}\right)=f^{-1}\left(B-\bigcap_{\alpha\in\mathcal{A}}U_{\alpha}\right)=f^{-1}\left(\bigcup_{\alpha\in\mathcal{A}}\left(B-U_{\alpha}\right)\right)$

(where the first equality is gotten noticing that $A=f^{-1}\left(B\right)$ and using property d) from the previous problem). Then, recalling the last problem we can see that this is equal to

$\displaystyle \bigcup_{\alpha\in\mathcal{A}}f^{-1}\left(B-U_{\alpha}\right)=\bigcup_{\alpha\in\mathcal{A}}\left(f^{-1}\left(B\right)-f^{-1}\left(U_{\alpha}\right)\right)=A-\bigcap_{\alpha\in\mathcal{A}}f^{-1}\left(U_{\alpha}\right)$

and so comparing the LHS of the first equality with the RHS of the last leads to the desired result.

f) Let $\displaystyle f(x)\in f\left(\bigcup_{\alpha\in\mathcal{A}}U_{\alpha}\right)$. Then, $f(x)=f(y)$ for some  $\displaystyle y\in\bigcup_{\alpha\in\mathcal{A}}U_{\alpha}$. Thus, $f(x)=f(y)$ for some $y\in U_{\alpha_0}$ for some $\alpha_0\in \mathcal{A}$. Thus, $f(x)\in f\left(U_{\alpha_0}\right)$ and so $\displaystyle f(x)\in\bigcup_{\alpha\in\mathcal{A}}f\left(U_{\alpha}\right)$.

Conversely, let $\displaystyle f(x)\in\bigcup_{\alpha\in\mathcal{A}}f\left(U_{\alpha}\right)$. Then, $f(x)\in f\left(U_{\alpha_0}\right)$ for some $\alpha_0\in\mathcal{A}$. It follows that $f(x)=f(y)$ for some $y\in U_{\alpha_0}$ and so $f(x)=f(y)$ for some $\displaystyle y\in\bigcup_{\alpha\in\mathcal{A}}U_{\alpha}$. Therefore, $\displaystyle f(x)\in f\left(\bigcup_{\alpha\in\mathcal{A}}U_{\alpha}\right)$

g) Let $\displaystyle f(x)\in f\left(\bigcap_{\alpha\in\mathcal{A}}U_{\alpha}\right)$. Then, $f(x)=f(y)$ for some $\displaystyle y\in\bigcap_{\alpha\in\mathcal{A}}U_{\alpha}$. It follows that $f(x)=f(y)$ for some $y\in U_{\alpha}$ for every $\alpha\in\mathcal{A}$. Thus, $f(x)\in f\left(U_{\alpha}\right)$ for every $\alpha\in\mathcal{A}$ and so $\displaystyle f(x)\in\bigcap_{\alpha\in\mathcal{A}}f\left(U_{\alpha}\right)$

4.

Problem: Let $f:A\to B$ and $g:B\to C$

a) If $C_0\subseteq C$, show that $\left(g\circ f\right)^{-1}\left(C_0\right)=f^{-1}\left(g^{-1}\left(C_0\right)\right)$

b) If $f$ and $g$ are injective, show that $g\circ f$ is injective.

c) If $g\circ f$ is injective, what can you say about the injectivity of $f$ and $g$?

d) If $f$ and $g$ are surjective, prove that $g\circ f$ is surjective

e) If $g\circ f$ is surjective, what can you say about the surjectivity of $f$ and $g$?

Proof:

a) Let $x\in \left(g\circ f\right)^{-1}\left(C_0\right)$, then $\left(g\circ f \right)(x)=g(f(x))\in C_0$ but this clearly implies that $x\in f^{-1}\left(g^{-1}\left(C_0\right)\right)$. Conversely, if $x\in f^{-1}\left(g^{-1}\left(C_0\right)\right)$ we see that $f(x)\in g^{-1}\left(C_0\right)$ and so $g(f(x))=(g\circ f)(x)\in C_0$. Therefore, $x\in \left(g\circ f\right)^{-1}\left(C_0\right)$

b) If $x\ne y$ then by the injectivity of $f$ we see that $f(x)\ne f(y)$ and so by the injectivity of $g$ it follows that $g(f(x))\ne g(f(y))$.

c) If $g\circ f$ is injective then $f$ is injective. To see this, suppose not; then there exists $x,y\in A$ such that $x\ne y$ but $f(x)=f(y)\implies g(f(x))=g(f(y))$ which contradicts $g\circ f$‘s injectivity.

d) We note that $g\left(f\left(A\right)\right)=g\left(B\right)=C$ from where the conclusion follows.

e) It must be true that $g$ is surjective. To see this suppose not. Then,

$g\left(f\left(A\right)\right)\subseteq g\left(B\right)\subsetneq C$

which contradicts the surjectivity of $g\circ f$

5.

Problem: In general, let us denote the identity function on a set $C$ by $\text{id}_C$. That is, define

$\text{id}_C:C\to C:x\mapsto x$

Given $f:A\to B$ we say that a function  $g:B\to A$ is a left inverse for $f$ if $g\circ f=\text{id}_A$; and we see that $h:B\to A$ is a right inverse for $f$ if $f\circ h=\text{id}_B$.

a) Show that if $f$ has a left inverse, $f$ is injective; and if $f$ has a right inverse, $f$ is surjective

b) Give an example of a function that has a left inverse but no right inverse

c) Give an example of a function which has a right inverse but no left inverse.

d) Can a function have more than one left inverse? More than one right inverse?

e) Show that if $f$ has both a left inverse $g$ and a right inverse $h$, then $f$ is bijective and $g=h=f^{-1}$

Proof:

a) If $f(x)=f(y)$ we see that $x=g(f(x))=g(f(y))=y$ and so $f$ is injective. Conversely, let $y\in B$ be arbitrary, we know that $f(h(y))=y$ and so $h(y)\in A$ is the required element which maps to $y$ under $f$

b) The inclusion map $\iota:\mathbb{N}\hookrightarrow\mathbb{Z}:x\mapsto x$. Clearly it has no right inverse or it’d be surjective. But, the mapping

$j:\mathbb{Z}\to\mathbb{N}:x\mapsto\begin{cases}x & \mbox{if}\quad x\in\mathbb{N}\\ 0 & \mbox{if}\quad x\notin\mathbb{N}\end{cases}$

surely satisfies the left inverse requirement since

$j(\iota(x))=j(x)=x,\text{ }\forall x\in\mathbb{N}$.

c) Consider the function $f:\mathbb{R}\to\{1\}:x\mapsto 1$. Clearly this possesses no left inverse, otherwise it’d be injective. But, $j:\{1\}\to\mathbb{R}:1\mapsto1$ has the quality that $f(j(1))=f(j(1))=f(1)=1$ and so $j$ is a suitable right inverse

d) Yes, in the first example we could have taken $j(x)=-1,\text{ }x\notin\mathbb{N}$ and the second example we could have had $j(1)=0$, yet both are left and right inverses respectively.

e) Clearly by a) and b) we see that $f$ is bijective. Furthermore, let $f(x)\in B$ be arbitrary (we can say $f(x)$ since it’s surjective), then

$f^{-1}(f(x))=g(f(x))\implies f^{-1}=g$

and

$f(f^{-1}(x))=x=f(h(x))\implies f(f^{-1}(x))=f(h(x))\overset{*}{\implies}f^{-1}(x)=h(x)$

where $*$ is furnished by $f$‘s injectivity. Thus, putting the two together gives $g=f^{-1}=h$.

6.

Problem: Let $f:\mathbb{R}\to\mathbb{R}:x\mapsto x^3-x$. By restricting the domain and range of $f$ obtain from $f$ a bijective function $g$.

Proof: Merely take $\text{Dom }g=\{1\}$ and $\text{Ran }g=\{0\}$ and so then the function $g:\{1\}\to\{0\}:x\mapsto x^3-x$ is bijective and a restriction of $f$ (in fact, you could take the vaccuous function $f:\varnothing\to\varnothing$)

September 25, 2010

## Munkres Chapter one Section one

1.

Problem: Check the distributive laws for union and intersection. Also,  check DeMorgan’s laws.

Proof: Our first goal is to check that $A\cup\left(B\cap C\right)=\left(A\cup B\right)\cap\left(A\cap C\right)$ and $A\cap\left(B\cup C\right)=\left(A\cap B\right)\cup\left(A\cap C\right)$. To verify the fist we notice that

$x\in A\cup\left(B\cap C\right)\Leftrightarrow x\in A\text{ or }x\in\left(B\cap C\right)\Leftrightarrow x\in A\text{ or }\left(x\in B\text{ and }x\in C\right)$

which by the distributive laws of  ‘and’ and ‘or’ (easily checked by a truth table) we see that this is equivalent to

$\left(x\in A\text{ or }x\in B\right)\text{ and }\left(x\in A\text{ or }x\in C\right)\Leftrightarrow x\in\left(A\cup B\right)\cap\left(A\cup C\right)$

from where the conclusion follows. The other case is completely analogous.

Now, we aim to prove that $U-\left(A\cup B\right)=\left(U-A\right)\cap\left(U-B\right)$ and $U-\left(A\cap B\right)=\left(U-A\right)\cup\left(U-B\right)$. To do this we let $x\in U-\left(A\cup B\right)$. Now, if $x\in A,B$ we’d have that $x\in A\cup B$ and so $x\notin A\text{ and }x\notin B$ and so $x\in U-A\text{ and }x\in U-B$ and so $x\in\left(U-A\right)\cap\left(U-B\right)$. Conversely, if $x\in\left(U-A\right)\cap\left(U-B\right)$ we see that $x\in U-A\text{ and }x\in U-B$. If $x\in A$ or $x\in B$ we’d have that $x\notin U-A$ or $x\notin U-B$, and either way $x\notin\left(U-A\right)\cap\left(U-B\right)$. It follows that $x$ is in neither, i.e. $x\notin A\cup B\implies x\in U-\left(A\cup B\right)$. The conclusion follows.

To get the other DeMorgan’s law we merely note that (changing $U-$ to $\prime$)

$\left(A\cap B\right)\prime=\left(\left(A\prime\right)\prime\cap\left(B\prime\right)\prime\right)\prime=\left(\left(A\prime\cup B\prime\right)\prime\right)\prime=A\prime\cup B\prime$

2.

Problem: Determine which of the following statements are true for all sets $A,B,C$ and $D$. If a double implication fails, determine whether one or the other of the possible implications holds. IF an equality fails, determine whether the statement becomes true if the equals sign is replaced by an inclusion.

a) $A\subseteq B$ and $A\subseteq C$ iff $A\subseteq B\cup C$

b) $A\subseteq B$ or $A\subseteq C$ iff $A\subseteq B\cup C$

c) $A\subseteq B$ and $A\subseteq C$ iff $A\subseteq B\cap C$

d) $A\subseteq B$ or $A\subseteq C$ iff $A\subseteq B\cap C$

e) $A-\left(A-B\right)=B$

f) $A-\left(B-A\right)=A-B$

g) $A\cap\left(B-C\right)=\left(A\cap B\right)-\left(A\cap C\right)$

h) $A\cup\left(B-C\right)=\left(A\cup B\right)-\left(A\cup C\right)$

i) $\left(A\cap B\right)\cup\left(A-B\right)=A$

j) $A\subseteq C$ and $B\subseteq D$ implies $\left(A\times B\right)\subseteq \left(C\times D\right)$

k) The converse of j)

l) The converse of j) assuming that $A$ and $B$ are non-empty

m) $\left(A\times B\right)\cup\left(C\times D\right)=\left(A\cup C\right)\times\left(B\cup D\right)$

n) $\left(A\times B\right)\cap\left(C\times D\right)=\left(A\cap C\right)\times\left(B\cap D\right)$

o) $A\times\left(B-C\right)=\left(A\times B\right)-\left(A\times C\right)$

p) $\left(A-B\right) \times\left(C-D\right)=\left(A\times C-B\times C\right)-\left(A\times D\right)$

q) $\left(A\times B\right)-\left(C\times D\right)=\left(A-C\right)\times\left(B-D\right)$

Proof:

a)

$\implies$: Let $x\in A$ then $x\in B$ and thus $x\in B\cup C$ and so $A\subseteq B\cup C$

$\Leftarrow$: This isn’t necessarily true. Consider $A=\{1,2,3\},B=\{1,2\},C=\{3\}$.

b)

$\implies$: Let $x\in A$ then $x\in B$ or $x\in C$ and so $x\in B\cup C$. Thus, $A\subseteq B\cup C$.

$\Leftarrow$: This is also wrong, the same counterexample being applicable.

c)

$\implies$: Let $x\in A$ then $x\in B$ and $x\in C$ and so $x\in B\cap C$. Thus, $A\subseteq B\cap C$.

$\Leftarrow$: Clearly since $A\subseteq B\cap C\subseteq B,C$

d)

$\implies$: Not true. Take $A=B\ne\varnothing$ and $C=\varnothing$

$\Leftarrow$: This is true, considering it’s a weaker formulation of the second direction in the last problem.

e)  Clearly not since the LHS is a subset of $A$. For example, $\{1,2\}-\left(\{1,2\}-\{3\}\right)=\varnothing$.

f) No. If $A=B\ne\varnothing$ then the LHS is non-empty and the RHS is.

g) This is true. The RHS may be manipulated into the LHS as follows

$\left(A\cap B\right)-\left(A\cap C\right)=A\cap B\cap\left(A\cap C\right)\prime=B\cap A\cap\left(A\prime\cup C\prime\right)$

(the last part by DeMorgan’s) which upon distribution of $A$ gives

$B\cap\left[\left(A\cap A\prime\right)\cup\left(A\cap C\prime\right)\right]=B\cap A\cap C\prime=A\cap\left(B\cap C\prime\right)=A\cap\left(B-C\right)$

h) Not true. Note that if $A=B=C\ne\varnothing$ this says that

$A=A\cup\varnothing=A\cup\left(B-C\right)=\left(A\cup B\right)-\left(A\cup C\right)=A-A=\varnothing$

i) This makes sense, if you read it it would say “Everything that’s in both $A$ and $B$ plus everything that is in $A$ but not in $B$ gives everything in $A$.” Formally

$\left(A\cap B\right)\cup\left(A-B\right)=A\cap B\cup\left(A\cap B\prime\right)$

and so by distribution this equals

$A\cap\left(\left(B\cup A\right)\cap\left(B\cup B\prime\right)\right)=A\cap \left(B\cup A\right)=A$

j) This is true. Let $(x,y)\in A\times B$, then $x\in A\implies x\in C$ and $x\in B\implies x\in D$, thus $(x,y)\in C\times D$.

k) This is not always true. Let $A=C=\varnothing$ and $B=\mathbb{R},D=\{1\}$ then the LHS and RHS of the inclusion are empty yet $B\not\subseteq D$.

l) Then, this is true. Let $x\in A$, since $B\ne\varnothing$ there exists some $y\in B$, and thus $(x,y)\in A\times B$. Thus, $(x,y)\in C\times D$ and so $x\in C$. A similar procedure shows $B\subseteq D$

Remark: This shows why sometimes it’s perilous to just go “let $x\in$…”

m) This is false. Let $A=\{1,2\},B=\{3,4\},C=\{5,6\},D=\{7,8\}$. Evidently $(1,8)\in \left(A\cup C\right)\times \left(B\cup D\right)$ but $(1,8)\notin \left(A\times B\right)\cup\left(C\times D\right)$

n) This is true. Let $(x,y)\in \left(A\times B\right)\cap\left(C\times D\right)$ then

$(x,y)\in A\times B\text{ and }(x,y)\in C\times D$

or

$\left(x\in A\text{ and }y\in B\right)\text{ and }\left(x\in C\text{ and }y\in D\right)$.

or equivalently

$\left(x\in A\text{ and} x\in C\right)\text{ and }\left(y\in B\text{ and }x\in D\right)$

or equivalently

$x\in A\cap C\text{ and }y\in B\cap D\Leftrightarrow (x,y)\in\left(A\cap C\right)\times\left(B\cap D\right)$

o) This is true. Let $(x,y)\in A\times\left(B-C\right)$, then:

$x\in A\text{ and }y\in B-C\implies x\in A\text{ and }\left(y\in B\text{ and }y\notin C\right)$

or (and this looks weird)

$x\in A\text{ and }x\in A\text{ and }y\in B\text{ and }y\notin C$

or

$\left(x\in A\text{ and }y\in B\right)\text{ and }\left(x\in A\text{ and }y\notin C\right)$

or

$(x,y)\in A\times B\text{ and }(x,y)\notin A\times C$

or

$(x,y)\in \left(A\times B\right)-\left(A\times C\right)$

And so $\text{LHS}\subseteq\text{RHS}$. Conversely, let $(x,y)\in \left(A\times B\right)-\left(A\times C\right)$. Then,

$(x,y)\in A\times B\text{ and }(x,y)\notin A\times C$

(now, $(x,y)\notin A\times C$ if $x\notin A\text{ and }y\in C$, $x\in A\text{ and }y\notin C$, and $x\notin A\text{ and }y\notin C$. But, since the first and the last are impossible since the first part of the statement says $x\in A$ we may conclude that the second is true) thus

$\left(x\in A\text{ and }y\in B\right)\text{ and }\left(x\in A\text{ and }y\notin C\right)$

or

$x\in A\text{ and }y\in B\text{ and }y\notin C$

or

$x\in A\text{ and }y\in B-C\implies (x,y)\in A\times \left(B-C\right)$

p) This is true. Note that $(a,b)\in \left(A-B\right)\times\left(C-D\right)$ iff $a\in A-B\text{ and }b\in C-D$ which is true iff $\left(a\in A\text{ and } a\notin B\right)\text{ and }\left(b\in C\text{ and}b\notin D\right)$. But, this is true iff $\left(a\in A\text{ and }b\in C\right)\left(a\notin B\text{ or }b\notin C\right)\text{ and }\left(a\notin A\text{ or }b\notin D\right)$. (this ilast part was trick since we know that $a\in A$.) which is true iff $(a,b)\in A\times C\text{ and }(a,b)\notin B\times C\text{ and }(a,b)\notin A\times D$.

q) This isn’t true in general. Try $\mathbb{R}\supseteq A=B=C\ne\varnothing$ and $D=\mathbb{R}-A$

Remark: I did these in more detail earlier, if you search for them.

3.

Problem:

a) Write the contrapositive and converse of the statement “If $x>0$, then $x^2-x>0$” and determine which of the statements are true

b) Do the same for the statement “If $x>0$, then $x^2-x>0$

Proof:

a) The contrapositive is ” If $x^2-x\leqslant 0$ then $x\geqslant 0$ which is false $x=\frac{-1}{2}$. The converse is “If $x^2-x>0$, then $x<0$” which is false “$x=2$

b) The contrapositive is “If $x^2-x\leqslant 0$, then $x\leqslant 0$ which is true. The converse is ” if $x^2-x>0$, then $x>0$” which is not true $x=-1$

4.

