# Abstract Nonsense

## Halmos Section 36: Inverses (Pt. III)

Point of post: This is a continuation of this post.

December 4, 2010

## Halmos Sections 32 and 33: Linear Transformations and Transformations as Vectors (Pt. I)

Point of post: In this post we complete the problems that appear at the end of Halmos, sections 32 and 33.

November 22, 2010

## Halmos Section 28: Parity

Point of post: In this post I will solve the, very few, problems in the twenty-eighth section of Halmos. This section is on the parity (sign) of a permutation.

November 10, 2010

## Halmos Section 26 and 27: Permutations and Cycles cont.

Point of post: This is a continuation of this post.

November 10, 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

## Halmos Chaper One, Section 1: Fields

1.

Problem: Amost all the laws of elementary arithmetic are consequences of the axioms defining a field. Prove, in particular, that if $\mathfrak{F}$ is a field, and if $\alpha,\beta$ and $\gamma$ belong to $\mathfrak{F}$, then the following relations hold.

a) $0+\alpha=\alpha$

b) If $\alpha+\beta=\alpha+\gamma$ then $\beta=\gamma$

c) $\alpha+\left(\beta-\alpha\right)=\beta$

d) $\alpha0=0\alpha=0$

e) $(-1)\alpha=-\alpha$

f) $(-\alpha)(-\beta)=\alpha\beta$

g) If $\alpha\beta=0$ then either $\alpha=0$ or $\beta=0$

Proof:

a) By axiom 3 (A3) we know that $\alpha+0=\alpha$ and by the commutativity described in A1 we conclude that $0+\alpha=\alpha+0=\alpha$

b) We see that if $\alpha+\beta=\alpha+\gamma$ then $\left(\alpha+\beta\right)+-\alpha=\left(\alpha+\gamma\right)+-\alpha$ which by associativity and commutativity says that $\gamma+(\alpha+-\alpha)=\beta+(\alpha+-\alpha)$ which then implies that $\gamma=\gamma+0=\beta+0=\beta$.

c) We use associativity and commutativity to rewrite our equations as $\beta+(\alpha+-\alpha)=\beta+0=\beta$

d) By commutativity of the multiplication it suffices to note that $\alpha0=\alpha(0+0)=\alpha0+\alpha0$ and thus $\alpha0+-\alpha0=\left(\alpha0+\alpha0\right)+-\alpha0$ and  by associativity we arrive at $0=\alpha0$.

e) We merely note that $\alpha+(-1)\alpha=(1+-1)\alpha=0$ and thus $-\alpha=(-1)\alpha$.

f) We use e) to say that $(-\alpha)(-\beta)=(-1)\alpha(-1)\beta=(-1)(-1)\alpha\beta$. Then, we notice that $(-1)(-1)+(-1)=(-1)(-1+1)=0$ from where it follows that $-(-1)(-1)=-1$ and thus $(-1)(-1)=1$ and the conclusion follows.

g) Suppose that $\alpha,\beta\ne0$ then since $\alpha\ne0$ we see that $\alpha\beta=0\implies \beta=0\alpha^{-1}=0$ which contradicts our choice of $b$

2.

Problem:

a) Is the set of all positive integers a field?

b) What about the set of all integers?

c) Can the answers to both these question be changed by re-defining addition or multiplication (or both)?

Proof:

a) No, we merely note that there is no additive identity for $1$

b) No, there is no multiplicative identity for $2$

c) Yes. But first before we justify let us prove a lemma (which is useful),

Lemma: Let $\left(\mathfrak{F},+,\cdot\right)$ be a field with $\text{card }\mathfrak{F}=\mathfrak{n}$. Then, given any set $F$ with $\text{card }F=\mathfrak{n}$ there are operations $\oplus,\odot:F\times F\to F$ for which $\left(F,\oplus,\odot\right)$ is a field.