Problem: Let $A$ and $B$ be sets of real numbers. Write the negation of each of the following statements

a) For every $a\in A$, it is true that $a^2\in B$

b) For at least one $a\in A$, it is true that $a^2\in B$

c) For every $a\in A$, it is true that $a^2\notin B$

d) For at least one $a\notin A$, it is true that $a^2\in B$

Proof:

I’m going to write these in set notation

a) $\exists a\in A\backepsilon a^2\notin B$

b) $\forall a\in A, a^2\notin B$

c) $\exists a\in A\backepsilon a^2\in B$

d) $\forall a\notin A,a^2\notin B$

5.

Problem: Let $\left\{U_{\alpha}\right\}_{\alpha\in\mathcal{A}}$  be a nonempty class of sets. Determine the truth of each of the following statements and their converses

a) $\displaystyle x\in\bigcup_{\alpha\in\mathcal{A}}U_{\alpha}\implies x\in U_{\alpha_0},\text{ }\text{ for some }\alpha_0\in\mathcal{A}$

b) $\displaystyle x\in \bigcup_{\alpha\in\mathcal{A}}U_{\alpha}\implies x\in U_\alpha,\text{ }\forall \alpha\in\mathcal{A}$

c) $\displaystyle x\in\bigcap_{\alpha\in\mathcal{A}}U_{\alpha}\implies x\in U_{\alpha_0}\text{ for some }\alpha_0\in\mathcal{A}$

d) $\displaystyle x\in\bigcap_{\alpha\in\mathcal{A}}U_{\alpha}\implies x\in U_{\alpha},\text{ }\forall \alpha\in\mathcal{A}$

Proof:

a) This implication is true, as well as it’s converse (this is the definition of union)

b) The implication is false. Consider $\left\{\{1\},\{2\}\right\}$, then clearly $1\in\{1\}\cup\{2\}$ but $1\notin\{2\}$

c) The implication is true. The converse is false though. In the above, class of sets we see that $1\in\{1\}$ but $1\notin \{1\}\cap\{2\}=\varnothing$

d) The implication, and it’s converse are true. This is the definition of intersection.

6.

Problem: Write the contrapositive of each statements in exercise 5

Proof:

a) If $x\notin U_{\alpha}$ for every $\alpha\in\mathcal{A}$ then $\displaystyle x\notin\bigcup_{\alpha\in\mathcal{A}}U_{\alpha}$

b) If there exists $\alpha_0\in\mathcal{A}$ such that $x\notin U_{\alpha_0}$, then $\displaystyle x\notin\bigcup_{\alpha\in\mathcal{A}}U_{\alpha}$

c) If for every $\alpha\in\mathcal{A}$ it is true that $\displaystyle x\notin U_{\alpha}$ then $x\notin\bigcap_{\alpha\in\mathcal{A}}U_{\alpha}$

d) If there exists some $\alpha_0\in\mathcal{A}$ such that $x\notin U_{\alpha_0}$, then $\displaystyle x\notin\bigcap_{\alpha\in\mathcal{A}}U_{\alpha}$

7.

Problem: Given sets $A,B$ and $C$, express each of the following sets in terms of $A,B,$ and $C$, using the smbols $\cup,\cap,$ and $-$.

$D=\left\{x:x\in A\text{ and }\left(x\in B\text{ or }x\in C\right)\right\}$

$E=\left\{x:\left(x\in A\text{ and }x\in B\right)\text{ or }x\in C\right\}$

$F=\left\{x:x\in A\text{ and }\left(x\in B\implies x\in C\right)\right\}$

Proof:

a) $D= A\cap\left(B\cup C\right)$

b) $E=\left(A\cap B\right)\cup C$

c) $F=A\cap B\cap C$

8.

Problem: If a set $A$ has two elements, show that $\mathcal{P}(A)$ has four elements. How many elements does $\mathcal{P}(A)$ have if $A$ has one element? Three elements? No elements? Why is $\mathcal{P}\left(A\right)$ called the power set of $A$?

Proof:

Lemma: Let $Y^X=\left\{f\mid f:X\to Y\right\}$. Then, $\mathcal{P}\left(X\right)\cong \{0,1\}^{X}$

Proof: Let $\varphi:\mathcal{P}(X)\to\{0,1\}^{X}:E\mapsto \Psi_E$ where

$\Psi_E(x)=\begin{cases} 0 \quad\text{if}\quad x\notin E\\1\quad\text{if}\quad x\in E\end{cases}$

To see that $\varphi$ is injective suppose that $A\ne B\subseteq X$, then WLOG there is some $x\in A-B$ and so $\Psi_A(x)=1\ne0=\psi_B(x)$, thus $\Psi_A\ne\Psi_B$.  Next, to prove surjectivity let $\Psi:X\to\{0,1\}$ and let $E=\Psi^{-1}\left(\{1\}\right)$, evidently $\varphi\left(E\right)=\Psi$. It follows that $\varphi$ is a bijection, and the lemma is proved. $\blacksquare$

It follows that if $\text{card }X=n$ then $\text{card }\mathcal{P}(X)=\text{card }\{0,1\}^n=2^n$.

Remark: Alternatively, one may find a recurrence relation solution to this problem. Namely, let $n=\text{card }\mathcal{P}\left(\{1,\cdots,m\}\right)$. Then, if one paritions $\{1,\cdots,m+1\}$ into those sets which contain $m+1$ and those that don’t one can see that each block of the partition has $n$ elements, and thus

$\text{card }\mathcal{P}\left(\{1,\cdots,m\}\right)=2n=2\text{card }\mathcal{P}\left(\{1,\cdots,n\}\right)$

and thus noting that $\text{card }\mathcal{P}\left(\varnothing\right)=1$ one may conclude that $\text{card }\mathcal{P}\left(\{1,\cdots,m\}\right)=2^m$

9.

Problem: Formulate and prove DeMorgan’s laws for arbitrary unions and intersections.

a)

Claim: Let $\left\{U_{\alpha}\right\}_{\alpha\in\mathcal{A}},\text{ }\mathcal{A}\ne\varnothing$ be a class of subsets of some universal set $U$, then

$\displaystyle U-\bigcup_{\alpha\in\mathcal{A}}U_{\alpha}=\bigcap_{\alpha\in\mathcal{A}}\left(U-U_{\alpha}\right)$

Proof: If $\displaystyle x\in\bigcup_{\alpha\in\mathcal{A}}U_{\alpha}$ then $x\in U_{\alpha_0}$ for some $\alpha_0\in\mathcal{A}$. Thus, $x\notin U-U_{\alpha_0}$ and so $\displaystyle x\notin\bigcap_{\alpha\in\mathcal{A}}\left(U-U_{\alpha}\right)$. Conversely, suppose that $x\notin\bigcap_{\alpha\in\mathcal{A}}\left(U-U_{\alpha}\right)$ then there exists some $\alpha_0\in\mathcal{A}$ such that $x\notin U-U_{\alpha_0}\implies x\in U_{\alpha_0}$. It follows that $\displaystyle x\in\bigcup_{\alpha\in\mathcal{A}}U_{\alpha}$ and so finally $x\notin U-\bigcup_{\alpha\in\mathcal{A}}U_{\alpha}$. The conclusion follows.

Corollary:

$\displaystyle U-\bigcap_{\alpha\in\mathcal{A}}U_{\alpha}=\bigcup_{\alpha\in\mathcal{A}}\left(U-U_{\alpha}\right)$

Proof: We merely note that

$\displaystyle U-\bigcap_{\alpha\in\mathcal{A}}U_{\alpha}=U-\bigcap_{\alpha\in\mathcal{A}}\left(U-\left(U-U_{\alpha}\right)\right)$

which by our last proven claim is

$\displaystyle U-\left(U-\bigcup_{\alpha\in\mathcal{A}}\left(U-U_{\alpha}\right)\right)=\bigcup_{\alpha\in\mathcal{A}}\left(U-U_{\alpha}\right)$

10.

Problem: Let $\mathbb{R}$  denote the set of real numbers. For each of the following subsets of $\mathbb{R}$, determine whether it is equal to the cartesian product of two subsets of $\mathbb{R}$

a) $A=\left\{(x,y):x\in\mathbb{Z}\right\}$

b) $B=\left\{(x,y):0

c) $C=\left\{(x,y):y>x\right\}$

d) $D=\left\{(x,y):x\notin\mathbb{Z}\text{ and }y\in\mathbb{Z}\right\}$

e) $E=\left\{(x,y):x^2+y^2<1\right\}$

Proof:

a) This is, namely $A=\mathbb{Z}\times\mathbb{R}$

b) This is, namely $B=\mathbb{R}\times(0,1]$

c) This is not. To see this suppose that $C=X\times Y$. We see that $(0,1)\in X\times Y$ and thus $1\in Y,0\in X$. Also, $(2,3)\in X\times Y$ and so $2\in X,3\in Y$. But, $(2,1)\notin C$ and thus $C\ne\left\{(x,y):x\in X\text{ and }y\in Y\right\}$, which is a contradiction.

d) This is, namely $D=\left(\mathbb{R}-\mathbb{Z}\right)\times\mathbb{Z}$

e) This is not, using the same argument as in c), first noting that $(0,.85),(.85,0)\in E$

September 23, 2010

## Munkres Chapter 1 Section 2

1.

Problem: Let $f:A\to B$. Let $A_0\subseteq A$ and $B_0\subseteq B$.

a) Show that $A_0\subseteq f^{-1}\left(f\left(A_0\right)\right)$ and that equality holds if $f$ is injective

b) Show that $f\left(f^{-1}\left(B_0\right)\right)\subseteq B_0$ and equality holds if and only if $f$ is surjective

Proof:

a) The first part is obvious enough. If $x\in A_0$ then $f(x)\in f\left(A_0\right)$ and so $x\in f^{-1}\left(f\left(A_0\right)\right)$ from where it follows that $A_0\subseteq f^{-1}\left(f\left(A_0\right)\right)$.

Now, suppose that $f$ is injective but $A_0\subsetneq f^{-1}\left(f\left(A_0\right)\right)$. Then, there is some $x\notin A_0$ such that $f(x)\in f\left(A_0\right)$. But, by definition since $f(x)\in f\left(A_0\right)$ there exists some $a\in A_0$ such that $f(x)=f(a)$ but since $a\in A_0$ and $x\notin A_0$ it follows that $x\ne a$ and this contradicts injectivity.

Conversely, suppose that $A_0=f^{-1}\left(f\left(A_0\right)\right)$. Then, if $x\ne y$ then $y\notin \{x\}$ and so $\{y\}\notin f^{-1}\left(f\left(\{x\}\right)\right)$ which means that $f(y)\ne f(x)$ as desired.

b) The first part is clear again. If $f(x)\in f\left(f^{-1}\left(B_0\right)\right)$ then $x\in f^{-1}\left(B_0\right)$ and so $f(x)\in B_0$ from where it follows that $f\left(f^{-1}\left(B_0\right)\right)\subseteq B_0$.

Now, suppose that $f$ is surjective. By the previous part it suffices to show the reverse inclusion. So, let $f(x)\in B_0$ (we know that any element of $B_0$ is of the form $f(x)$ by surjectivity), then $x\in f^{-1}\left(B_0\right)$ and so $f(x)\in f\left(f^{-1}\left(B_0\right)\right)$ and so $B_0\subseteq f\left(f^{-1}\left(B_0\right)\right)$ from where the conclusion follows.

Conversely, suppose that $B_0=f\left(f^{-1}\left(B_0\right)\right)$ and let $y_0\in B$. Then, $\{y\}=f\left(f^{-1}\left(\{y\}\right)\right)$ in particular $y\in f\left(A\right)$ from where it follows that $B\subseteq f(A)$ and so surjectivity is guaranteed.

2.

Problem: Let $f:A\to B$ and let $A_i\subseteq B$ and $B_i\subseteq B$ for $i=0,1$. Show that $f^{-1}$ preserves inclusions, unions, intersections, and differences of sets:

a) $B_0\subseteq B_1\implies f^{-1}\left(B_0\right)\subseteq f^{-1}\left(B_1\right)$

b) $f^{-1}\left(B_0\cup B_1\right)=f^{-1}\left(B_0\right)\cup f^{-1}\left(B_1\right)$

c) $f^{-1}\left(B_0\cap B_1\right)=f^{-1}\left(B_0\right)\cap f^{-1}\left(B_1\right)$

d) $f^{-1}\left(B_0-B_1\right)=f^{-1}\left(B_0\right)-f^{-1}\left(B_1\right)$

Show that $f$ preserves inclusions and unions only:

e) $A_0\subseteq A_1\implies f\left(A_0\right)\subseteq f\left(A_1\right)$

f) $f\left(A_0\cup A_1\right)=f\left(A_0\right)\cup f\left(A_1\right)$

g) $f\left(A_0\cap A_1\right)\subseteq f\left(A_0\right)\cap f\left(A_1\right)$ and show that equality holds precisely when $f$ is injective

h) $f\left(A_0-A_1\right)\supseteq f\left(A_0\right)-f\left(A_1\right)$ and show that equality holds precisely when $f$ is surjective.

Proof:

We assume in all cases that the sets are non-empty and in the case of intersections and set differences, intersecting. For, if not this is trivial.

a) Let $x\in f^{-1}\left(B_0\right)$ then $f(x)\in B_0$ and so $f(x)\in B_1$ and so $x\in f^{-1}\left(B_1\right)$, from where it follows that $f^{-1}\left(B_0\right)\subseteq f^{-1}\left(B_1\right)$

b) Let $x\in f^{-1}\left(B_0\cup B_1\right)$ then $f(x)\in B_0\cup B_1$ and so $f(x)\in B_0\text{ or }f(x)\in B_1$ and so $x\in f^{-1}\left(B_0\right)\text{ or }x\in f^{-1}\left(B_1\right)$ and so $x\in f^{-1}\left(B_0\right)\cup f^{-1}\left(B_1\right)$. Noting that all the “and so”s above were actually “if and only if”s shows the reverse inclusion.

c) We note that $x\in f^{-1}\left(B_0\cap B_1\right)$ if and only if $f(x)\in B_0\cap B_1$ which is true if and only if $f(x)\in B-0\text{ and }f(x)\in B_1$ which is true if and only if $x\in f^{-1}\left(B_0\right)\text{ and }x\in f^{-1}\left(B_1\right)$ which is true if and only if $x\in f^{-1}\left(B_0\right)\cap f^{-1}\left(B_1\right)$

d) We note that $x\in f^{-1}\left(B_0-B_1\right)$ if and only if $f(x)\in B_0-B_1$ which is true if and only if $f(x)\in B_0\text{ and }f(x)\notin B_1$. Now, clearly the first part is true if and only if $x\in f^{-1}\left(B_0\right)$ and (noticing that $x\in f^{-1}\left(B_1\right)\implies f(x)\in B_1$ )the second part is true if and only if $x\notin f^{-1}\left(B_1\right)$ and so putting these together we get the statement in the previous sentence is true if and only if $x\in f^{-1}\left(B_0\right)\text{ and }x\notin f^{-1}\left(B_1\right)$ which is true if and only if $x\in f^{-1}\left(B_0\right)-f^{-1}\left(B_1\right)$

e) Let $f(x)\in f\left(A_0\right)$ then $f(x)=f(a)$ for some $a\in A_0$. But, since $A_0\subseteq A_1$ it follows that $f(x)=f(a)$ for some $a\in A_1$ and so $f(x)\in f\left(A_1\right)$ from where it follows that $f\left(A_0\right)\subseteq f\left(A_1\right)$.

f) Let $f(x)\in f\left(A_0\cup A_1\right)$ then $f(x)=f(a)$ for some $a\in A_0\cup A_1$ and thus $f(x)=f(a)$ for some $a\in A_0\text{ or }a\in A_1$ and so either $f(x)=f(a)$ for some $a\in A_0$ or $f(x)=f(a)$ for some $a\in A_1$ and so $f(x)\in f\left(A_0\right)\text{ or }f(x)\in f\left(A_1\right)$ which is true if and only if $f(x)\in f\left(A_0\right)\cup f\left(A_1\right)$

g) Let $f(x)\in f\left(A_0\cap A_1\right)$ then $f(x)=f(a)$ for some $a\in A_0\cap A_1$ and so $f(x)=f(a)$ for some $a\in A_0$ and $f(x)=f(a)$ for some $a\in A_1$ and so $f(x)\in f\left(A_0\right)\text{ and }f(x)\in f\left(A_1\right)$ and so $f(x)\in f\left(A_0\right)\cap f\left(A_1\right)$ from where it follows that $f\left(A_0\cap A_1\right)$.

Now, suppose that $f$ is injective. By the last part to show equality it suffices to show the reverse inclusion. To do this we let $f(x)\in f\left(A_0\right)\cap f\left(A_1\right)$ then $f(x)\in f\left(A_0\right)\text{ and }f(x)\in f\left(A_1\right)$ and thus (by injectivity) $x\in A_0\text{ and }x\in A_1$ and so $x\in A_0\cap A_1$ and so $f(x)\in f\left(A_0\cap A_1\right)$. The conclusion follows by previous comment.

Conversely, suppose that $x\ne y$ then $\{x\}\cap\{y\}=\varnothing$ and so

$\varnothing=f\left(\varnothing\right)=f\left(\{x\}\cap\{y\}\right)=f\left(\{x\}\right)\cap f\left(\{y\}\right)=\{f(x)\}\cap\{f(y)\}$

from where it follows that $f(x)\ne f(y)$

Remark: I technically took $A_0$ to be any subset of $A$

h) Let $f(x)\in f\left(A_0\right)-f\left(A_1\right)$ then $f(x)\in f\left(A_0\right)$ and $f(x)\notin f\left(A_1\right)$ and so $f(x)=f(a)$ for some $a\in A_0$ but this $a\notin A_1$ and so $f(x)=f(a)$ for some $a\in A_0-A_1$ and so $f(x)\in f\left(A_0-A_1\right)$

Now, suppose that $f$ is injective. From the last part to prove equality it suffices to show the reverse inclusion. So, let $f(x)\in f\left(A_0-A_1\right)$ then $x\in A_0-A_1$ (by injectivity) and so $x\in A_0\text{ and }x\notin A_1$ and so $f(x)\in f\left(A_0\right)\text{ and }f(x)\notin f\left(A_1\right)$ (second part by injectivity) and thus $f(x)\in f\left(A_0\right)-f\left(A_1\right)$

Conversely, if $f(x)=f(y)$ then

$\varnothing=\{f(x)\}-\{f(y)\}=f\left(\{x\}\right)-f\left(\{y\}\right)=f\left(\{x\}-\{y\}\right)$

from where it follows that

$\{x\}-\{y\}=\varnothing\implies \{x\}\subseteq\{y\}\implies x\in\{y\}\implies x=y$

Injectivity follows.

Remark: The same remark applies.

3.

Problem: Show that b), c), f) and g) of the last exercise hold for arbitrary unions and intersections.

Proof:

We once again assume that all the following are non-empty since the proof otherwise is trivial.

b) Let $\displaystyle x \in f^{-1}\left(\bigcup_{\alpha\in\mathcal{A}}U_\alpha\right)$ then $\displaystyle f(x)\in\bigcup_{\alpha\in\mathcal{A}}U_\alpha$ and so in particular $f(x)\in U_{\alpha_0}$ for some $\alpha_0\in\mathcal{A}$. Thus, $x\in f^{-1}\left(U_{\alpha_0}\right)$ and so $\displaystyle x\in\bigcup_{\alpha\in\mathcal{A}}f^{-1}\left(U_\alpha\right)$. It follows that

$\displaystyle f^{-1}\left(\bigcup_{\alpha\in\mathcal{A}}U_\alpha\right)\subseteq\bigcup_{\alpha\in\mathcal{A}}f^{-1}\left(U_\alpha\right)$

Conversely, let $\displaystyle x\in\bigcup_{\alpha\in\mathcal{A}}f^{-1}\left(U_\alpha\right)$ then $x\in f^{-1}\left(U_{\alpha_0}\right)$ for some $\alpha_0\in\mathcal{A}$. Thus, $f(x)\in U_{\alpha_0}$ and so $\displaystyle f(x)\bigcup_{\alpha\in\mathcal{A}}U_\alpha$ so that $\displaystyle x\in f^{-1}\left(\bigcup_{\alpha\in\mathcal{A}}U_\alpha\right)$. It follows that

$\displaystyle \bigcup_{\alpha\in\mathcal{A}}f^{-1}\left(U_\alpha\right)\subseteq f^{-1}\left(\bigcup_{\alpha\in\mathcal{A}}U_\alpha\right)$

The problem follows.

c) Let $\displaystyle x\in f^{-1}\left(\bigcap_{\alpha\in\mathcal{A}}U_\alpha\right)$ then $\displaystyle f(x)\in\bigcap_{\alpha\in\mathcal{A}}U_\alpha$ and so $f(x)\in U_\alpha$ for every $\alpha\in\mathcal{A}$. Thus, $x\in f^{-1}\left(U_\alpha\right)$ for every $\alpha\in\mathcal{A}$ and thus $\displaystyle x\in\bigcap_{\alpha\in\mathcal{A}}f^{-1}\left(U_\alpha\right)$

$\displaystyle f^{-1}\left(\bigcap_{\alpha\in\mathcal{A}}U_\alpha\right)\subseteq \bigcap_{\alpha\in\mathcal{A}}f^{-1}\left(U_\alpha\right)$

Conversely, if $\displaystyle x\in\bigcap_{\alpha\in\mathcal{A}}f^{-1}\left(U_\alpha\right)$ we have that $x\in f^{-1}\left(U_\alpha\right)$ for every $\alpha\in\mathcal{A}$ and so $f(x)\in U_\alpha$ for every $\alpha\in\mathcal{A}$. Thus, $\displaystyle f(x)\in\bigcap_{\alpha\in\mathcal{A}}U_\alpha$ and so $\displaystyle x\in f^{-1}\left(\bigcap_{\alpha\in\mathcal{A}}U_\alpha\right)$. It follows that

$\displaystyle \bigcap_{\alpha\in\mathcal{A}}f^{-1}\left(U_\alpha\right)\subseteq f^{-1}\left(\bigcap_{\alpha\in\mathcal{A}}U_\alpha\right)$

from where the problem follows.

f) Let $\displaystyle f(x)\in f\left(\bigcup_{\alpha\in\mathcal{A}}U_\alpha\right)$ then $f(x)=f(a)$ where $\displaystyle a\in\bigcup_{\alpha\in\mathcal{A}}U_\alpha$. So, $f(x)=f(a)$ where $a\in U_{\alpha_0}$ for some $\alpha_0\in\mathcal{A}$. It follows that $f(x)\in f\left(U_{\alpha_0}\right)$ and thus $\displaystyle f(x)\in\bigcup_{\alpha\in\mathcal{A}}f\left(U_\alpha\right)$ from where it follows that

$\displaystyle f\left(\bigcup_{\alpha\in\mathcal{A}}U_\alpha\right)\subseteq\bigcup_{\alpha\in\mathcal{A}}f\left(U_\alpha\right)$

Conversely, let $\displaystyle f(x)\in\bigcup_{\alpha\in\mathcal{A}}f\left(U_\alpha\right)$ then $f(x)\in f\left(U_{\alpha_0}\right)$ for some $\alpha_0\in\mathcal{A}$. Thus, $f(x)=f(a)$ for some $a\in U_{\alpha_0}$ thus $f(x)=f(a)$ for some $\displaystyle a\in\bigcup_{\alpha\in\mathcal{A}}U_{\alpha}$ and so $\displaystyle f(x)\in f\left(\bigcup_{\alpha\in\mathcal{A}}U_\alpha\right)$ from where it follows that

$\displaystyle \bigcup_{\alpha\in\mathcal{A}}f\left(U_\alpha\right)\subseteq f\left(\bigcup_{\alpha\in\mathcal{A}}U_\alpha\right)$

and so the problem follows.

g) Let $\displaystyle f(x)\in f\left(\bigcap_{\alpha\in\mathcal{A}}U_\alpha\right)$, then $f(x)=f(a)$ for some $\displaystyle a\in\bigcap_{\alpha\in\mathcal{A}}U_\alpha$. It follows that $f(x)=f(a)$ for some $U_\alpha$ for every $\alpha\in\mathcal{A}$. Thus, $\displaystyle f(x)\in\bigcap_{\alpha\in\mathcal{A}}f\left(U_\alpha\right)$ from where it follows that

$\displaystyle f\left(\bigcap_{\alpha\in\mathcal{A}}U_\alpha\right)\subseteq \bigcap_{\alpha\in\mathcal{A}}f\left(U_\alpha\right)$

Now, assume that $f$ is injective and let $\displaystyle x\in\bigcap_{\alpha\in\mathcal{A}}f\left(U_\alpha\right)$. Then, $f(x)\in f\left(U_\alpha\right)$ for every $\alpha\in\mathcal{A}$. Thus, by injectivity we have that $x\in U_\alpha$ for every $\alpha\in\mathcal{A}$ and so $\displaystyle x\in\bigcap_{\alpha\in\mathcal{A}}U_\alpha$. Thus, $\displaystyle f(x)\in f\left(\bigcap_{\alpha\in\mathcal{A}}U_\alpha\right)$ and thus

$\bigcap_{\alpha\in\mathcal{A}}f\left(U_\alpha\right)\subseteq f\left(\bigcap_{\alpha\in\mathcal{A}}U_\alpha\right)$

from where the problem follows.

4.

Problem: Let $f:A\to B$ and $g:B\to C$.

a) If $C_0\subseteq C$ show that $\left(g\circ f\right)^{-1}\left(C_0\right)=f^{-1}\left(g^{-1}\left(C_0\right)\right)$

b) If $f$ and $g$ are injective show that $g\circ f$ is injective.

c) If $g\circ f$ is injective what can you say about the injectivity $g$ or $f$?

d) If $f$ and $g$ are surjective prove that $g\circ f$ is surjective.

e) If $g\circ f$ is surjective what can you say about $g$ and $f$‘s surjectivity?

f) Summarize your answers to b-e in the form of a theorem.

Proof:

a) Let $x\in \left(g\circ f\right)^{-1}\left(C_0\right)$, then $(g\circ f)(x)\in C_0$ and so $g(f(x))\in C_0$. So, $f(x)\in g^{-1}\left(C_0\right)$ and so finally $x\in f^{-1}\left(g^{-1}\left(C_0\right)\right)$.

Conversely, if $x\in f^{-1}\left(g^{-1}\left(C_0\right)\right)$ then $f(x)\in g^{-1}\left(C_0\right)$ and so $g(f(x))=(g\circ f)(x)\in C_0$ so that $x\in \left(g\circ f\right)^{-1}\left(C_0\right)$.

The conclusion follows.

b) If $x\ne y\in A$ then $f(x)\ne f(y)\in B$ (by $f$‘s injectivity) and so $g(f(x))\ne g(f(y))\in C$ (by $g$‘s injectivity). The conclusion follows.

c) We claim that $g\circ f$ is injective implies that $f$ is injective. To do this we prove the contrapositive. Suppose that $f$ was not injective, then there is $x\ne y\in A$ such that $f(x)=f(y)$ and so clearly then $g(f(x))=g(f(y))$ and so $g\circ f$ is not injective. The conclusion follows.

d) This follows since $g\left(f\left(A\right)\right)=g\left(B\right)=C$ as desired.

e) We claim that $g\circ f$ is surjective implies that $g$ is surjective. One again, we do this by proving the contrapositive. We note that $g\left(f\left(A\right)\right)\subseteq g\left(B\right)\subsetneq C$ from where the conclusion follows.

f) I’m not sure what’s desired but I guess it would be that b) and d) imply that if $f,g$ are bijections then $g\circ f$ is a bijection. Also, c) and e) prove that if $g\circ f$ is a bijection then $f$ is injective and $g$ surjective.

5.

Problem: In genereal, let us denote the identity function for a set $C$ by $\text{id}_C$. That is define $\text{id}_C:C\to C:x\mapsto x$.

Given $f:A\to B$ we say the function $g:B\to A$ is a left inverse for $f$ if $g\circ f=\text{id}_A$. Also, we say $h:B\to A$ is a right inverse if $f\circ h=\text{id}_B$.

a) Prove that if $f$ has a left inverse, $f$ is injective, and if $f$ has a right inverse, $f$ is surjective.

b) Give an example of a function has a left inverse but no right inverse

c) Give an example of a function that has a right inverse but no left inverse.

d) Can a function have more than left inverse? More than one right inverse?

e) Show that if $f$ has  both a left $g$ and a right inverse $h$ then $f$ is bijective and $g=h=f^{-1}$

Proof:

a) Suppose that $f$ has a left inverse $g$ and let $\ne y\in A$. We know then that $x=g(f(x))\ne g(f(y))=y$ which would be impossible (since $g$ is a function$if $f(x)=f(y)$. Conversely, let $f$ have right inverse $h$ and let $y\in B$ be arbitrary. We know then that $f(h(y))=y$ and since $h(y)\in A$ it is the desired value whose image is $y$. b) Consider the function $f:[0,1]\to\mathbb{R}:x\mapsto x$. Clearly this has a left inverse but no right inverse. c) How about $f:\mathbb{R}\to\{1\}:x\mapsto 1$? d) A left inverse need not be unique by this definition. For example, consider the mapping $f:[0,1]\to\mathbb{R}:x\mapsto x$. Then, we can define $g:\mathbb{R}\to[0,1]$ by $\displaystyle g(x)=\begin{cases}x\quad\text{if}\quad x\in[0,1] \\ c\quad\text{if}\quad x\notin[0,1]\end{cases}$ Clearly then $g(f(x))=x$ for every $x\in[0,1]$. But, we may take $c$ to be anything so that $g$ is not unique. Right inverses are not unique either. For example, consider $\displaystyle f:\{1,2,3,4\}\to\{1,2\}:x\mapsto\begin{cases}1\quad\text{if}\quad x=1,2\\2\quad\text{if}\quad x=3,4\end{cases}$ We then consider two function $h_1:\{1,2\}\to\{1,2,3,4\}:x\mapsto\begin{cases}1\quad\text{if}\quad x=1\\3\quad\text{if}\quad x=2\end{cases}$ We then note that $f\left(h(1)\right)=f(1)=1$ and $f\left(h(2)\right)=f(3)=2$ So that $f\circ h_1=\text{id}_{\{1,2\}}$ But, a similar analysis shows that if $h_2$ is defined by $h_2:\{1,2\}\to\{1,2,3,4\}:x\mapsto\begin{cases} 2\quad\text{if}\quad x=1\\ 4\quad\text{if}\quad x=2\end{cases}$ also has the property that $f\circ h_2=\text{id}_{\{1,2\}}$ And since $h_1\ne h_2$ it follows that right inverses need not be unique. e) Clearly by our previous problems we have that $f$ having a left and right inverse respectively implies that$laetx f$is injective and surjective respectively and thus bijective. Now, Let $y\in B$ be arbitrary and suppose that $h(y)\ne f^{-1}(y)$ then we see by $f$‘s injectivity that $y=f(h(y))\ne f(f^{-1}(y))=y$ which is clearly absurd. Also, let $y\in B$ be arbitrary, then by surjectivity we have that $y=f(x)$ for some $x\in A$. Then, if $g(y)\ne f^{-1}(y)$ this implies that $x=g(f(x))=g(y)\ne f^{-1}(y)=f^{-1}(f(x))=x$ which is also a contradiction. It follows that $g=f^{-1}$ and $h=f^{-1}$ and thus $g=h=f^{-1}$ as desired. 6. Problem: Let $f:\mathbb{R}\to\mathbb{R}:x\mapsto x^3-x$. By restricting the domain and range of $f$ appropriately obtain from $f$ a bijective function $g$. Proof: We note that $f'(x)=3x^2-1$ and so $f$ is increasing on $[1,2]$. Thus, $f\left([1,2]\right)\subseteq \left[f(1),f(2)\right]=\left[2,11\right]$ but since $f$ is continuous it follows from the intermediate value theorem that $f\left([1,2]\right)=[2,11]$ and thus noting that $f$ is injective on $[1,2]$ (since it’s continuous and monotone) we may conclude that $f\mid_{[1,2]}:[1,2]\to[2,11]$ is bijective. June 15, 2010 ## Munkres Chapter 1 Section 1 1. Problem: Prove the distributive laws for union and intersection, and prove DeMorgan’s laws. Proof: a) We claim that $A\cup\left(B\cap C\right)=\left(A\cup B\right)\cap \left(B\cup C\right)$. To see this we denote the right side by $P$ and the left side by $Q$, then $x\in Q$ if and only if $x\in A\text{ or }\left(x\in B\text{ and }x\in C\right)$ but by the distributive laws of the “and” and “or” operators (just draw a simple truth table) this is equivalent to $\left(x\in A\text{ or }x\in B\right)\text{ and }\left(x\in A\text{ or }x\in C\right)$ but this is just the logical statement that $x\in\left(A\cup B\right)\cap\left(A\cup C\right)=P$. The exact same, (but reversing the arrows…haha not category theory) gives that $x\in P\implies x\in Q$. It follows that $P=Q$ as desired. b) We now claim that $A\cap\left(B\cup C\right)=\left(A\cap B\right)\cup\left(A\cap C\right)$. Once again we denote the LHS by $Q$ and the RHS by $P$. Then, $x\in Q$ implies that $x\in A\text{ and }x\in B\cup C$, which implies that $x\in A\text{ and }\left(x\in B\text{ or }x\in C\right)$. Once again, constructing a truth table if necessary, we use the distributivity laws of the logical operators “and” and “or” to get the above implies $x\in \left(x\in A\text{ and }x\in B\right)\text{ or }\left(x\in A\text{ and }x\in C\right)$ which says that $x\in \left(A\cap C\right)\cup \left(A\cap C\right)=P$. Once again, using the exact same logic in reverse we get that $x\in P\implies x\in Q$ from where it follows that $P=Q$. c) We claim in general that if $\left\{U_{\alpha}\right\}_{\alpha\in\mathcal{A}}$ are subsets of some universal set $U$, then $\displaystyle U-\bigcup_{\alpha\in\mathcal{A}}U_\alpha=\bigcap_{\alpha\in\mathcal{A}}\left(U-U_{\alpha}\right)$ To see this we once again denote the LHS by $Q$ and the RHS by $P$. Then, if $x\notin Q$ then $\displaystyle x\in\bigcup_{\alpha\in\mathcal{A}}U_\alpha$ and so $x\in U_{\alpha_0}$ for some $\alpha_0\in \mathcal{A}$. It follows that $x\notin U-U_{\alpha_0}$ and thus $\displaystyle x\notin P$. Conversely, if $x\notin P$ then $x\notin U-U_{\alpha_0}$ f0r some $\alpha_0\in\mathcal{A}$ and thus $x\in U_{\alpha_0}$ and so $\displaystyle x\in\bigcup_{\alpha\in\mathcal{A}}U_{\alpha}$ and so $x\notin Q$. The conclusion follows. We now claim that $\displaystyle U-\bigcap_{\alpha\in\mathcal{A}}U_\alpha=\bigcup_{\alpha\in\mathcal{A}}\left(U-U_{\alpha}\right)$ To do this we note that $\displaystyle U-\bigcup_{\alpha\in\mathcal{A}}\left(U-U_{\alpha}\right)=\bigcap_{\alpha\in\mathcal{A}}\left(U-\left(U-U_\alpha\right)\right)=\bigcap_{\alpha\in\mathcal{A}}U_\alpha$ And thus if we keep to our LHS-RHS notation we see that $U-P=U-Q$ from where it follows from elementary set theory that $P=Q$. 2. Problem: Determine which of he following statements are true for all sets $A,B,C,D$ (contained in some universal set $U$). If a double implications fails, determine whether one or the other statement of the possible implication holds, If an equality fails, determine whether the statement becomes true if the “equals” is related by one or the other of the inclusion symbols $\subseteq$ or $\supseteq$. a) $A\subseteq B\text{ and }A\subseteq C\Leftrightarrow A\subseteq\left(C\cup B\right)$ b) $A\subseteq B\text{ or }A\subseteq C\Leftrightarrow A\subseteq\left(B\cup C\right)$ c) $A\subseteq B\text{ and }A\subseteq C\Leftrightarrow A\subseteq\left(B\cap C\right)$ d) $A\subseteq B\text{ or }A\subseteq C\Leftrightarrow A\subseteq\left(B\cap C\right)$ e) $A-\left(A-B\right)=B$ f) $A-\left(B-A\right)=A-B$ g) $A\cap\left(B-C\right)=\left(A\cap B\right)-\left(A\cap C\right)$ h) $A\cup\left(B-C\right)=\left(A\cup B\right)-\left(A\cup C\right)$ i) $\left(A\cap B\right)\cup\left(A-B\right)=A$ j) $A\subseteq C\text{ and }B\subseteq D\implies \left(A\times B\right)\subseteq\left(C\times D\right)$ k) The converse of j) l) The converse of j), assuming that $A$ and $B$ are non-empty. m) $\left(A\times B\right)\cup\left(C\times D\right)=\left(A\cup C\right)\times\left(B\cup D\right)$ n) $\left(A\times B\right)\cap\left(C\times D\right)=\left(A\cap C\right)\times\left(B\cap D\right)$ o) $A\times\left(B-C\right)=\left(A\times B\right)-\left(A\times C\right)$ p) $\left(A-B\right)\times\left(C-D\right)=\left(A\times C-B\times C\right)-A\times D$ q) $\left(A\times B\right)-\left(C\times D\right)=\left(A-C\right)\times\left(B-D\right)$ Proof: a) The right hand implication is clearly true since $A\subseteq B\text{ and }A\subseteq C\implies A\subseteq B\cap C\subseteq A\cup B$. That said, the left hand implication is false. Clearly given any non-empty set $B$ we have that $B\subseteq \varnothing\cup B$ but it is not true that $B\subseteq B\text{ and }B\subseteq A$. b) If the LHS statement is true then for each $x\in A$ we have that $x\in A\text{ or }x\in B$ and so $x\in A\cup B$, i.e. $A\subseteq B$. The other direction is not correct though. Consider the sets $A=\{1,3\},B=\{1,2\},C=\{3,4\}$ then $A\subseteq B\cup C$ but it is not true that $A\subseteq B\text{ or }A\subseteq C$. c) This two-sided implication is true. To see the right implications we note that if $x\in A$ then $x\in B$ and $x\in C$ so that $x\in B\cap C$ and so it follows that $A\subseteq B\cap C$. Conversely, if $A\subseteq B\cap C$ then for each $x\in A$ we have that $x\in B\text{ and }x\in C$ and so $A\subseteq B\text{ and }A\subseteq C$. d) The right hand implications is not false. Take for example $A=B=\{1\}$ and $C=\varnothing$. Then it is true that $A\subseteq B\text{ or }A\subseteq C$ but it is not true that $A\subseteq B\cap C=\varnothing$. The converse is true since it is weaker than the second part of the last statement. e) The equality is not always true. Consider for example $\mathbb{R}=U$ as the universal set. Then, if $A=\{1\}$ and $B=\{2\}$ then $A-\left(A-B\right)=A\cap\left(U-\left(A\cap\left(U-B\right)\right)\right)$ which is equal to $\varnothing\ne B$. The $\subseteq$ inclusion is true thought since $x\in A-\left(A-B\right)$ means that $x\in A\text{ and }x\notin A-B$ or that $x\in A\text{ and }\left(x\notin A\text{ or }x\in B\right)$ but since $x\in A$ it follows that $x\in A\cap B$. Thus, $A-\left(A-B\right)=A\cap B\subseteq B$. f) Let’s just see what the LHS is equal to. $A-\left(B-A\right)=A\cap\left(U-\left(B\cap\left(U-A\right)\right)\right)$ but this is equal to $A\cap\left(\left(U-B\right)\cup\left(A\right)\right)$ (DeMorgan’s law) but this is equal $\left(A\cap\left(U-B\right)\right)\cup\left(A\cap A\right)$ but this is equal to $\left(A\cap\left(U-B\right)\right)\cup A=A$ since the left set is a subset of $A$. It clearly follows that $A-B\subseteq A$ but the inclusion may be strict by considering $\varnothing\subsetneq B\subseteq A$ g) The LHS is equal to $A\cap\left(B\cap\left(U-C\right)\right)$ and the RHS is $A\cap B\cap\left(\left(U-A\right)\cup \left(U-C\right)\right)$ which is equal to $B\cap\left(\left(A\cap U-A\right)\cup\left(A\cap U-C\right)\right)=B\cap A\cap U-C$. Thus, they are, in fact, equal. h) The RHS is $\left(A\cup B\right)\cap\left(A'\cap C'\right)$ (prime denotes complement in the universal set) which is equal to by distributivity laws $\left(\left(A\cap\left(A'\cap C\right)\right)\right)\cup\left(B\cap\left(A'\cap C'\right)\right)=B\cap A'\cap C'$. The LHS is $A\cup\left(B\cap C'\right)=\left(A\cup B\right)\cap\left(A\cup C'\right)$. Clearly the neither inclusion holds (in general) since the LHS is always contains elements of $A$ and the RHS is contains only elements of $A'$ (if it contains any). i) The LHS is $\left(A\cap B\right)\cup\left(A-B\right)=\left(A\cap B\right)\cup\left(A\cap B'\right)$ which distributing gives $\left(A\cup\left(A\cap B'\right)\right)\cap\left(B\cup\left(A\cap B'\right)\right)$ which is equal to $A\cap\left(\left(B\cup A\right)\right)\cup\left(B\cup B'\right)$ which is equal $A\cap\left(B\cup A\right)=A$. So, yes, they are equal. j) This implication is true. If $(a,b)\in A\times B$ then $a\in A\text{ and }b\in B$ and thus $a\in C\text{ and }b\in D$ and so $(a,b)\in C\times D$ k) No, the converse is not true. Consider $A=\varnothing, B=\{1\},C=D=\{2\}$ then $A\times B=\varnothing\subseteq C\times D=\{(2,2)\}$. But, clearly $A\subseteq C\text{ and }B\subseteq D$ is not true. l) In this case it is true. Let $a\in A$ then for some (we know there exists some, since $B\ne\varnothing$) $b\in B$ we have that $(a,b)\in A\times B\subseteq C\times D$ and so $a\in C$. The other direction works the same. m) This equality is false. Consider the intervals $A=[0,1]=B=[0,1],C=D=[0,2]$. Note then that $\left(A\times B\right)\cup\left(C\times D\right)\subsetneq \left(A\cup C\right)\times\left(B\cup D\right)$ (once can picture this as taking the square of side length two whose bottom left hand corner is on the origin. Then, if we carve this square into four smaller squares in the obvious way the LHS is just the upper right and lower left ones whereas the RHS is the full square). In general $\left(A\times B\right)\cup\left(C\times D\right)\subseteq\left(A\cup C\right)\times\left(B\cup D\right)$. To see this we note that if $(a,b)\in\left(A\times B\right)\cup\left(C\times D\right)$ then we may assume WLOG that $(a,b)\in A\times B$ and thus $a\in A\text{ and }b\in B$ and so $a\in A\cup C,b\in B\cup D$ and so $(a,b)\in \left(A\cup C\right)\times\left(B\cup D\right)$ n) This is true. To see this we note that if $(a,b)\in\left(A\cap C\right)\times \left(B\cap D\right)$ then $a\in A\cap C \text{ and }b\in B\cap D$ and so $\left(a\in A\text{ and }a\in C\right)\text{ and }\left(b\in B\text{ and }b\in D\right)$ and so in particular $\left(a\in A\text{ and }b\in B\right)\text{ and }\left(a\in C\text{ and }b\in D\right)$ or $(a,b)\in \left(A\times B\right)\cap\left(C\times D\right)$. Conversely, if $(a,b)\in\left(A\times B\right)\cap\left(C\times D\right)$ then $(a,b)\in A\times B$ and $(a,b)\in C\times D$ which says that $a\in A\text{ and }b\in B$ and $a\in C\text{ and }b\in D$ and so $a\in A\cap C$ and $b\in B\cap D$ so that $(a,b)\in\left(A\cap C\right)\times\left(B\cap D\right)$ o) This is true. Note that $(a,b)\in \left(A\times B\right)-\left(A\times C\right)$ if and only if $(a,b)\in A\times B\text{ and }(a,b)\notin A\times C$ which is true if and only if $\left(a\in A\text{ and }b\in B\right)\text{ and }\left(a\notin A\text{ or }b\notin C\right)$ which is true if and only if $a\in A\text{ and }b\in B\text{ and }b\notin C$ or $a\in A\text{ and }b\in B-C$ and so $(a,b)\in A\times\left(B-C\right)$ p) This is true. Note that $(a,b)\in\left(A-B\right)\times\left(C-D\right)$ if and only if $a\in A-B\text{ and }b\in C-D$ which is true if and only if $\left(a\in A\text{ and }a\notin B\right)\text{ and }\left(b\in C\text{ and }b\notin D\right)$. But, this is true if and only if $\left(a\in A\text{ and }b\in C\right)\text{ and }\left(a\notin B\text{ or }b\notin C\right)\text{ and }\left(a\notin A\text{ and }b\notin D\right)$ (because by the first part we must have $a\in A$ and so the last two are just “tricks”) which is true if and only if $(a,b)\in A\times C\text{ and }(a,b)\notin B\times C\text{ and }(a,b)\notin A\times D$ which is just a fancy way of saying $(a,b)\in A\times C-B\times C-A\times D$ q) This equality does not hold in general. If $a\in A,b\in B,a\in C,b\notin D$ then $(a,b)\in A\times B-C\times D$ but $(a,b)\notin\left(A-C\right)\times\left(B-D\right)$. The, $\supseteq$ inclusion does hold though. To see this let $(a,b)\in \left(A-C\right)\times\left(B-D\right)$ then $a\in A-C\text{ and }b\in B-D$ and so in particular $(a,b)\in A\times B$ but $(a,b)\notin C\times D$ and so $(a,b)\in A\times B-C\times D$ Remark: God that was horrendous… 3. Problem: a)Write the contrapostive and converse of the following statement: “If $x<0, then$latex x^2-x>0$” and determine which (if any) of the three statements are true.