Proof: By virtue of their equal cardinalities there exists some bijection $\theta:F\to\mathfrak{F}$. Then, for $\alpha,\beta\in F$ define

$\alpha\oplus\beta=\theta^{-1}\left(\theta(\alpha)+\theta(\beta)\right)$

and

$\alpha\odot\beta=\theta^{-1}\left(\theta(\alpha)\cdot\theta\left(\beta\right)\right)$

We prove that with these operations $\left(F,\oplus,\odot\right)$ is a field. We first note that $\oplus,\odot:F\times F\to F$ and so they are legitimate binary operations. We now begin to show that all the field axioms are satisfied

1) Addition is commutative- This is clear since

$\alpha\oplus\beta=\theta^{-1}\left(\theta(\alpha)+\theta(\beta)\right)=\theta^{-1}\left(\theta(\beta)+\theta(\alpha)\right)=\beta+\alpha$

2) Addition is associative- This is also clear since

$\alpha\oplus\left(\beta\oplus\gamma\right)=\theta^{-1}\left(\theta(\alpha)+\theta\left(\beta\oplus\gamma\right)\right)=\theta^{-1}\left(\theta(\alpha)+\theta\left(\theta^{-1}\left(\theta(\beta)+\theta(\gamma)\right)\right)\right)$

which is equal to

$\theta^{-1}\left(\theta(\alpha)+\left(\theta(\beta)+\theta(\gamma)\right)\right)=\theta^{-1}\left(\left(\theta(\alpha)+\theta(\beta)\right)+\theta(\gamma)\right)$

which finally is equal to

$\theta^{-1}\left(\theta^{-1}\left(\theta\left(\theta(\alpha)+\theta(\beta)\right)\right)+\theta(\gamma)\right)=\theta^{-1}\left(\theta\left(\alpha\oplus\beta\right)+\theta(\gamma)\right)=\left(\alpha\oplus\beta\right)\oplus\gamma$

3) There exists a zero element- Let $0$ be the zero element of $\mathfrak{F}$ then $\theta^{-1}(0)$ is clearly the zero element of $F$. To see this we note that

$\alpha\oplus\theta^{-1}(0)=\theta^{-1}\left(\theta(\alpha)+\theta\left(\theta^{-1}\left(0\right)\right)\right)=\theta^{-1}\left(\theta(\alpha)+0\right)=\theta^{-1}\left(\theta\left(\alpha\right)\right)=\alpha$

for every $\alpha\in F$.

4) Existence of inverse element- If $\alpha\in F$ we note that

$\alpha\oplus\theta^{-1}\left(-\theta(\alpha)\right)=\theta^{-1}\left(\theta(\alpha)+\theta\left(\theta^{-1}\left(-\theta(\alpha)\right)\right)\right)$

which equals

$\theta^{-1}\left(\theta(\alpha)+-\theta(\alpha)\right)=\theta^{-1}(0)$

which is the identity element of $F$

5-8 are the analogous axioms for multiplication, which are (for the most part) the exact same as the above.

9) Distributivity- We note that

$\alpha\odot\left(\beta\oplus\gamma\right)=\theta^{-1}\left(\theta(\alpha)\cdot\theta\left(\beta\oplus\gamma\right)\right)$

which equals

$\theta^{-1}\left(\theta(\alpha)\cdot\theta\left(\theta^{-1}\left(\theta(\beta)+\theta(\gamma)\right)\right)\right)=\theta^{-1}\left(\theta(\alpha)\cdot\left(\theta(\beta)+\theta(\gamma)\right)\right)$

from where the rest is obvious.

This completes the lemma $\blacksquare$

Now, we may answer the question. Since $\mathbb{Q}$ is a field and $\mathbb{Q}\cong\mathbb{N}\cong\mathbb{Z}$ the above lemma implies there exists addition and multiplications on $\mathbb{N}$ and $\mathbb{Z}$ which make them into fields.

3.

Problem: Let $m\in\mathbb{N}-\{1\}$ and let $\mathbb{Z}_m$ denote the integers $\text{mod }m$.

a) Prove this is a field precisely when $m$ is prime

b) What is $-1$ in $\mathbb{Z}_5$?

c) What is $\tfrac{1}{3}$ in $\mathbb{Z}_7$?

Proof:

a) We appeal to the well-known fact that $ax=1\text{ mod }m$ is solvable precisely when $(a,m)=1$. From there we may immediately  disqualify non-primes since the number of multiplicatively invertible elements of $\mathbb{Z}_m$ is $\varphi(m)$ and $\varphi(m) when $m$ is not a prime. When $m$ is a prime the only thing worth noting is that every non-zero element of $\mathbb{Z}_m$ has a multiplicative inverse. The actual work of showing the axioms hold is busy work, and I’ve done it before.

b) It’s clearly $4$. Since $1+4=5=0$

c) It’s $5$. To see this we note that $5\cdot3=15=1$

4.

Problem : Let $\mathfrak{F}$ be a field and define

$c:\mathbb{N}\to\mathfrak{F}:n\mapsto\underbrace{1+\cdots+1}_{n\text{ times}}$

show that either there is no $n$ such that $c(n)=0$ or that if there is, the smallest such $n$ is prime

Proof: Assume that $c^{-1}\left(\{0\}\right)\ne\varnothing$ and $p=\min c^{-1}\left(\{0\}\right)$. Now, suppose that $p=ab$ where $1. We see then that

$c(a)c(b)=(\underbrace{1+\cdots+1}_{a\text{ times}})c(b)=\underbrace{c(b)+\cdots+c(b)}_{a\text{ times}}$

which upon expansion equals

$\underbrace{(\underbrace{1+\cdots+1}_{b\text{ times}})+\cdots+(\underbrace{1+\cdots1}_{b\text{ times}}}_{a\text{ times}})$

which by associativity and grouping is equal to

$\underbrace{1+\cdots+1}_{ab\text{ times}}=\underbrace{1+\cdots+1}_{p\text{ times}}=0$

which by concatenation of the equations yields

$c(a)c(b)=0$

but since $\mathfrak{F}$ is a field it follows that $c(a)=0$ or $c(b)=0$, either way the minimality of $p$ is violated.

5.

Problem: Let $\mathbb{Q}(\sqrt{2})=\left\{a+b\sqrt{2}:a,b\in\mathbb{Q}\right\}$

a) Is $\mathbb{Q}(\sqrt{2})$ a field?

b) What if $\alpha,\beta$ are required to be integers?

Proof:

a) This is a classic yet tedious exercise, I will not do it here.

b) No. For example, consider $1+\sqrt{2}$. Then, we have that

$\displaystyle \left(1+3\sqrt{2}\right)^{-1}=\frac{1-3\sqrt{2}}{1-18}=\frac{-1}{17}+\frac{3}{17}\sqrt{2}\notin\mathbb{Z}(\sqrt{2})$

6.

Problem:

a) Doest the set of all polynomials with integer coefficients ($\mathbb{Z}[x]$) form a field?

b) What about $\mathbb{R}[x]$?

Proof:

a) No.

b) No. I’ll let you figure these out (it’s really easy)

7.

Problem:

Let $\mathfrak{F}$ be the set of all ordered pairs $(a,b)$ of real numbers

a) Is $\mathfrak{F}$ a field if addition and multiplication are done coordinate wise?

b) If addition and multiplication are done as one multiplies complex numbers?

Proof:

a) No. Consider that $(0,1)$ is a not the additive identity but it has no multiplicative inverse.

b) Yes, this is just field isomorphic to $\mathbb{C}$

September 21, 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 2

1.

Problem: Show that if $Y$ is a subspace of $X$, and $A$ is a subspace of $Y$, the the topology it inherits as a subspace of $Y$ is the same as it inherits as a subspace of $X$.

Proof: Let $\mathfrak{J}_X$ and $\mathfrak{J}_Y$ be the induced topologies on $A$ as a subspace of $X,Y$ respectively. Let $U\in\mathfrak{J}_X$ then $U=A\cap V$ for some open set $V$ in $X$. But, $A\cap V=Y\cap A\cap V=A\cap (V\cap Y)$ and since $V\cap Y$ is open in $Y$ it follows that $U\in\mathfrak{J}_Y$. Conversely, suppose that $U\in\mathfrak{J}_Y$ then $U=A\cap V$ for some open set $V$ in $Y$. But, since $Y$ is a subspace of $X$ we know that $V=W\cap Y$ for some open set $W$ in $X$. Thus, $U=A\cap V=A\cap W\cap Y=A\cap W$ and thus $A\in\mathfrak{J}_X$. The conclusion follows. $\blacksquare$

2.

Problem: If $\mathfrak{J},\mathfrak{J}'$ are topologies on $X$ and $\mathfrak{J}'$ is strictly finer than $\mathfrak{J}$, what can you say about the corresponding subspace topologies on the subset $Y$ of $X$?