b) Do the same for the statement “if $x>0$ then $x^2-x>0$.

Proof:

a) Contrapositive says $if x^2-x\leqslant 0$ then $x\geqslant 0$. The negation says that if $x<0$ then $x^2-x\geqslant 0$

We note that $x^2\leqslant x$ then $0\leqslant x$. Thus, the contrapositive is true. The negation is not true. The actual statement, being logically equivalent to the contrapositive, is true.

b) This is exactly the same idea.

4.

Problem: Let $A$ and $laetx B$ be sets of real numbers. Write the negation of each of the following statements:

a) For every  $a\in A$ it is true that $a^2\in A$

b) For at least one $a\in A$ it is true that $a^2\in B$

c) For every $a\in A$ it is true that $a^2\notin B$

d) For at least one $a\notin A$, it is true that $a^2\in B$

Proof:

a) There exists some $a\in A$ such that $a^2\notin B$

b) For every $a\in A$ it is true that $a^2\notin B$

c) There exists some $a\in A$ such that $a^2\in B$

d) For every $a\notin A$ it is true that $a^2\notin B$

5.

Problem: Let $\mathcal{A}$ be a non-empty collection of sets. Determine the truth of each of the following statements about their converses:

a) $\displaystyle x\in\bigcup_{A\in\mathcal{A}}A\implies x\in A\text{ for some }A\in\mathcal{A}$

b) $\displaystyle x\in\bigcup_{A\in\mathcal{A}}A\implies x\in A\text{ for every }x\in A$

c) $\displaystyle x\in\bigcap_{A\in\mathcal{A}}A\implies x\in A\text{ for some }A\in\mathcal{A}$

d) $\displaystyle x\in\bigcap_{A\in\mathcal{A}}A\implies x\in A\text{ for every }A\in\mathcal{A}$

Proof:

a) Of course this is true, it’s the definition.

b) Clearly not. Take the elements of $\mathcal{A}$ to be disjoint.

c) Of course! By definition if it’s in the intersections it’s in all of them

d) What I said in the last one.

6.

Problem: Write the contrapositive of each of the statements of Exercise 5.

a) If for every $A\in\mathcal{A}$ it is true that $x\notin A$ then $\displaystyle x\notin\bigcup_{A\in\mathcal{A}}A$

b) If for some $A\in\mathcal{A}$ it is true that $x\notin A$ then $\displaystyle x\notin\bigcup_{A\in\mathcal{A}}A$

c) If $x\notin A$ for some $A\in\mathcal{A}$ then $\displaystyle x\notin\bigcap_{A\in\mathcal{A}}A$

d) IF $x\notin A$ for every $A\in\mathcal{A}$ then $\displaystyle x\notin\bigcap_{A\in\mathcal{A}}A$

7.

Problem: Given sets $A,B$ and $C$, express each of the following sets in terms of $A,B$ and $C$ using the symbols $\cup,\cap$ and $-$.

Proof:

a) The set is $D=\left\{x\in U:x\in A\text{ and}\left(x\in B\text{ or }x\in c\right)\right\}=A\cap\left(B\cup C\right)$

b) The set is $E=\left\{x\in U:\left(x\in A\text{ and }x\in B\right)\text{ or }x\in C\right\}=\left(A\cap B\right)\cup C$

c) The set is $F=\left\{x\in U:x\in A\text{ and }\left(x\in B\implies x\in C\right)\right\}=A\cap\left(\left(U-B\right)\cup\left(B\cap c\right)\right)$

8.

Problem: If a set $A$ has two elements, show that $\mathcal{P}(A)$ has four elements. How may elements does $\mathcal{P}(A)$ have if $A$ has one element? Three elements? No elements? Why is $\mathcal{P}(A)$ called the power set of $A$?

Proof: We prove that $\text{card }A=n\implies\text{card }\mathcal{P}(A)=2^n$. To do this we let $a_n=\text{card }\mathcal{P}(\{1,\cdots,n\})$.

Partition $\mathcal{P}\left(\{1,\cdots,n+1\}\right)$ into two blocks, namely

$B_1=\left\{S\in\mathcal{P}\left(\{1,\cdots,n+1\}\right):n+1\notin S\right\}$

and

$B_2=\mathcal{P}\left(\{1,\cdots,n+1\}\right)-B_1$.

Note, that clearly $\text{card }B_2=\text{card }\mathcal{P}\left(\{1,\cdots,n\}\right)=a_n$. Now, given $S\in B_1$ we have two choices, either $S=\{n+1\}$ or $S\cap\{1,\cdots,n\}\ne\varnothing$. Thus, excluding $\{n+1\}$ for a second each element of $B_1$ is paired with an element of $\mathcal{P}\left(\{1,\cdots,n\}\right)$ is a unique way. It follows that $\text{card }B_2-\{\{n+1\}\}\leqslant a_n$. But, noting that in fact for each $S\in\mathcal{P}\left(\{1,\cdots,n+1\}\right)-\{\varnothing\}$ we have that $S\cup\{n+1\}\in B_1$ it follows that $\text{card }B_2-\{\{n+1\}\}=a_n-1\implies \text{card }B_2=a_n$. Thus,

$a_{n+1}=\text{card }\mathcal{P}\left(\{1,\cdots,n+1\}\right)=\text{card }B_1+\text{card }B_2=2a_n$

Thus, noting that $a_0=\text{card }\mathcal{P}(\varnothing)=1$ we can easily prove by induction that $a_n=2^n$

9.

Problem: Formulate and prove a generalized version of DeMorgan’s theorem.

Proof: Already did, cf. problem one.

10.

Problem: Let $\mathbb{R}$ denote the set of real numbers. For each of the following subsets of $\mathbb{R}\times\mathbb{R}$, determine whether it is equal to the Cartesian product of two subsets of $\mathbb{R}$

a) $A=\left\{(x,y):x\in\mathbb{Z}\right\}$

b) $B= \left\{(x,y):0

c) $C=\left\{(x,y):y>x\right\}$

d) $D=\left\{(x,y):x\notin\mathbb{Z}\text{ and }y\in\mathbb{Z}\right\}$

e) $E=\left\{(x,y):x^2+y^2<1\right\}$

Proof:

a) Yes, $A=\mathbb{Z}\times\mathbb{R}$

b) Yes, $B=\mathbb{R}\times(0,1]$

c) No, notice that $(1,2)\in C$ as well as $(3,4)$ but $(3,1)\notin C$.

d) Yes, $D=\left(\mathbb{R}-\mathbb{Z}\right)\times\mathbb{Z}$

e) No, notice that $(1,0),(0,1)\in E$ but $(1,1)\notin E$.

June 13, 2010

## Munkres Chapter 2 Section 19 (Part II)

9.

Problem: Show that the axiom of choice (AOC) is equivalent to the statement that for any indexed family $\left\{U_{\alpha}\right\}_{\alpha\in\mathcal{A}}$ of non-empty sets, with $\mathcal{A}\ne\varnothing$, the product $\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha\ne\varnothing$

Proof: This is pretty immediate when one writes down the actual definition of the product, namely:

$\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha=\left\{\bold{x}:\mathcal{A}\to\bigcup_{\alpha\in\mathcal{A}}U_{\alpha}:\bold{x}(\alpha)\in U_\alpha,\text{ for every }\alpha\in\mathcal{A}\right\}$

So, if one assumes the AOC then one must assume the existence of a choice function

$\displaystyle c:\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}\to\bigcup_{\alpha\in\mathcal{A}},\text{ such that }c\left(U_\alpha\right)\in U_\alpha\text{ for all }\alpha\in\mathcal{A}$

So, then if we consider $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}\overset{\text{def.}}{=}\Omega$ as just a class of sets, the fact that we have indexed them implies there exists a surjective “indexing function$$i:\mathcal{A}\to\Omega$ where clearly since we have already indexed out set we have that $i:\alpha\mapsto U_\alpha$. So, consider $c\circ i:\mathcal{A}\to\bigcup_{\alpha\in\mathcal{A}}U_\alpha$ This is clearly a well-defined mapping and $\left(c\circ i\right)(\alpha)=c\left(U_\alpha\right)\in U_\alpha$ and thus $\displaystyle c\circ i\in\prod_{\alpha\in\mathcal{A}}U_\alpha$ from where it follows that $\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha\ne\varnothing$ Conversely, let $\Omega$ be a class of sets and let $i:\mathcal{A}\to\Omega$ be an indexing function. We may then index $\Omega$ by $\Omega=\left\{U_{\alpha}\right\}_{\alpha\in\mathcal{A}}$. Then, by assumption $\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha\ne\varnothing$ Thus there exists some $\displaystyle \bold{x}\in\prod_{\alpha\in\mathcal{A}}U_\alpha$ Such that $\displaystyle \bold{x}:\mathcal{A}\to\bigcup_{\alpha\in\mathcal{A}},\text{ such that }\bold{x}(\alpha)\in U_\alpha$ Thus, we have that $\displaystyle \bold{x}\circ i^{-1}:\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}\to\bigcup_{\alpha\in\mathcal{A}}$ is a well-defined mapping with $\displaystyle \bold{x}\left(U_\alpha\right)\in U_\alpha$ For each $\alpha\in\mathcal{A}$. It follows that we have produced a choice function for $\Omega$ and the conclusion follows. $\blacksquare$ Remark: We have assumed the existence of a bijective indexing function $i:\mathcal{A}\to\Omega$, but this is either A) a matter for descriptive set theory or B) obvious since $\text{id}:\Omega\to\Omega$ satisfies the conditions. This depends on your level of rigor. 10. Problem: Let $A$ be a set; let $\left\{X_\alpha\right\}_{\alpha\in\mathcal{A}}$ be an indexed family of spaces; and let $\left\{f_\alpha\right\}_{\alpha\in\mathcal{A}}$ be an indexed family of functions $f_\alpha:A\to X_\alpha$ a) Prove there is a unique coarsest topology $\mathfrak{J}$ on $A$ relative to whish each of the functions $f_\alpha$ is continuous. b) Let $\mathcal{S}_\beta=\left\{f_\beta^{-1}\left(U_\beta\right):U_\beta\text{ is ope}\text{n in}X_\beta\right\}$ and let $\displaystyle \mathcal{S}=\bigcup_{\alpha\in\mathcal{A}}$. Prove that $\mathcal{S}$ is a subbasis for $\mathfrak{J}$. c) Show that the map $g:Y\to A$ is continuous relative to $\mathfrak{J}$ if and only if each map $f_\alpha\circ g:Y\to X_\alpha$ is continuous. d) Let $\displaystyle f:A\to\prod_{\alpha\in\mathcal{A}}X_\alpha$ be defined by the equation $f(x)=\left(f_\alpha(x)\right)_{\alpha\in\mathcal{A}}$ Let $Z$ denote the subspace of $f\left(A\right)$ of the product space $\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha$. Prove taht the image under $f$ of each element of $\mathfrak{J}$ is an open set in $Z$. Proof: a) We first prove a lemma Lemma: Let $\mathfrak{J}$ be a topology on $A$, then all the mappings $f_\alpha:A\to X_\alpha$ are continuous if and only if $\mathcal{S}\subseteq\mathfrak{J}$ where $\mathcal{S}$ is defined in part b). Proof:Suppose that all the mappings $f_\alpha:A\to X_\alpha$ are continuous. Then, given any open set $U_\alpha\in X_\alpha$ we have that $f_\alpha$ is continuous and so $f_\alpha^{-1}\left(U_\alpha\right)$ is open and thus $f_{\alpha}^{-1}\left(U_\alpha\right)\in\mathfrak{J}$ from where it follows that $\mathcal{S}\subseteq\mathfrak{J}$. Conversely, suppose that $\mathcal{S}\subseteq\mathfrak{J}$. It suffices to prove that $f_\alpha:A\to X_\alpha$ for a fixed but arbitrary $\alpha\in\mathcal{A}$. So, to do this let $U$ be open in $X_\alpha$ then $f_{\alpha}^{-1}\left(U\right)\in\mathfrak{J}$ and thus by assumption $f_\alpha^{-1}\left(U\right)\in\mathfrak{J}$; but this precisely says that $f_\alpha^{-1}\left(U\right)$ is open in $A$. By prior comment the conclusion follows. $\blacksquare$ So, let $\mathcal{C}=\left\{\mathfrak{T}:\mathfrak{J}\text{ is a topology on }A\text{ and }\mathcal{S}\subseteq\mathfrak{J}\right\}$ and let $\displaystyle \mathfrak{J}=\bigcap_{\mathfrak{I}\in\mathcal{C}}\mathfrak{T}$ By previous problem $\mathfrak{J}$ is in fact a topology on $A$, and by our lemma we also know that all the mappings $f_\alpha:A\to X_\alpha$ are continuous since $\mathcal{S}\subseteq\mathfrak{J}$. To see that it’s the coarsest such topology let $\mathfrak{U}$ be a topology for which all of the $f_\alpha:A\to X_\alpha$ are continuous. Then, by the other part of our lemma we know that $\mathcal{S}\subseteq\mathfrak{U}$ and thus $\mathfrak{U}\in\mathcal{C}$. So, $\displaystyle \mathfrak{J}=\bigcap_{\mathfrak{T}\in\mathcal{C}}\mathfrak{T}\subseteq\mathfrak{U}$ And thus $\mathfrak{J}$ is coarser than $\mathfrak{U}$. The uniqueness is immediate. b) It follows from the previous problem that we must merely show that $\mathcal{S}$ is a subbasis for the topology $\mathfrak{J}$. The conclusion will follow from the following lemma (which was actually an earlier problem, but we reprove here for referential reasons): Lemma: Let $X$ be a set and $\Omega$ be a subbasis for a topology on $X$. Then, the topology generated by $\Omega$ equals the intersection of all topologies which contain $\Omega$. Proof: Let $\mathcal{C}\left\{\mathfrak{T}:\mathfrak{T}\text{ is a topology on }X\text{ and }\Omega\subseteq\mathfrak{T}\right\}$ and $\displaystyle \mathfrak{J}=\bigcap_{\mathfrak{T}\in\mathcal{C}}\mathfrak{T}$ Also, let $\mathfrak{G}$ be the topology generated by the subbasis $\Omega$. Clearly since $\Omega\subseteq\mathfrak{G}$ we have that $\mathfrak{J}\subseteq\mathfrak{G}$. Conversely, let $U\in\mathfrak{G}$. Then, by definition to show that $U\in\mathfrak{J}$ it suffices to show that $U\in\mathfrak{T}$ for a fixed but arbitrary $\mathfrak{T}\in\mathcal{C}$. To do this we first note that by definition that $\displaystyle U=\bigcup_{\alpha\in\mathcal{A}}U_\alpha$ where each $U_\alpha=O_1\cap\cdots\cap O_{m_\alpha}$ for some $O_1,\cdots,O_{m_\alpha}\in\Omega$. Now, if $\mathfrak{T}\in\mathcal{C}$ we know (since $\Omega\subseteq\mathfrak{T}$) that $O_1,\cdots,O_{m_\alpha}\in\mathfrak{T}$ and thus $O_1\cap\cdots\cap O_{m_\alpha}=U_\alpha\in\mathfrak{T}$ for each $\alpha\in\mathcal{A}$. It follows that $U$ is the union of sets in $\mathfrak{T}$ and thus $U\in\mathfrak{T}$. It follows from previous comment that $\mathfrak{G}\subseteq\mathfrak{J}$. The conclusion follows. $\blacksquare$ The actual problem follows immediately from this. c) So, let $g:Y\to A$ be some mapping and suppose that $f_\alpha\circ g:Y\to X_\alpha$ is continuous for each $\alpha\in\mathcal{A}$. Then, given a subbasic open set $U$ in $A$ we have that $U=f_{\alpha_1}^{-1}\left(U_{\alpha_1}\right)\cap\cdots\cap f_{\alpha_n}^{-1}\left(U_{\alpha_n}\right)$ for some $\alpha_1,\cdots,\alpha_n$ and for some open sets $U_{\alpha_1},\cdots,U_{\alpha_n}$ in $X_{\alpha_1},\cdots,X_{\alpha_n}$ respectively. Thus $g^{-1}(U)$ may be written as $\displaystyle g^{-1}\left(\bigcup_{j=1}^{n}f_{\alpha_j}^{-1}\left(U_{\alpha_j}\right)\right)=\bigcup_{j=1}^{n}g^{-1}\left(f_{\alpha_j}^{-1}\left(U_{\alpha_j}\right)\right)=\bigcup_{j=1}^{n}\left(f_{\alpha_j}\circ g\right)^{-1}\left(U_{\alpha_j}\right)$ but since each $f_{\alpha_j}\circ g:Y\to X_{\alpha_j}$ we see that $g^{-1}\left(U\right)$ is the finite union of open sets in $Y$ and thus open in $Y$. It follows that $g$ is continuous. Conversely, suppose that $g$ is continuous then $f_\alpha\circ g:Y\to X_{\alpha}$ is continuous since it’s the composition of continuous maps. d) First note that $\displaystyle f^{-1}\left(\prod_{\alpha\in\mathcal{A}}U_\alpha\right)=\bigcap_{\alpha\in\mathcal{A}}f\alpha^{-1}\left(U_\alpha\right)$ from where it follows that the initial topology under the class of maps $\{f_\alpha\}$ on $A$ is the same as the initial topology given by the single map $f$. So, in general we note that if $X$ is given the initial topology determined by $f:X\to Y$ then given an open set $f^{-1}(U)$ in $X$ we have that $f\left(f^{-1}(U)\right)=U\cap f(X)$ which is open in the subspace $f(X)$. June 9, 2010 ## Munkres Chapter 2 Section 19 (Part I) 1. Problem: Suppose that for each $\alpha\in\mathcal{A}$ the topology on $X_\alpha$ is given by a basis $\mathfrak{B}_\alpha$. The collection of all sets of the form $\displaystyle \prod_{\alpha\in\mathcal{A}}B_\alpha$ Such that $B_\alpha\in\mathfrak{B}_\alpha$ is a basis for $\displaystyle X=\prod_{\alpha\in\mathcal{A}}X_\alpha$ with the box topology, denote this collection by $\Omega_B$. Also, the collection $\Omega_P$ of all sets of the form $\displaystyle \prod_{\alpha\in\mathcal{A}}B_\alpha$ Where $B_\alpha\in\mathfrak{B}_\alpha$ for finitely many $\alpha$ and $B_\alpha=X_\alpha$ otherwise is a basis for the product topology on $X$. Proof: To prove the first part we let $U\subseteq X$ be open. Then, by construction of the box topology for each $(x_\alpha)_{\alpha\in\mathcal{A}}\overset{\text{def.}}{=}(x_\alpha)\in U$ we may find some $\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha$ such that $U_\alpha$ is open in $X_\alpha$ and $\displaystyle (x_\alpha)\in \prod_{\alpha\in\mathcal{A}}U_\alpha\subseteq U$. So, then for each $x_\alpha$ we may find some $B_\alpha\in\mathfrak{B}_\alpha$ such that $x_\alpha\in B_\alpha\subseteq U_\alpha$ and thus $\displaystyle (x_\alpha)\in\prod_{\alpha\in\mathcal{A}}B_\alpha\subseteq\prod_{\alpha\in\mathcal{A}}U_\alpha\subseteq U$ Noticing that $\displaystyle \prod_{\alpha\in\mathcal{A}}B_\alpha\in\Omega_B$ and every element of $\Omega_B$ is open finishes the argument. Next, we let $U\subseteq X$ be open with respect to the product topology. Once again for each $(x_\alpha)\in U$ we may find some $\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha$ such that $U_\alpha$ is open in $X_\alpha$ for each $\alpha\in\mathcal{A}$ and $U_\alpha=X_\alpha$ for all but finitely many $\alpha$, call them $\alpha_1,\cdots,\alpha_m$. So, for each $\alpha_k,\text{ }k=1,\cdots,m$ we may find some $B_k\in\mathfrak{B}_k$ such that $x\in B_k\subseteq U_k$ and so $\displaystyle (x_\alpha)\in\prod_{\alpha\in\mathcal{A}}V_\alpha\subseteq\prod_{\alpha\in\mathcal{A}}U_\alpha\subseteq U$ Where $\displaystyle V_\alpha=\begin{cases}B_k\quad\text{if}\quad \alpha=\alpha_k ,\text{ }k=1,\cdots,m\\ X_\alpha\quad\text{if}\quad x\ne \alpha_1,\cdots,\alpha_m\end{cases}$ Noting that $\displaystyle \prod_{\alpha\in\mathcal{A}}V_\alpha\in\Omega_P$ and $\Omega_P$ is a collection of open subsets of $X$ finishes the argument. $\blacksquare$ 2. Problem: Let $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ be a collection of topological spaces such that $U_\alpha$ is a subspace of $X_\alpha$ for each $\alpha\in\mathcal{A}$. Then, $\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha=Y$ is a subspace of $\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha=X$ if both are given the product or box topology. Proof: Let $\mathfrak{J}_S,\mathfrak{J}_P$ denote the topologies $Y$ inherits as a subspace of $X$ and as a product space respectively. Note that $\mathfrak{J}_S,\mathfrak{J}_P$ are generated by the bases $\mathfrak{B}_S=\left\{B\cap Y:B\in\mathfrak{B}\right\}$ (where $\mathfrak{B}$ is the basis on $X$ with the product topology), and $\displaystyle \mathfrak{B}_P=\left\{\prod_{\alpha\in\mathcal{A}}O_\alpha:O_\alpha\text{ is open in }U_\alpha,\text{ and equals }U_\alpha\text{ for all but finitely many }\alpha\right\}$ So, let $(x_\alpha)\in B$ where $B\in\mathfrak{B}_S$ then $\displaystyle B=Y\cap\prod_{\alpha\in\mathcal{A}}V_\alpha=\prod_{\alpha\in\mathcal{A}}\left(V_\alpha\cap U_\alpha\right)$ Where $V_\alpha$ is open in $X_\alpha$, and thus $V_\alpha\cap Y$ is open in $U_\alpha$. Also, since $V_\alpha=X_\alpha$ for all but finitely many $\alpha$ it follows that $V_\alpha\cap U_\alpha=U_\alpha$ for all but finitely many $\alpha$. And so $B\in\mathfrak{B}_P$. Similarly, if $B\in\mathfrak{B}_P$ then $\displaystyle B=\prod_{\alpha\in\mathcal{A}}O_\alpha$ Where $O_\alpha$ is open in $U_\alpha$, but this means that $O_\alpha=V_\alpha\cap U_\alpha$ for some open set $V_\alpha$ in $X_\alpha$ and so $\displaystyle B=\prod_{\alpha\in\mathcal{A}}O_\alpha=\prod_{\alpha\in\mathcal{A}}\left(V_\alpha\cap U_\alpha\right)=\prod_{\alpha\in\mathcal{A}}V_\alpha\cap Y\in\mathfrak{B}_S$ From where it follows that $\mathfrak{B}_S,\mathfrak{B}_P$ and thus $\mathfrak{J}_S,\mathfrak{J}_P$ are equal. The case for the box topology is completely analgous. $\blacksquare$ 3. Problem: Let $\left\{X_\alpha\right\}_{\alpha\in\mathcal{A}}$ be a collection of Hausdorff spaces, then $\displaystyle X=\prod_{\alpha\in\mathcal{A}}X_\alpha$ is Hausdorff with either the box or product topologies Proof: It suffices to prove this for the product topology since the box topology is finer. So, let $(x_\alpha),(y_\alpha)\in X$ be distinct. Then, $x_\beta\ne y_\beta$ for some $\beta\in\mathcal{A}$. Now, since $X_\beta$ is Hausdorff there exists disjoint open neighborhoods $U,V$ of $x_\beta,y_\beta$ respectively. So, $\pi_\beta^{-1}(U),\pi_\beta^{-1}(V)$ are disjoint open neighborhoods of $(x_\alpha),(y_\alpha)$ respectively. The conclusion follows. $\blacksquare$ 4. Problem: Prove that $\left(X_1\times\cdots\times X_{n-1}\right)\times X_n\overset{\text{def.}}{=}X\approx X_1\times\cdots\times X_n\overset{\text{def.}}{=}Y$. Proof: Define $\displaystyle \varphi:X\to Y:((x_1,\cdots,x_{n-1}),x_n)\mapsto (x_1,\cdots,x_n)$ Clearly this is continuous since $\pi_{\beta}\circ\varphi=\pi_\beta$ 5. Problem: One of the implications holds for theorem 19.6 even in the box topology, which is it? Proof: If $\displaystyle f:A\to\prod_{\alpha\in\mathcal{A}}X_\alpha$ where the latter is given the box topology then we have that each $\pi_\alpha$ is continuous and thus so is each $\pi_\alpha\circ f:A\to X_\alpha$. $\blacksquare$# 6. Problem: Let $\left\{\bold{x}_n\right\}_{n\in\mathbb{N}}$ be a sequence of points in the product space $\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha=X$. Prove that $\left\{\bold{x}_n\right\}_{n\in\mathbb{N}}$ converges to $\bold{x}$ if and only if the sequences $\left\{\pi_\alpha(\bold{x}_n)\right\}_{n\in\mathbb{N}}$ coverge to $\pi_\alpha(\bold{x})$ for each $\alpha\in\mathcal{A}$. Is this fact true if one uses the box topology? Proof: Suppose that $U$ is a neighborhood of $\pi_{\alpha}(\bold{x})$ such that $\left(X_\alpha-U\right)\cap\left\{\bold{x}_n:n\in\mathbb{N}\right\}=K$ is infinite. Notice then that if $\pi_{\alpha}(\bold{x}_n)\in K$ that $\bold{x}_n\notin\pi_{\alpha}^{-1}\left(U\right)$ from where it follows that $\pi_{\alpha}^{-1}\left(U\right)$ is a neighborhood of $\bold{x}$ which does not contain all but finitely many values of $\left\{\bold{x}_n:n\in\mathbb{N}\right\}$ contradicting the fact that $\bold{x}_n\to\bold{x}$ in $X$. Conversely, suppose that $\pi_{\alpha}(\bold{x}_n)\to\pi_{\alpha}(\bold{x})$ for each $\alpha\in\mathcal{A}$ and let $\displaystyle \prod_{\alpha\in\mathcal{A}}U_{\alpha}$ be a basic open neighborhood of $\bold{x}$. Then, letting $\alpha_1,\cdots,\alpha_m$ be the finitely many indices such that $U_{\alpha_k}\ne X_k$. Since each $\pi_{\alpha_k}(\bold{x}_n)\to\pi_{\alpha_k}(\bold{x})$ there exists some $n_\ell\in\mathbb{N}$ such that $n_\ell\leqslant n\implies \pi_{\alpha_k}(\bold{x}_k)\in U_k$. So, let $N=\max\{n_1,\cdots,n_k\}$. Now, note that if $\displaystyle \bold{x}_n\notin\prod_{\alpha\in\mathcal{A}}U_\alpha$ then $\pi_{\alpha}(\bold{x}_n)\notin U_\alpha$ for some $\alpha\in\mathcal{A}$. But, since clearly $\pi_{\alpha}(\bold{x}_n)\in X_\alpha$ we must have that $\pi_{\alpha}(\bold{x}_n)\notin U_{\alpha_1}\cup\cdots\cup U_{\alpha_m}$ and thus $n\leqslant N$. It follows that for every $N\leqslant n$ we have that $\displaystyle \bold{x}_n\in\prod_{\alpha\in\mathcal{A}}U_\alpha$. Then, since every neighborhood of $\bold{x}$ contains a basic open neighborhood and the above basic open neighborhood was arbitrary the conclusion follows. Clearly the first part holds in the box topology since there was no explicit mention of the product topology or it’s defining characteristics. That said, the second does not hold. Consider $\displaystyle \bold{x}_n=\prod_{m\in\mathbb{N}}\left\{\frac{1}{n}\right\}$. Clearly each coordinate converges to zero, but $\displaystyle \prod_{m\in\mathbb{N}}\left(\frac{-1}{m},\frac{1}{m}\right)=U$ is a neighborhood of $\bold{0}$ in the product topology. But, if one claimed that for every $n\geqslant N$ (for some $N\in\mathbb{N}$ that $\bold{x}_n\in U$ they’d be wrong. To see this merely note $\displaystyle \frac{1}{N}\notin\left(\frac{-1}{N+1},\frac{1}{N+1}\right)$ and so $\pi_{N+1}\left(\bold{x}_N\right)\notin\pi_{N+1}(U)$ and thus $\bold{x}_{N}\notin U$. 