Proof: Let $\mathfrak{I}$ and $\mathfrak{I}'$ be the subspace topologies on $Y$ inherited from $\mathfrak{J},\mathfrak{J}'$ respectively. Let $U\in\mathfrak{I}$ then $U=Y\cap V$ for some $V\in\mathfrak{J}\subseteq\mathfrak{J}'$ and thus $U\in\mathfrak{I}'$. Thus, $\mathfrak{I}\subseteq\mathfrak{I}'$. The inclusion need not be strict. For example consider the set $X=\{a,b,c\}$ with the two topologies

$\mathfrak{J}=\left\{\varnothing,X\right\}$

$\mathfrak{J}'=\left\{\varnothing,\{a\},X\right\}$

Now, consider $Y=\{b\}\subseteq X$ then

$\mathfrak{I}=\left\{\varnothing\cap\{b\},X\cap\{b\}\right\}=\left\{\varnothing,Y\right\}$

$\mathfrak{I}'=\left\{\varnothing\cap\{b\},\{a\}\cap\{b\},X\cap\{b\}\right\}=\left\{\varnothing,Y\right\}$

And thus $\mathfrak{I}=\mathfrak{I}'$. The inclusion will be strict if $Y$ intersects every element of $\mathfrak{J}$. $\blacksquare$

3.

Problem: Consider the set $Y=[-1,1]$ as a subspace of $\mathbb{R}$. Which of the following sets are open in $Y$?

$A=\left\{x:\frac{1}{2}<|x|<1\right\}$

$B=\left\{\frac{1}{2}<|x|\leq1\right\}$

$C=\left\{x:\frac{1}{2}\leq|x|\leq1\right\}$

$D=\left\{x:\frac{1}{2}\leq|x|\leq1\right\}$

$E=\left\{x:0<|x|<1\text{ and }\frac{1}{x}\notin\mathbb{N}\right\}$

Proof:

We only bother to state which are open in $Y$ since the results should be clear for $\mathbb{R}$ to any student in pre-algebra.

$A$ is open sine $A=Y\cap \left((-1,\frac{-1}{2})\cup(\frac{1}{2},1)\right)$.

$B$ is open since $B=Y\cap\left( (-\infty,\frac{-1}{2})\cup(\frac{1}{2},\infty)\right)$

$C$ is not open. To see this suppose that it was then $Y-C=\left(\frac{-1}{2},\frac{1}{2}\right)\cup\{-1,1\}$ is closed in $Y$, but since $Y$ is closed in $\mathbb{R}$ this would imply that $(\frac{-1}{2},\frac{1}{2})\cup\{-1,1\}$ was closed in $\mathbb{R}$.

$D$ is not open since if it were it’s complement $Y-D=(\frac{-1}{2},\frac{1}{2})$ would be closed in $Y$ and since $Y$ is closed in $X$ it follows $(\frac{-1}{2},\frac{1}{2})$ is closed in $\mathbb{R}$.

$E$: This is open since $E=Y\cap (-1,1)\cap\left(\mathbb{R}-K\right)$ the last of which is open. $\blacksquare$

4.

Problem: A map $f:X\to Y$ is said to be an open map if for every open set $U$ of $X$, the set $f(U)$ is open in $Y$. Show that $\pi_1:X\times Y\to X$ and $\pi_2:X\times Y\to Y$ are open maps.

Proof: Assuming that $X\times Y$ has the product topology we know that given an open set $W\subseteq X\times Y$ we know that $\displaystyle W=\bigcup_{\alpha\in\mathcal{A}}\left(U_\alpha\times V_\alpha\right)$ where $U_\alpha,V_\alpha$ are open in $X,Y$ respectively. Thus,

$\displaystyle \pi_1\left(W\right)=\pi_1\left(\bigcup_{\alpha\in\mathcal{A}}\left(U_\alpha\times V_\alpha\right)\right)=\bigcup_{\alpha\in\mathcal{A}}\pi_1\left(U_\alpha\times V_\alpha\right)=\bigcup_{\alpha\in\mathcal{A}}U_\alpha$

$\displaystyle \pi_2\left(W\right)=\pi_2\left(\bigcup_{\alpha\in\mathcal{A}}\left(U_\alpha\times V_\alpha\right)\right)=\bigcup_{\alpha\in\mathcal{A}}\pi_2\left(U_\alpha\times V_\alpha\right)=\bigcup_{\alpha\in\mathcal{A}}V_\alpha$

Which are open in $X,Y$ respectively. $\blacksquare$

5.