7. Problem: Let $\mathbb{R}^{\infty}$ be the subset of $\mathbb{R}^{\omega}$ consisting of all eventually zero sequences. What is $\overline{\mathbb{R}^{\infty}}$ in the box and product topology? Proof: We claim that in the product topology $\overline{\mathbb{R}^{\infty}}=\mathbb{R}^{\omega}$. To see this let $\displaystyle \prod_{n\in\mathbb{N}}U_n$ be a basic non-empty open set in $\mathbb{R}^{\omega}$ with the product topology. Since we are working with the product topology we know there are finitely many indices $n_1,\cdots,n_m$ such that $U_{n_k}\ne \mathbb{R}$. So, for each $n_1,\cdots,n_m$ select some $x_{n_k}\in U_{n_k}$ and consider $(x_n)_{n\in\mathbb{N}}$ where $\displaystyle \begin{cases} x_{n_k}\quad\text{if}\quad n=n_1,\cdots,n_m \\ 0 \text{ } \quad\text{if}\quad \text{otherwise}\end{cases}$ Clearly then $\displaystyle (x_n)_{n\in\mathbb{N}}\in \prod_{n\in\mathbb{N}}U_n\cap\mathbb{R}^{\infty}$ and thus every non-empty open set in $\mathbb{R}^{\omega}$ intersects $\mathbb{R}^{\infty}$ and the conclusion follows. Now, we claim that with the box topology that $\overline{\mathbb{R}^{\infty}}=\mathbb{R}^{\infty}$. To see this let $(x_n)_{n\in\mathbb{N}}\notin\mathbb{R}^{\infty}$. Then, there exists some subsequence $\{x_{\varphi(n)}\}$ of the sequence $\{x_n\}$ which is non-zero. For each $\varphi(n)$ form an interval $I_{\varphi(n)}$ such that $0\notin I_{\varphi(n)}$. Then, consider $\displaystyle \prod_{n\in\mathbb{N}}U_n,\text{ }U_n=\begin{cases} I_{\varphi(m)}\quad\text{if}\quad n=\varphi(m)\\ \mathbb{R}\quad\text{ }\quad\text{if}\quad \text{otherwise}\end{cases}$ Clearly then $\displaystyle \prod_{n\in\mathbb{N}}U_n$ is a neighborhood of $(x_n)_{n\in\mathbb{N}}$ and since each clearly has an infinite subsequence of non-zero values it is disjoint from $\mathbb{R}^{\infty}$. It follows that in $\mathbb{R}^{\omega}$ with the box topology that $\mathbb{R}^{\infty}$ is closed and thus $\overline{\mathbb{R}^{\infty}}=\mathbb{R}^{\infty}$ as desired. $\blacksquare$ 8. Problem: Given sequences $(a_n)_{n\in\mathbb{N}}$ and $(b_n)_{n\in\mathbb{N}}$ of real numbers with $a_n>0$ define $\varphi:\mathbb{R}^{\omega}\to\mathbb{R}^{\omega}$ by the equation $\varphi:(x_n)_{n\in\mathbb{N}}\mapsto \left(a_n x_n+b_n\right)_{n\in\mathbb{N}}$ Show that if $\mathbb{R}^{\omega}$ is given the product topology that $\varphi$ is a homeomorphism. What happens if $\mathbb{R}^{\omega}$ is given the box topology? Proof: Let us first prove that $\varphi$ is a bijection. To do this we prove something more general… Lemma: Let $\left\{X_\alpha\right\}_{\alpha\in\mathcal{A}}$ and $\left\{Y_\alpha\right\}_{\alpha\in\mathcal{A}}$ be two classes of untopologized sets and $\left\{f_\alpha\right\}_{\alpha\in\mathcal{A}}$ a collection of bijections $f_\alpha:X_\alpha\to Y_\alpha$. Then, if $\displaystyle \prod_{\alpha\in\mathcal{A}}f_\alpha\overset{\text{def.}}{=}\varphi:\prod_{\alpha\in\mathcal{A}}X_\alpha\to\prod_{\alpha\in\mathcal{A}}Y_\alpha:(x_\alpha)_{\alpha\in\mathcal{A}}\mapsto\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}$ we have that $\varphi$ is a bijection. Proof: To prove injectivity we note that if $\varphi\left((x_\alpha)_{\alpha\in\mathcal{A}}\right)=\varphi\left((y_\alpha)_{\alpha\in\mathcal{A}}\right)$ Then, $\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}=\left(f_\alpha(y_\alpha)\right)_{\alpha\in\mathcal{A}}$ And by definition of an $\alpha$-tuple this implies that $f_\alpha(x_\alpha)=f_\alpha(y_\alpha)$ for each $\alpha\in\mathcal{A}$. But, since each $f_\alpha:X_\alpha\to Y_\alpha$ is injective it follows that $x_\alpha=y_\alpha$ For each $\alpha\in\mathcal{A}$. Thus, $(x_\alpha)_{\alpha\in\mathcal{A}}=(y_\alpha)_{\alpha\in\mathcal{A}}$ as desired. To prove surjectivity we let $\displaystyle (y_\alpha)_{\alpha\in\mathcal{A}}\in\prod_{\alpha\in\mathcal{A}}Y_\alpha$ be arbitrary. We then note that for each fixed $\alpha\in\mathcal{A}$ we have there is some $x_\alpha\in X_\alpha$ such that $f_\alpha(x_\alpha)=y_\alpha$. So, if $\displaystyle (x_\alpha)_{\alpha\in\mathcal{A}}\in\prod_{\alpha\in\mathcal{A}}X_\alpha$ is the corresponding $\alpha$-tuple of these values we have that $\varphi\left((x_\alpha)_{\alpha\in\mathcal{A}}\right)=\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}=(y_\alpha)_{\alpha\in\mathcal{A}}$ from where surjectivity follows. Combining these two shows that $\varphi$ is indeed a bijection. $\blacksquare$ Lemma: Let $\left\{X_\alpha\right\}_{\alpha\in\mathcal{A}}$ and $\left\{Y_\alpha\right\}_{\alpha\in\mathcal{A}}$ be two classes of non-empty topological spaces and $\left\{f_\alpha\right\}_{\alpha\in\mathcal{A}}$ a corresponding class of continuous functions such that $f_\alpha:X_\alpha\to Y_\alpha$. Then, if $\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha$ and $\displaystyle \prod_{\alpha\in\mathcal{A}}Y_\alpha$ are given the product topologies the mapping $\displaystyle \prod_{\alpha\in\mathcal{A}}f_\alpha\overset{\text{def.}}{=}\varphi:\prod_{\alpha\in\mathcal{A}}X_\alpha\to\prod_{\alpha\in\mathcal{A}}Y_\alpha:(x_\alpha)_{\alpha\in\mathcal{A}}\mapsto\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}$ is continuous. Proof: Since the codomain is a product space it suffices to show that $\displaystyle \pi_\beta\circ\varphi:\prod_{\alpha\in\mathcal{A}}X_\alpha\to Y_{\beta}$ is continuous for each $\beta\in\mathcal{A}$. We claim though that the diagram  commutes where $\pi^Y_\beta$ and $\pi^X_\beta$ denote the canonical projections from $\displaystyle \prod_{\alpha\in\mathcal{A}}Y_\alpha$ and $\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha$ to $Y_\beta$ and $X_\beta$ respectively. To see this we merely note that $\displaystyle \pi^Y_\beta\left(\varphi\left((x_\alpha)_{\alpha\in\mathcal{A}}\right)\right)=\pi^Y_\beta\left(\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}\right)=f_{\beta}(x_\beta)$ and $f_\beta\left(\pi^X_\beta\left((x_\alpha)_{\alpha\in\mathcal{A}}\right)\right)=f_\beta\left(x_\beta\right)$ which confirms the commutativity of the diagram. But, the conclusion follows since $f_\beta\circ\pi_\beta$ is the composition of two continuous maps (the projection being continuous since $\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha$ is a product space). The lemma follows by previous comment. $\blacksquare$ We come to our last lemma before the actual conclusion of the problem. Lemma: Let $\left\{X_\alpha\right\}_{\alpha\in\mathcal{A}}$ and $\left\{Y_\alpha\right\}_{\alpha\in\mathcal{A}}$ be two classes of non-empty topological spaces and $\left\{f_\alpha\right\}_{\alpha\in\mathcal{A}}$ a set of homeomorphisms with $f_\alpha:X_\alpha\to Y_\alpha$. Then, $\displaystyle \prod_{\alpha\in\mathcal{A}}f_\alpha\overset{\text{def.}}{=}\varphi:\prod_{\alpha\in\mathcal{A}}X_\alpha\to\prod_{\alpha\in\mathcal{A}}Y_\alpha:(x_\alpha)_{\alpha\in\mathcal{A}}\mapsto\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}$ is a homeomorphism if $\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha$ and $\displaystyle \prod_{\alpha\in\mathcal{A}}Y_\alpha$ are given the product topology. Proof: Our last two lemmas show that $\varphi$ is bijective and continuous. To prove that it’s inverse is continuous we note that $\displaystyle \left(\left(\prod_{\alpha\in\mathcal{A}}f_{\alpha}^{-1}\right)\circ\varphi\right)\left((x_\alpha)_{\alpha\in\mathcal{A}}\right)=\prod_{\alpha\in\mathcal{A}}\left\{f_\alpha^{-1}\left(f_\alpha(x_\alpha)\right)\right\}=(x_\alpha)_{\alpha\in\mathcal{A}}$ And similarly for the other side. Thus, $\displaystyle \varphi^{-1}=\prod_{\alpha\in\mathcal{A}}f_{\alpha}^{-1}$ Which is continuous since each $f_{\alpha}^{-1}:Y_\alpha\to X_\alpha$ is continuous and appealing to our last lemma again. Thus, $\varphi$ is a bi-continuous bijection, or in other words a homeomorphism. The conclusion follows. $\blacksquare$ Thus, getting back to the actual problem we note that if we denote $T_n:\mathbb{R}\to\mathbb{R}:x\mapsto a_nx+b_n$ that each $T_n$ is a homeomorphism. Thus, since it is easy to see that $\displaystyle \varphi=\prod_{n\in\mathbb{N}}T_n$ we may conclude by our last lemma (since we are assuming that we are giving $\mathbb{R}^{\omega}$ in both the domain and codomain the product topology) that $\varphi$ is a homeomorphism. This is also continuous if we give $\mathbb{R}^{\omega}$ the box topology. To see this we merely need to note that $\displaystyle \left(\prod_{\alpha\in\mathcal{A}}f_\alpha\right)^{-1}\left(\prod_{\alpha\in\mathcal{A}}U_\alpha\right)=\prod_{\alpha\in\mathcal{A}}f_\alpha^{-1}\left(U_\alpha\right)$ and thus if all of the $U_\alpha$ are open then so are (since each $f_\alpha$ is continuous) is each $f_\alpha^{-1}(U_\alpha)$ and thus the inverse image of a basic open set is the unrestricted product of open sets and thus basic open. A similar argument works for the inverse function. $\blacksquare$ June 7, 2010 ## Munkres Chapter 2 Section 18 1. Problem: Show that the normal $\varepsilon-\delta$ formulation of continuity is equivalent to the open set version. Proof: Suppose that $\left(\mathcal{M},d\right),\left(\mathcal{N},d'\right)$ are metric spaces and for every $\varepsilon>0$ and every $f(x)\in\mathcal{N}$ there exists some $\delta>0$ such that $f\left(B_{\delta}(x)\right)\subseteq B_{\varepsilon}(f(x))$. Then, given an open set $U\subseteq \mathcal{N}$ we have that $f^{-1}(U)$. To see this let $x\in f^{-1}(U)$ then $f(x)\in U$ and since $U$ is open by hypothesis there exists some open ball $B_\varepsilon(f(x))$ such that $B_{\varepsilon}(f(x))\subseteq U$ and thus by assumption of $\varepsilon-\delta$ continuity there is some $\delta>0$ such that $\displaystyle f\left(B_{\delta}(x)\right)\subseteq B_{\varepsilon}(f(x))\subseteq U$ and so $B_{\delta}\subseteq f^{-1}(U)$ and thus $x$ is an interior point of $f^{-1}(U)$. Conversely, suppose that the preimage of an open set is always open and let $f(x)\in\mathcal{N}$ and $\varepsilon>0$ be given. Clearly $B_{\varepsilon}(f(x))$ is open and thus $f^{-1}\left(B_{\varepsilon}(f(x))\right)$ is open. So, since $x\in f^{-1}\left(B_{\varepsilon}(f(x))\right)$ there exists some $\delta>0$ such that $B_{\delta}(x)\subseteq f^{-1}\left(B_{\varepsilon}(f(x))\right)$ and so $f\left(B_{\delta}(x)\right)\subseteq f\left(f^{-1}\left(B_{\varepsilon}(f(x))\right)\right)\subseteq B_{\varepsilon}(f(x))$ $\blacksquare$ 2. Problem: Suppose that $f:X\to Y$ is continuous. If $x$ is a limit point of the subset $A$ of $X$, is it necessarily true that $f(x)$ is a limit point of $f(A)$? Proof: No. Consider $(-1,0)\cup(0,1)$ with the suspace topology inherited from $\mathbb{R}$ with the usual topology. Define $f:(-1,0)\cup(0,1)\to D:x\mapsto\begin{cases}0\quad\text{if}\quad x\in(-1,0)\\ 1\quad\text{if}\quad x\in(0,1)\end{cases}$ This is clearly continuous since $f^{-1}(\{1\})=(-1,0)$ and $f^{-1}(\{1\})=(0,1)$ which are obviously open. But, notice that $\frac{-1}{2}$ is a limit point for $(-1,0)$ since given a neighborhood $N$ of $\frac{-1}{2}$ we must have that there is some $(a,b)\cap \left((-1,0)\cup(0,1)\right)\cap \subseteq N$ which contains it. But, $f\left(\frac{-1}{2}\right)=\{0\}$ is not a limit point for $f\left((-1,0)\right)=\{0\}$ since that set has no limit points. $\blacksquare$ 3. Problem: Let $X$ and $X'$ denote a singlet set in the two topologies $\mathfrak{J}$ and $\mathfrak{J}'$ respectively. Let $\text{id}:X'\to X$ be the identity function. Show that a) $\text{id}$ is continuous if and only if $\mathfrak{J}'$ is finer than $\mathfrak{J}$ b) $\text{id}$ is a homeomorphism if and only if $\mathfrak{J}=\mathfrak{J}'$ Proof: a) Assume that $\text{id}$ is continuous then given $U\in\mathfrak{J}$ we have that $\text{id}^{-1}(U)=U\in\mathfrak{J}'$. Conversely, if $\mathfrak{J}'$ is finer than $\mathfrak{J}$ we have that given $U\in\mathfrak{J}$ that $\text{id}^{-1}(U)=U\in\mathfrak{J}'$ b) If $\text{id}$ is a homeomorphism we see that both it and $\text{id}^{-1}=\text{id}:X\to X'$ are continuous and so mimicking the last argument we see that $\mathfrak{J}\subseteq\mathfrak{J}'$ and $\mathfrak{J}'\subseteq\mathfrak{J}$. Conversely, if $\mathfrak{J}=\mathfrak{J}'$ then we now that $U\in\mathfrak{J}\text{ iff }U\in\mathfrak{J}'$ or equivalently that $U\text{ is open in }X\text{ iff }\text{id}(U)=U\text{ is open in }X'$ which defines the homeomorphic property. $\blacksquare$ 4. Problem: Given $x_0\in X$ and $y_0\in Y$ show that the maps $f:X\to X\times Y$ and $g:X\times Y\to Y$ given by $f:x\mapsto (x,y_0)$ and $g:y\mapsto (x_0,y)$ are topological embeddings. Proof: Clearly $f$ and $g$ are continuous since the projection functions are the identity and constant functions. They are clearly injective for if, for example, $f(x)=(x,y_0)=(x',y_0)=f(x')$ then by definition of an ordered pair we must have that $x=x'$. Lastly, the inverse function is continuous since $f^{-1}:X\times \{y_0\}\to X:(x,y_0)\mapsto x$ is the restriction of the projection to $X\times\{y_0\}$. The same is true for $g$. $\blacksquare$ 5. Problem: Show that with the usual subspace topology $[0,1]\approx[a,b]$ and $(0,1)\approx(a,b)$. Proof: Define $f:[0,1]\to[a,b]:x\mapsto (b-a)+a$ and $g:(0,1)\to(a,b):x\mapsto (b-a)+a$. These are easily both proven to be homeomorphisms. $\blacksquare$ 6. Problem: Find a function $f:\mathbb{R}\to\mathbb{R}$ which is continuous at precisely one point. Proof: Define $f:\mathbb{R}\to\mathbb{R}:\begin{cases}x\quad\text{if}\quad x\in\mathbb{Q}\\ 0\quad\text{if}\quad x\notin\mathbb{Q}\end{cases}$ Suppose that $f$ is continuous at $x_0$, then choosing sequences $\{q_n\}_{n\in\mathbb{N}},\{i_n\}_{n\in\mathbb{N}}$ of rational and irrationals numbers respectively both converging to $x_0$. We see by the limit formulation of metric space continuity that $x_0=\lim\text{ }q_n=\lim\text{ }f(q_n)=f(x_0)=\lim\text{ }f(i_n)=\lim\text{ }0=0$ And so if $f$ were to be continuous anywhere it would have to be at $0$. To show that it is in fact continuous at $0$ we let $\varepsilon>0$ be given then choosing $\delta=\varepsilon$ we see that $|x|<\delta\implies |f(x)|\leqslant |x|<\delta=\varepsilon$ from where the conclusion follows since this implies that $\displaystyle \lim_{x\to 0}f(x)=0=f(0)$. $\blacksquare$ 7. Problem: a) Suppose that $f:\mathbb{R}\to\mathbb{R}$ is “continuous from the right”, that is, $\displaystyle \lim_{x\to a^+}f(x)=f(a)$ for each $a\in\mathbb{R}$. Show that $f$ is continuous when considered as a function from $\mathbb{R}_\ell$ to $\mathbb{R}$. b) Can you conjecture what kind of functions $f:\mathbb{R}\to\mathbb{R}$ are continuous when considered as maps as $\mathbb{R}\to\mathbb{R}_\ell$. As maps from $\mathbb{R}_\ell$ to $\mathbb{R}_\ell$? Proof: a) Note that by the assumption that $\displaystyle \lim_{x\to a^+}f(x)=f(a)$ we know that for every $\varepsilon>0$ there exists some $\delta>0$ such that $0\leqslant x-a<\delta$ implies that $|f(x)-f(a)|<\varepsilon$. So, let $U\subseteq\mathbb{R}$ be open and let $a\in f^{-1}(U)$. Then, $f(a)\in U$ and since $U$ is open we see that there is some $\varepsilon>0$ such that $B_{\varepsilon}(f(a))\subseteq U$. But, by assumption there exists some $\delta>0$ such that $0\leqslant x-a<\delta\implies f(x)\in B_{\varepsilon}(f(a))$. But, $\left\{x: 0\leqslant x-a<\delta\right\}=[a,a+\delta)$ and thus $f\left([a,a+\delta)\right)\subseteq B_{\varepsilon}(f(a))\subseteq U$ and thus $[a,a+\delta)\subseteq f^{-1}(U)$ and so $a$ is an interior point for $f^{-1}(U)$ from where it follows that $f^{-1}(U)$ is open and thus $f$ is continuous. b) I’m not too sure, and not too concerned right now. My initial impression is that if $f:\mathbb{R}\to\mathbb{R}_\ell$ is continuous then $f^{-1}([a,b))$ is open which should be hard to do. Etc. 8. Problem: Let $Y$ be an ordered set in the order topology. Let $f,g:X\to Y$ be continuous. a) Show that the set $\Omega=\left\{x\in X:f(x)\leqslant g(x)\right\}$ is closed in $X$ b) Let $h:X\to Y:x\mapsto \max\{f(x),g(x)\}$. Show that $h$ is continuous. Proof: a) Let $x_0\notin\Omega$ then $f(x_0)>g(x_0)$. Suppose first that there is no $g(x_0)<\xi and consider $f^{-1}\left(g(x_0),\infty)\right)\cap g^{-1}\left((-\infty,f(x_0)\right)=U$ This is clearly open in $X$ by the continuity of $f,g$ and $x_0$ is contained in it. Now, to show that $U\cap \Omega=\varnothing$ let $z\in U$ then $f(z)\in f\left(f^{-1}\left(g(x_0),\infty)\right)\cap g^{-1}\left(-\infty,f(x_0)\right)\right)$ which with simplification gives the important part that $f(z)\in (g(x_0),\infty)$ and so $f(z)>g(x_0)$ but since there is no $\xi$ such that $g(x_0)<\xi this implies that $f(z)\geqslant f(x_0)$. Similar analysis shows that $g(z)\in (-\infty,f(x_0))$ and since there is no $\xi$ as was mentioned above this implies that $g(z)\leqslant g(x_0)$. Thus, $g(z)\leqslant g(x_0) and thus $z\notin\Omega$. Now, suppose that there is some $\xi$ such that $g(x_0)<\xi then letting $V=f^{-1}(\xi,\infty)\cap g^{-1}(-\infty,\xi)$ we once again see that $V$ is open and $x_0\in V$. Furthermore, a quick check shows that if $z\in V$ that $f(z)\in(\xi,\infty)$ and so $f(z)>\xi$ and $g(z)\in(-\infty,\xi)$ and so $g(z)<\xi$ and so $f(z)>g(z)$ so that $z\notin\Omega$. The conclusion follows b) Let $\Omega_f=\left\{x\in X:f(x)\geqslant g(x)\right\}$ and $\Omega_g=\left\{x\in X:g(x)\geqslant f(x)\right\}$. As was shown in a) both $\Omega_f,\Omega_g$ are closed and thus define $f\sqcup g:X=\left(\Omega_f\cup\Omega_g\right)\to Y:x\mapsto\begin{cases}f(x)\quad\text{if}\quad x\in\Omega_f\\ g(x)\quad\text{if}\quad x\in\Omega_g\end{cases}$ Notice that since $f,g$ are both assumed continuous and $f\mid_{\Omega_g\cap\Omega_f}=g\mid_{\Omega_f\cap\Omega_g}$ that we may conclude by the gluing lemma that $f\sqcup g$ is in fact continuous. But, it is fairly easy to see that $f\sqcup g=\max\{f(x),g(x)\}$ $\blacksquare$ 9. Problem: Let $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ be a collection of subset of $X$; let $\displaystyle X=\bigcup_{\alpha\in\mathcal{A}}U_\alpha$. Let $f:X\to Y$ and suppose that $f\mid_{U_\alpha}$ is continuous for each $\alpha\in\mathcal{A}$ a) Show that if the collection $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ is finite each set $U_\alpha$ is closed, then $f$ is continuous. b) Find an example where the collection $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ is countable and each $U_\alpha$ is closed but $f$ is not continuous. c) An indexed family of sets $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ is said to be locally finite if each point of $X$ has a neighborhood that intersects only finitely many elements of $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$. Show that if the family $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ is locally finite and each $U_\alpha$ is closed then $f$ is continuous.$

Proof:

a) This follows since if $V\subseteq Y$ is closed then it is relatively easy to check that $\displaystyle f^{-1}(V)=\bigcup_{\alpha\in\mathcal{A}}\left(f\mid_{U_\alpha}\right)^{-1}(V)$ but since each $f\mid_{U_\alpha}$ is continuous we see that $\left(f\mid_{U_\alpha}\right)^{-1}(V)$ is closed in $U_\alpha$. But, since each $U_\alpha$ is closed in $X$ it follows that each $\left(f\mid_{U_\alpha}\right)^{-1}(V)$ is closed in $X$. Thus, $f^{-1}(V)$ is the finite union of closed sets in $X$, and thus closed.

b) Give $[0,1]$ the subspace topology inherited from $\mathbb{R}$ with the usual topology and consider $\left\{f_n\right\}_{n\in\mathbb{N}-\{1,2\}}$ with

$f_n=\iota_{[0,1-\frac{1}{n}]}:\left[0,1-\tfrac{1}{n}\right]\to[0,1]:x\mapsto x$

Clearly each $f_n$ i

Lemma: Let $Y$ be any topological space and $\left\{V_\beta\right\}_{\beta\in\mathcal{B}}$ be a locally finite collection of subsets of $Y$. Then, $\displaystyle \bigcup_{\beta\in\mathcal{B}}\overline{V_\beta}=\overline{\bigcup_{\beta\in\mathcal{B}}V_\beta}$

Proof: The left hand inclusion is standard, so it suffices to prove the right inclusion. So, let $\displaystyle x\in\overline{\bigcup_{\beta\in\mathcal{B}}V_\beta}$ since the collection of sets is locally finite there exists some neighborhood $N$ of $x$ such that it intersects only finitely many, say $V_{\beta_1},\cdots,V_{\beta_n}$, elements of the collection. So, suppose that $x\notin \left(\overline{V_{\beta_1}}\cup\cdots\cup \overline{V_{\beta_n}}\right)$ then $N\cap-\left(\overline{V_{\beta_1}}\cup\cdots\cup\overline{V_{\beta_n}}\right)$ is a neighborhood of $x$ which does not intersect $\displaystyle \bigcup_{\beta\in\mathcal{B}}V_\beta$ contradicting the assumption it is in the closure of that set. $\blacksquare$

Now, once again we let $V\subseteq Y$ be closed and note that $\displaystyle f^{-1}(V)=\bigcup_{\alpha\in\mathcal{A}}\left(f\mid_{U_\alpha}\right)^{-1}(V)$ and each $\left(f\mid_{U_\alpha}\right)^{-1}(V)$ is closed in $U_\alpha$ and since $U_\alpha$ is closed in $X$ we see that $\left(f\mid_{U_\alpha}\right)^{-1}(V)$ is closed in $X$. So, noting that $\left(f\mid_{U_\alpha}\right)^{-1}(V)\subseteq U_\alpha$ it is evident from the assumption that $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ is locally finite in $X$ that so is $\left\{\left(f\mid_{U_\alpha}\right)^{-1}(V)\right\}_{\alpha\in\mathcal{A}}$ and thus (for notational convenience) letting $F_\alpha=\left(f\mid_{U_\alpha}\right)^{-1}(V)$ the above lemma implies that

$\displaystyle \overline{f^{-1}(V)}=\overline{\bigcup_{\alpha\in\mathcal{A}}F_\alpha}=\bigcup_{\alpha\in\mathcal{A}}\overline{F_\alpha}=\bigcup_{\alpha\in\mathcal{A}}F_\alpha=f^{-1}(V)$

From where it follows that the preimage of a closed set under $f$ is closed. The conclusion follows. $\blacksquare$

10.

Problem: Let $f:A\to B$ and $g:C\to D$ be continuous functions. Let us define a map $f\times g:A\times C\to B\times D$ by the equation $(f\times g)(a\times c)=f(a)\times g(c)$. Show that $f\times g$ is continuous.

Proof: This follows from noting the two projections of $f\times g$ are $\pi_1\circ(f\times g):A\times B\to C:a\times b\mapsto f(a)$ and $\pi_2\circ(f\times g):A\times B\to D:a\times b\mapsto f(b)$. But, both of these are continuous since $\left(\pi_1(f\times g)\right)^{-1}(U)=f^{-1}(U)\times B$. To see this we note that $x\in f^{-1}(U)\times B$ if and only if $x\in f^{-1}(U)$ which is true if and only if $f(x)=\left(\pi_1\circ(f\times g)\right)(x)\in U$ or in other words $x\in \left(\pi_1\circ(f\times g)\right)^{-1}(U)$. Using this we note that the preimage an open set in $C$ will be the product of open sets by the continuity of $f$. It clearly follows both projections, and thus the function itself are continuous. $\blacksquare$

11.

Problem: Let $F:X\times Y\to Z$. We say that $F$ is continuous in eahc variable separately if for each $y_0\in Y$, the map $h:X\to Z:x\mapsto F(x\times y_0($ is continuous and for each $x_0\in X$ the map $j:Y\to Z:y\mapsto F(x_0\times y)$ is continuous. Show that if $F$ is continuous then $F$ is continuous in each variable separately.

Proof: If $F$ is continuous then clearly it is continuous in each variable since if we denote by $G_{y_0}$ the mapping $G_{y_0}:X\to Z:x\mapsto F(x\times y_0)$ we see that $G_{y_0}=H_{y_0}\circ(F\mid_{X\times\{y_0\}})$ where $H_{y_0}:X\to X\times Y:x\mapsto x\times y_0$ but the RHS is the composition of continuous maps and thus continuous. A similar analysis holds for the other variable.

12.

Problem: Let $F:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ be given by

$\displaystyle F(x\times y)=\begin{cases} \frac{xy}{x^2+y^2}&\mbox{if}\quad x\times y\ne 0\times 0\\ 0 &\mbox{if} \quad x\times y=0\times0\end{cases}$

a) Show that $F$ is continuous in each variable separately.

b) Compute $g:\mathbb{R}\to\mathbb{R}:x\mapsto F(x\times x)$.

c) Show that $F$ is not continuous

Proof:

a) Clearly both $F(x\times y_0)$ and $F(x_0\times y)$ are continuous for $x,y\ne 0$ since they are the quotient of polynomials and the denominator is non-zero. Lastly, they are both continuous at $x,y=0$ since it is trivial to check that $$\displaystyle 0=F(0\times y_0)=F(x_0\times 0)=\lim_{x\to 0}F(x\times y_0)=\lim_{y\to 0}F(x_0\times y)$ b) Evidently $\displaystyle g(x)=F(x\times x)=\begin{cases}\frac{1}{2}\quad\text{if}\quad x\times x\ne 0\\ 0\quad\text{if}\quad x\times x=0\end{cases}$ c) This clearly proves that $F(x\times y)$ is not continuous with $\mathbb{R}^2$ is not continuous since if $\Delta$ is the diangonal we have that $\displaystyle \lim_{(x,y)\in\Delta\to (0,0)}F(x\times y)=\frac{1}{2}\ne F(0\times 0)$ and so in particular $\displaystyle \lim _{(x,y)\to(0,0)}F(x\times y)\ne F(0\times 0)$ 13. Problem: Let $A\subseteq X$; let $f:A\to X$ be continuous and let $Y$ be Hausdorff. Prove that if $f$ may be extended to a continuous function $\overset{\sim}{f}:\overline{A}\to Y$, then $\overset{\sim}{f}$ is uniquely determined by $f$. Proof: I am not too sure what this question is asking, but assuming it’s asking that if this extension existed it’s unique we can do this two ways Way 1(fun way!): Lemma: Let $X$ be any topological space and $Y$ a Hausdorff space. Suppose that $\varphi,\psi:X\to Y$ are continuous and define $A(\varphi,\psi)=\left\{x\in X:\varphi(x)=\psi(x)\right\}$. Then, $A(\varphi,\psi)$ is closed in $X$ Proof: Note that $\varphi\oplus\psi:X\to Y\times Y:x\mapsto (\varphi(x),\psi(x))$ is clearly continuous since $\pi_1\circ(\varphi\oplus\psi)=\varphi$ and $\pi_2\circ(\varphi\oplus\psi)=\psi$. It is trivial then to check that $\displaystyle A(\varphi,\psi)=\left(\varphi\oplus\psi\right)^{-1}(\Delta_Y)$ and since $Y$ is Hausdorff we have that $\Delta_Y\subseteq Y\times Y$ is closed and the conclusion follows. $\blacksquare$ From this we note that if $\varphi,\psi$ agree on $D\subseteq X$ such that $\overline{D}=X$ we have that $X\supseteq A(\varphi,\psi)=\overline{A(\varphi,\psi)}\supseteq\overline{D}=X$ From where it follows that $A(\varphi,\psi)=X$ and so $\varphi=\psi$. So, thinking of $\overline{A}$ as a subspace of $X$ we see that $\text{cl}_{\overline{A}}\text{ }A=Y\cap\text{cl}_{X}\text{ }A=\overline{A}$ and thus clearly $A$ is dense in $\overline{A}$. So, the conclusion readily follows by noting that if $\overset{\sim}{f_1},\overset{\sim}{f_2}$ are two continuous extensions then by definition $A\left(\overset{\sim}{f_1},\overset{\sim}{f_2}\right)\supseteq A$. Way 2(unfun way): Let $\overset{\sim}{f_1},\overset{\sim}{f_2}$ be two extensions of $f$ and suppose there is some $x\in\overline{A}-A(\varphi,\psi)$. Clearly $x\notin A$ and thus $x$ is a limit point of $A$. So, by assumption $\overset{\sim}{f_1}(x)\ne\overset{\sim}{f_2}(x)$ and so using the Hausdorffness of $Y$ we may find disjoint neighborhoods $U,V$ of them respectively. Thus, $\overset{\sim}{f_1}^{-1}(U),\overset{\sim}{f_2}^{-1}(V)$ are neighborhoods of $x$ in $X$. Thus, $\overset{\sim}{f_1}^{-1}(U)\cap\overset{\sim}{f_2}^{-1}(V)$ is a neighborhood of $x$. But, clearly there can be no $y\in A\cap\left(\overset{\sim}{f_1}^{-1}(U)\cap\overset{\sim}{f_2}^{-1}(V)\right)$ otherwise $\overset{\sim}{f_1}(y)=\overset{\sim}{f_2}(y)\in U\cap V$. It follows that $\overset{\sim}{f_1}^{-1}(U)\cap\overset{\sim}{f_2}^{-1}(V)$ is a neighborhood of $x$ disjoint from $A$ which contradicts the density of $A$ in $\overline{A}$. The conclusion follow. $\blacksquare$ May 28, 2010 ## Munkres Chapter 2 Section 17 Theorem 17.11 Problem: Prove that every simply ordered set $X$ with the order topology is Hausdorff. The product of two Hausdorff spaces is Hausdorff. A subspace of a Hausdorff space is Hausdorff. Proof: Let $X$ have the order topology and let $a,b\in X$ be distinct and assume WLOG that $a. If there does not exists a $c\in X$ such that $a then $(-\infty,b),(a,\infty)$ are disjoint neighborhoods of $a,b$ respectively. They are disjoint for to suppose that $x\in(-\infty,b)\cap(a,\infty)$ would imply that $a contradictory to our assumption. If there is some $a then $(-\infty,c),(c,\infty)$ are disjoint neighborhoods of $a,b$ respectively. Let $X,Y$ be Hausdorff and $(x,y),(x',y')\in X\times Y$ be distinct. Since $(x,y)\ne(x',y')$ are distinct it follows we may assume WLOG that $x\ne x'$. But, since $X$ is Hausdorff there exists disjoint neighborhoods $U,V$ of $x,y$. So, consider $U\times Y,V\times Y$ these are clearly neighborhoods of $(x,y),(x',y')$ in $X\times Y$ and to assume $(u,v)\in U\times Y\cap V\times Y$ would imply $\pi_1((u,v))-u\in\pi_1(U\times Y\cap V\times Y)\subseteq U\cap V$. Lastly, suppose that $X$ is Hausdorff and let $Y$ be a subspace of $X$. If $x,y\in Y$ are distinct there exists disjoint neighborhoods $U,V$ of them in $X$. Thus, $U\cap Y,V\cap Y$ are disjoint neighborhoods of them in $Y$. $\blacksquare$ 1. Problem: Let $\mathcal{C}$ be a collection of subsets of the set $X$. Suppose that $\varnothing,X\in\mathcal{C}$, and that the finite union and arbitrary intersection of elements of $\mathcal{C}$ are in $\mathcal{C}$. Prove that the collection $\mathfrak{J}=\left\{X-C:C\in\mathcal{C}\right\}$ is a topology on $X$. Proof: Clearly $X=X-\varnothing$ and $\varnothing=X-X$ are in $\mathfrak{J}$. Now, suppose that $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}\subseteq\mathfrak{J}$ then $\displaystyle X-\bigcup_{\alpha\in\mathcal{A}}U_\alpha=\bigcap_{\alpha\in\mathcal{A}}\left(X-U_\alpha\right)$ and since $\left\{X-U_\alpha\right\}_{\alpha\in\mathcal{A}}\subseteq\mathcal{C}$ and $\mathcal{C}$ is closed under arbitrary intersection it follows that $\displaystyle \bigcap_{\alpha\in\mathcal{A}}\left(X-U_\alpha\right)=X-\bigcup_{\alpha\in\mathcal{A}}U_\alpha\in\mathcal{C}$ and thus $\displaystyle \bigcup_{\alpha\in\mathcal{A}}U_\alpha\in\mathcal{C}$. Lastly, suppose that $\left\{U_1,\cdots,U_n\right\}\subseteq\mathcal{C}$ then $\displaystyle X-\left(U_1\cap\cdots\cap U_n\right)=\left(X-U_1\right)\cup\cdots\cup\left(X-U_n\right)$ and since $\left\{X-U_1,\cdots,X-U_n\right\}\subseteq\mathcal{C}$ and $\mathcal{C}$ is closed under finite union it follows that $X-\left(U_1\cap\cdots\cap U_n\right)\in\mathcal{C}$ and thus $U_1\cap\cdots\cap U_n\in\mathfrak{J}$. $\blacksquare$ 2. Problem: Show that if $A$ is a closed in $Y$ and $Y$ is closed in $X$ that $A$ is closed in $X$. Proof: Since $A$ is closed in $Y$ and $Y$ is a subspace of $X$ we have that $A=Y\cap G$ for some closed set $G\subseteq X$ but since $Y$ is closed it follows that $A$ is the intersection of two closed sets in $X$ and thus closed in $X$. $\blacksquare$ 3. Problem: Show that if $A$ is closed in $X$ and $B$ is closed in $Y$ then $A\times B$ is closed in $X\times Y$ Proof: This follows immediately from question 9. $\blacksquare$ 4. Problem: Show that if $U$ is open in $X$ and $A$ is closed in $X$, then $U-A$ is open in $X$ and $A-U$ is closed in $X$. Proof: This follows immediately from the fact that $U-A=U\cap\left(X-A\right)$ and $A-U=A\cap\left(X-U\right)$. $\blacksquare$ 5. Problem: Let $X$ be an ordered set in the order topology. Show that $\overline{(a,b)}\subseteq[a,b]$. Under what conditions does equality hold? Proof: Let $x\in(-\infty,a)\cup(b,\infty)$ then $(-\infty,a)\cup(b,\infty)$ is a neighborhood of $x$ disjoint from $(a,b)$ and thus $x\notin\overline{(a,b)}$. Equality will hold when $a,b$ are limit points of $(a,b)$ or said otherwise whenever $a we have that there are some$lated d,e$such that $a. $\blacksquare$ 6. Problem: Let $A,B$ and $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ denote subsets of $X$. Prove that if $A\subseteq B$ then $\overline{A}\subseteq\overline{B}$, $\overline{A\cup B}=\overline{A}\cup\overline{B}$, and $\displaystyle \overline{\bigcup_{\alpha\in\mathcal{A}}U_\alpha}\supseteq\bigcup_{\alpha\in\mathcal{A}}\overline{U_\alpha}$. Give an example where this last inclusion is strict. Proof: We choose to prove the second part first. Let $x\notin\overline{A\cup B}$ then there is a neighborhood $N$ of $x$ such that $N\cap (A\cup B)=(N\cap A)\cup (N\cap B)=\varnothing$ and thus $x\notin \overline{A}\text{ and }x\notin\overline{B}$ and so $x\notin\overline{A}\cup\overline{B}$. Conversely, suppose that $x\notin\overline{A}\cup\overline{B}$ then $x\notin \overline{A}$ and $x\notin\overline{B}$ and so there are neighborhoods $N,G$ of $x$ such that $N\cap A=\varnothing,G\cap B=\varnothing$ clearly then $N\cap G$ is a neighborhood of $x$ such that $(N\cap G)\cap(A\cup B)=\varnothing$ and thus $x\notin\overline{A\cup B}$. Using this, if $A\subseteq B$ then we have that $\overline{B}=\overline{B\cup A}=\overline{B}\cup\overline{A}$ from where it follows that $\overline{A}\subseteq\overline{B}$. Let $\displaystyle x\notin\overline{\bigcup_{\alpha\in\mathcal{A}}U_\alpha}$ then there is a neighborhood $N$ of $x$ such that $\displaystyle N\cap\bigcup_{\alpha\in\mathcal{A}}U_\alpha=\bigcup_{\alpha\in\mathcal{A}}\left(N\cap U_\alpha\right)=\varnothing$ and thus $N\cap U_\alpha=\varnothing$ for every $\alpha\in\mathcal{A}$ and thus $x\notin\overline{U_\alpha}$ for every $\alpha\in\mathcal{A}$ and so finally we may conclude that $\displaystyle x\notin\bigcup_{\alpha\in\mathcal{A}}\overline{U_\alpha}$. To see when inclusion can be strict consider $\mathbb{R}$ with the usual topology. Then, $\displaystyle \mathbb{R}=\overline{\bigcup{q\in\mathbb{Q}}\{q\}}\supsetneq\bigcup_{q\in\mathbb{Q}}\overline{\{q\}}=\bigcup_{q\in\mathbb{Q}}\{q\}=\mathbb{Q}$ $\blacksquare$ 7. Problem: Criticize proof (see book). Proof: There is no guarantee that the $A_\alpha$ for which $U$ intersected will b e the same $A_\alpha$ that $V$ will intersect if you pick another $V$. $\blacksquare$ 8. Problem: Let $A,B$ and $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ denote subsets of $X$. Determine whether the following equations hold, if any equality fails determine which inclusion holds. a) $\overline{A\cap B}=\overline{A}\cap\overline{B}$ b) $\displaystyle \overline{\bigcap_{\alpha\in\mathcal{A}}U_\alpha}=\bigcap_{\alpha\in\mathcal{A}}\overline{U_\alpha}$ c) $\overline{A-B}=\overline{A}-\overline{B}$ Proof: a) Equation does not hold. Consider that $\overline{(-1,0)\cap(0,1)}=\overline{\varnothing}=\varnothing$ but $\overline{(-1,0)}\cap\overline{(0,1)}=[-1,0]\cap[0,1]=\{0\}$. The $\subseteq$ inclusion always holds. b) This follows from a) that equality needn’t hold. Once again the $\subseteq$ inclusion is true. c) This need be true either $\overline{\mathbb{R}-\mathbb{Q}}=\mathbb{R}\ne\overline{\mathbb{R}}-\overline{\mathbb{Q}}=\varnothing$. The $\subseteq$ inclusion holds. $\blacksquare$ 9. Problem: Prove that if $A\subseteq X,B\subseteq Y$ then $\overline{A\times B}=\overline{A}\times\overline{B}$ in the product topology on $X\times Y$. Proof: Let $\displaystyle (x,y)\in\overline{A\times B}$. Let $U\times V$ be a basic open set in $X\times Y$ which contains $(x,y)$. Since $x\in \overline{A},y\in \overline{B}$ we can choose some point $x'\in U\cap A,y'\in V\cap B$. Then, $(x',y')\in U\times V\cap A\times B$. It follows that $\displaystyle (x,y)\in\overline{A\times B}$ Conversely, let $(x,y)\in\overline{A\times B}$. Let $U\subseteq X$ be any set such that $x\in U$. Since $\pi_1^{-1}(U)$ is open in $X\times Y$ it contains some point $(x',y')\in A\times B$. Then, $x'\in U\cap A$. It follows that $x\in\overline{A}$. A similar technique works for $Y$. $\blacksquare$ 10., 11,. 12 Covered in theorem stated and proved at the beginning of the post. 13. Problem: Prove that $X$ is Hausdorff if and only if the diagonal $\Delta=\left\{(x,x):x\in X\right\}$ is closed in $X\times X$ with the product topology. Proof: Suppose that $X$ is Hausdorff then given $x\ne y$ we may find disjoint neighborhoods $U,V$ of them. So, $(x,y)\in U\times V$ and $U\times V\cap \Delta=\varnothing$ since $U\cap V=\varnothing$. It follows that $\Delta$ is closed. Conversely, suppose $\Delta$ is closed in $X\times X$ and $x,y\in X$ are distinct. Then, $(x,y)\notin\Delta$ and so there exists a basic open neighborhood $U\times V$ of $(x,y)$ such that $U\times V\cap \Delta=\varnothing$ and so $U,V$ are neighborhoods of $x,y$ in $X$ which are disjoint. For, to suppose they were not disjoint would to assume that $z\in U\cap V\implies (z,z)\in U\times V$ contradicting the assumption that $U\times V\cap\Delta=\varnothing$. $\blacksquare$ 14. Problem: In the cofinite topology on $\mathbb{R}$ to what point or points does the sequence $\left\{\frac{1}{n}\right\}_{n\in\mathbb{N}}$ converge to? Proof: It converges to every point of $\mathbb{R}$. To see this let $x\in\mathbb{R}$ be arbitrary and let $U$ be any neighborhood of it. Then, $\mathbb{R}-U$ is finite and in particular $\left(\mathbb{R}-U\right)\cap K$ is finite. If it’s empty we’re done, so assume not and let $\frac{1}{n_0}=\min\left(\left(\mathbb{R}-U\right)\cap K\right)$ then clearly for all $n\in\mathbb{N}$ such that $n>n_0$ we have that $\frac{1}{n}<\frac{1}{n_0}$ and thus $\frac{1}{n}$ is in $U$. The conclusion follows. $\blacksquare$ 15. Problem: Prove that the $T_1$ axiom is equivalent to the condition that for each pair of points $x,y\in X$ there are neighborhoods of each which doesn’t contain the other. Proof: Suppose that $X$ is $T_1$ then given distinct $x,y\in X$ the sets $X-\{y\},X-\{x\}$ are obviously neighborhoods of $x,y$ respectively which don’t contain the other. Conversely, suppose the opposite is true and let $y\in X-\{x\}$ then there is a neighborhood $U$ of it such that $x\notin U\implies U\subseteq X-\{x\}$ and thus $X-\{x\}$ is open and $\{x\}$ is therefore closed. $\blacksquare$ 16. Problem: Consider the five topologies on $\mathbb{R}$ given in exercise 7 of section 13 (my section 2). a) Deterime the closure of $K$ under each of these topologies. b) Which of these topologies are Hausdorff? Which are $T_1$ Proof: a) As a reminder the topologies are $\mathfrak{J}_1=\text{usual topology}$ $\mathfrak{J}_2=\text{topology on }\mathbb{R}_K$ $\mathfrak{J}_3=\text{cofinite topology}$ $\mathfrak{J}_4=\text{upper limit topology}$ $\mathfrak{J}_5=\text{left ray topology}$ For the first one we easily see that $\overline{K}=K\cup\{0\}$. For the second one we can see that $\overline{K}=K$. To see this note that $\displaystyle \bigcup_{a is open by definition and thus $K$ being the complement of it is closed. Thus, $\overline{K}=K$ For the third one $\overline{K}=\mathbb{R}$. To see this note that we in a sense already proved this in 14, but for any $x\in\mathbb{R}$ and any neighborhood $N$ of it we have that $N=\mathbb{R}-\{x_1,\cdots,x_n\}$ and thus if $\displaystyle \frac{1}{n_0}=\min\left(\{x_1,\cdots,x_n\}\cap K\right)$ (assuming it’s non-empty) we see that $n>n_0\implies \frac{1}{n}\notin\{x_1,\cdots,x_n\}\implies \frac{1}{n}\in N$. Thus, given any point of $\mathbb{R}$ and any neighborhood $N$ of it we have that $N\cap K$ is infinite, and thus clearly $x\in\overline{K}$. The conclusion follows from that. For the fourth topology we note that $\displaystyle \mathbb{R}-K=(-\infty,0]\cup\bigcup_{n\in\mathbb{N}}\left(\frac{1}{n+1},\frac{1}{n}\right)\cup(1,\infty)$ which is open in $\mathbb{R}$ with the upper limit topology. Remember that $\displaystyle (a,b)=\bigcup_{a For the last one $\overline{K}=[0,\infty)$. Clearly $K\subseteq[0,\infty)$ and since $[0,\infty)$ is closed and $\overline{K}$ is the intersection of all closed supersets of $K$ it follows that $\overline{K}\subseteq[0,\infty)$. Now, suppose that $x\notin[0,\infty)$ then $x\in(-\infty,0)$ which is a neighborhood of $x$ which doesn’t intersect $K$ and thus $x\notin\overline{K}$. So, $[0,\infty)\subseteq\overline{K}$. The conclusion follows. b) Lemma: If $X$ is Hausdorff, then it is $T_1$ Proof: Let $x\in X$ and $y\in X-\{x\}$ by assumption there exists disjoint neighborhoods $U,V$ of $x,y$ respectively and so clearly $y\in V\subseteq X-\{x\}$ and thus $X-\{x\}$ is open and so $\{x\}$ is closed. $\blacksquare$ $\mathfrak{J}_1$: Clearly the first topology is Hausdorff since it is an order topology and we proved in the initial post that all order topologies are Hausdorff. Lemma: Let $X$ be a set and $\mathfrak{J},\mathfrak{J}'$ two topologies on $X$ such that $\mathfrak{J}'$ is finer than $\mathfrak{J}$. Then, if $X$ is Hausdorff with the $\mathfrak{J}$ topology it is Hausdorff with the $\mathfrak{J}'$ topology. Proof: Clearly if $x,y\in X$ are distinct we may find disjoint neighborhoods $U,V\in\mathfrak{J}$ of them in the topology given by $\mathfrak{J}$ and thus the same neighborhoods work in consideration of the topology given by $\mathfrak{J}'$. $\blacksquare$ $\mathfrak{J}$: From this lemma it follows that $\mathbb{R}_K$ having a finer topology than $\mathbb{R}$ with the usual topology is Hausdorff $\mathfrak{J}_3$: The cofinite topology is $T_1$ but not Hausdorff. To see that it’s $T_1$ it suffices to prove that $\{x\}$ is closed for any $x\in\mathbb{R}$. But, this is trivial since $\mathbb{R}-\{x\}$ being the complement of a finite set is open, thus $\{x\}$ closed. But, it is not Hausdorff since it cannot support two disjoint, non-empty open sets. To see this suppose that $U$ is open in $\mathbb{R}$ with the cofinite topology then $\mathbb{R}-U$ is finite and since a disjoint set $V$ would have to be a subset of $\mathbb{R}-U$ it follows that $V$ is finite and thus it’s complement not finite. Thus, $V$ is not open. $\mathfrak{J}_4$: Once again this topology finer than that of $\mathbb{R}$ with the usual topology since $\displaystyle (a,b)=\bigcup_{c $\mathfrak{J}_5$: This isn’t even $T_1$. To see this we must merely note that if $U$ is any set containing $1$ we must have that there is some basic open set $(-\infty,\alpha),\text{ }\alpha>1$ such that $1\in(-\infty,\alpha)\subseteq U$, but this means that $0\in U$. So, there does not exist a neighborhood of $1$ which does not contain $0$. 17. Problem: Consider the lower limit topology on $\mathbb{R}$ and the topology given by the basis $\mathcal{C}=\left\{[a,b):a. Determine the closure of the intervals $A=(0,\sqrt{2})$ and $B=(\sqrt{2},3)$ in these two topologies. Proof: We first prove more generally that if $(a,b)\subseteq\mathbb{R}_\ell$ then $\overline{(a,b)}=[a,b)$. To see this we first note that $[a,b)$ is open since $\mathbb{R}-[a,b)=(-\infty,a)\cup[b,\infty)$ which we claim is open. To see this we note that $\displaystyle (-\infty,a)\cup[b,\infty)=\bigcup_{xb}[b,y)$. So, since $\overline{(a,b)}$ is the intersection of all closed supersets of $(a,b)$ and $[a,b)\supseteq(a,b)$ is closed we see that $\overline{(a,b)}\subseteq[a,b)$. So, we finish the argument by showing that $\overline{(a,b)}\ne(a,b)$. To see this we show that $a$ is a limit point for $(a,b)$ from where the conclusion will follow. So, let $N$ be any neighborhood of $a$, then we may find some basic open neighborhood $[\alpha,\beta)$ such that $a\in[\alpha,\beta)\subseteq N$, but clearly $[\alpha,\beta)\cap(a,b)$ contains infinitely many points from where it follows that $a$ is a limit point of $(a,b)$. From this we may conclude for $\mathbb{R}_\ell$ that $\overline{(0,\sqrt{2})}=[0,\sqrt{2})$ and $\overline{(\sqrt{2},3)}=[\sqrt{2},3)$ We now claim that in the topology generated by $\mathcal{C}$ that $\overline{(0,\sqrt{2})}=[0,\sqrt{2}]$ and $\overline{(\sqrt{2},3)}=[\sqrt{2},3)$. More generall, let us prove that in this topology $\overline{(a,b)}=\begin{cases}[a,b]\quad\text{if}\quad b\notin\mathbb{Q}\\ [a,b)\quad\text{if}\quad b\in\mathbb{Q}\end{cases}$ Clearly if $\alpha then choosing some rational number $q<\alpha$ and some rational number $\alpha then $\alpha\in[q,p)$ and $[q,p)\cap(a,b)=\varnothing$ so that $\alpha\notin\overline{(a,b)}$. Also, if $\alpha>b$ then choosing some $p\in\mathbb{Q}$ such that $b we see that $\alpha\in[p,\alpha+1)$ and $[p,\alpha+1)\cap(a,b)=\varnothing$. Thus, the only possibilities for $\overline{(a,b)}$ are $(a,b),[a,b),[a,b]$. But, just as before if $N$ is any neighborhood $a$ we may find some open basic neighborhood$latex [p,q)\$ such that $a\in[p,q)\subseteq N$ but clearly $[p,q)\cap(a,b)\ne\varnothing$ and thus since $N$ was arbitrary it follows that $a\in\overline{(a,b)}$.

So, now we split into the two cases. First assume that $b\notin\mathbb{Q}$ then given any neighborhood $N$ of $b$ we may find some basic neighborhood $[p,q)$  such that $b\in[p,q)\subseteq N$, but since $b\notin\mathbb{Q}$ we see that $b\ne p$ from where it follows that $[p,q)\cap(a,b)\ne\varnothing$ and thus since $N$ was arbitrary it follows that $b\in\overline{(a,b)}$ and thus $\overline{(a,b)}=[a,b]$. Now suppose that $b\in\mathbb{Q}$, then $[b,b+1)$ is clearly a neighborhood of $b$ that doesn’t intersect $(a,b)$ and thus $b\notin{(a,b)}$ so that $\overline{(a,b)}=[a,b)$

The conclusion follows. $\blacksquare$

Problem: If $A\subseteq X$, we define the boundary of $A$ by $\partial A=\overline{A}-\overline{X-A}$.

a) Prove that $A^\circ$ and $\partial A$ are disjoint and $\overline{A}=A^\circ\cup\partial A$

b) Prove that $\partial A=\varnothing$ i and only if $A$ is both open and closed

c) Prove that $U$ is open if and only if $\partial U=\overline{U}-U$

d) If $U$ is open, is it true that $U=\left(\overline{U}\right)^{\circ}$?

Proof:

a) Clearly if $x\in A^{\circ}$ then there exists a neighborhood of $x$ whose intersection with the complement of $A$ is empty, thus $x\notin\partial A$. Now, let $x\in\overline{A}$ now if there exists a neighborhood $N$ of $x$ such that $N\subseteq A$ then $x\in A^{\circ}$ and if not then every neighborhood of $x$ contains points of $X-A$ and since $x\in D(A)$ it follows that it also contains points of $A$. Thus, either $x\in A^{\circ}$ or $x\in\partial A$, thus $x\in A^{\circ}\cup\partial A$. Conversely, if $x\in A^{\circ}\partial A$ then either there exists a neighborhood $N$ of $x$ such that $N\subseteq A$ and thus $x\in A$.  Conversely, if $x\in\partial A$ then for every neighborhood $N$ of $x$ we have that $N\cap A,N\cap (X-A)=\ne\varnothing$ and in particular $N\cap A\ne\varnothing$ and so $x\in\overline{A}$.

b) Suppose first that $\partial A=\varnothing$. If $A=\varnothing$ we’re done, so assume not and let $x\in A$. Since $x\notin\partial A$ there is a neighborhood $N$ of $x$ such that $N\cap A\varnothing\text{ or }N\cap (X-A)=\varnothing$ but since $x\in N\cap A$ it follows that $N\subseteq A$ and thus $A$ is open. Conversely, letting $x\in X-A$ we see that $x\notin\partial A$ and so by the same logic there exists a neighborhood $N$ of $x$ such that $N\subseteq X-A$ and thus $X-A$ is open and so $A$ closed.

Conversely, suppose that $A$ is both open and closed and suppose that $\partial A\ne\varnothing$. If $x\in\partial A\cap A$ then for every neighborhood $N$ of $x$ we must have that $N\cap (X-A)\ne\varnothing$ and thus $x\notin A^{\circ}=A$, which is a contradiction. Conversely, if $x\in (X-A)\cap\partial A$ then for every neighborhood $N$ of $x$ we must have that $N\cap A\ne\varnothing$ and thus $x\notin (X-A)^{\circ}=X-A$ which is a contradiction.

c) Suppose that $U$ is open and let $x\in\partial U$, then every neighborhood $N$of $x$ contains points of $U$ by definition, but since $x\notin U$ (it can’t be an interior point, thus $x\notin U^{\circ}=U$)  they must be points of $U-\{x\}$ and thus $x\in D(U)$. So, $x\in \overline{U}-U$. Conversely, if $x\in \overline{U}-U$ then $x$ must be a limit point of $U$ which is not in $U$ and thus every neighborhood $x$ contains points of $U$ and $X-U$ so $x\in\partial U$

d) No, it is not true. With the usual topology on $\mathbb{R}$ the set $(-1,0)\cup(0,1)$ is open, but $\left(\overline{(-1,0)\cup(0,1)}\right)^{\circ}=[-1,1]^\circ=(-1,1)$

We choose to omit 18, 19, 21 on the grounds that they are monotonous and easy, monotonous and easy, and way, way to long.

May 22, 2010