Problem: Let $X$ and $X'$ denote a simle set in the topologies $\mathfrak{J},\mathfrak{J}'$ respectively; let $Y$ and $Y'$ denote a singe set in the topologies $\mathfrak{U},\mathfrak{U}'$ respectively. Assume that these sets are non-empty.  Prove that if $\mathfrak{J}'\supseteq\mathfrak{J}$ and $\mathfrak{U}'\supseteq\mathfrak{U}$ then the product topology $X'\times Y'$ is finer than the product topology on $X\times Y$

Proof:It suffices to check that each basic element in $X'\times Y'$ has a basic element in $X\times Y$ contained within it which contains and arbitrary point. So, let $U'\times V'$ be open in $X'\times Y'$ and let $(x,y)\in U'\times V'$$. Then, we see by the previous problem that $x\in U'$ and $y\in V'$ and thus since the topologies on $X,Y$ are finer than those on $X',Y'$ there are basic open sets $U,V$ in $X,Y$ respectively such that $x\in U\subseteq U'$ and $y\in V\subseteq V'$. Thus, $(x,y)\subseteq U\times V\subseteq U'\times V'$ and since $U\times V$ is basic open in $X\times Y$ the conclusion follows. $\blacksquare$ Problem: Show that $\left\{(a,b)\times(c,d):a is an open base for $\mathbb{R}^2$ Proof: This follows from the fact that $\left\{(a,b):a is an open base for $\mathbb{R}$ and $\mathbb{R}^2$ has the product topology. $\blacksquare$ 7. Problem: Let $X$ be an ordered set. If $Y$ is a proper subset of $X$ that is convex in $X$ does it follow that $Y$ is an interval or a ray in $X$? Proof: No. Consider $\mathbb{Q}$ with the usual ordering and the set $\left\{q\in\mathbb{Q}:q^2 <2\right\}=U$. This is clearly convex since if $a,b\in U$ then for any $a we have that $a^2 and thus $c\in U$ or in other words $(a,b)\subseteq U$. But, it is not an interval or a ray. Clearly if it were either it must be a ray, so assume that $U=(-\infty,\alpha)$ or $U=(-\infty,\beta]$. In both cases we must have that $\alpha,\beta<\sqrt{2}$ and thus by the density of the rationals in $\mathbb{R}$ we may find $\gamma$ such that $|\alpha|,|\beta|<\gamma<\sqrt{2}$ or in other words that $\alpha^2,\beta^2<\gamma^2<2$ and thus $\gamma\notin(-\infty,\alpha),(-\infty,\beta]$. $\blacksquare$ 8. Problem: If $L$ is a straight line in the plane describe the topology $L$ inherets as a subspace of $\mathbb{R}\times\mathbb{R}_\ell$ and as a subspace of $\mathbb{R}_\ell\times\mathbb{R}_\ell$. In each case it is a familiar topology. Proof: Clearly as was shown since $\left\{[a,b):a and $\left\{(a,b):a are open bases for $\mathbb{R}_\ell$ and $\mathbb{R}$ we know that $\mathfrak{B}=\left\{(a,b)\times[c,d):a is an open base for $\mathbb{R}\times\mathbb{R}_\ell$. Now, if you’ll recall the subspace topology on $L$ can be described by $\mathfrak{B}'=\left\{L\cap B:B\in\mathfrak{B}\right\}$. So, we break this into three cases based on what kind of line $L$ is. So, firstly suppose that $L$ is a vertical line then $L=\{x_0\}\times\mathbb{R}$ for some $x_0\in\mathbb{R}$. So, we now claim $\mathfrak{B}'=\left\{\{x_0\}\times[c,d):c. So, let $U=L\cap \left((a,b)\times[c,d)\right)\in\mathfrak{B}'$ then we have that $U=L\cap\left((a,b)\times[c,d)\right)=\left(\{x_0\}\times\mathbb{R}\right)\cap\left((a,b)\times[c,d)\right)=\left(\{x_0\}\cap(a,b)\right)\times [c,d)$ Which, equals $\displaystyle \begin{cases}\varnothing\quad\text{if}\quad x_0\notin(a,b)\\ \{x_0\}\times[c,d)\quad\text{if}\quad x_0\in(a,b)\end{cases}$ Either way $U\in\left\{\{x_0\}\times[c,d):c Conversely, let $U\in\left\{\{x_0\}\times[c,d):c. Then, $\displaystyle \begin{cases} U=L\cap\left((x_0-2,x_0-1)\times[0,1)\right)\quad\text{if}\quad U=\varnothing\\ U=L\cap\left((x_0-1,x_0+1)\times[c,d)\right)\quad\text{if}\quad U=\{x_0\}\times[c,d)\end{cases}$ Either way we see that $U\in\mathfrak{B}'$. A similar analysis shows that if $L=\mathbb{R}\times\{y_0\}$ that $\mathfrak{B}'=\left\{(a,b)\times\{y_0\}:a We now break this into the last case where $L=\left\{(x,\alpha x):x\in\mathbb{R}\right\}$ for some $\alpha\in\mathbb{R}-\{0\}$. So, once again we know that the subspace topology on $L$ will be that generated by $\mathfrak{B}'=\left\{B\cap L:B\in\mathfrak{B}\right\}$. We now get slightly more informal. The key point is that since none of the corners of the rectangle $(a,b)\times[c,d)$ are included the line, if it intersects, it intersects an interval like region of the line. Thus, a quick breakdown into the cases where it intersects the solid line $(a,b)\times\{c\}$ and when it doesn’t should quickly convince you that the topology on $L$ will end up being the lower-limit topology. Now, for the case with $\mathbb{R}_\ell\times\mathbb{R}_\ell$ we have quite a different scenario. For example consider the anti-diagonal $-\Delta=\left\{(x,-x):x\in\mathbb{R}\right\}$ and consider for each fixed $x_0\in -\Delta$ the basic open set $[x_0,x_0+1)\times[x_0,x_0+1)$ clearly then $-\Delta\cap [x_0,x_0+1)\times[x_0,x_0+1)=\{x_0\}$ from where it follows that $-\Delta$ is in fact a discrete space. In fact for any line that is non-horizontal and non-vertical we may find for each point on the line a basic open set whose intersection with the line is that single point. In other words the lines in $\mathbb{R}_\ell\times\mathbb{R}_\ell$ inherit the discrete topology as subspaces. For horizontal and vertical lines they still have the lower limit topology. $\blacksquare$ 9. Problem: Show that the dictionary order topology on the set $\mathbb{R}\times\mathbb{R}$ is the same as the product topology on $\mathbb{R}_d\times\mathbb{R}$ where $\mathbb{R}_d$ denotes $\mathbb{R}$ with the discrete topology. Compare this topology to the usual topology on $\mathbb{R}^2$ Proof: Let $(x_0,y_0)\in \mathbb{R}_d\times\mathbb{R}$ and let $\{x_0\}\times(a,b)$ be a basic open neighborhood of it. Then, consider the set $\left(\left(x_0,\frac{a+y_0}{2}\right),(x_0,b)\right)$ clearly this is open in $\mathbb{R}\times\mathbb{R}$ with the lexicographic ordering and if $(x,y)\in\left(\left(x_0,\frac{a+y_0}{2}\right),(x_0,b)\right)$ we must have that $\left(x_0,\frac{y_0+a}{2}\right)<(x,y)<(x_0,b)$ which implies that $x=x_0\text{ and }a<\frac{y_0+a}{2} and thus $(x,y)\in\{x_0\}\times(a,b)$. Thus, noting $(x_0,y_0)\in\left(x_0,\frac{y_0+a}{2}\right)$ finishes that portion of the argument. Conversely, let $(x_0,y_0)\in\mathbb{R}\times\mathbb{R}$ and $\left((a,b),(c,d)\right)$ be some basic open neighborhood containing it. If $a=c$ then $\left((a,b),(c,d)\right)=\{a\}\times(b,d)$ and we’re done. So, assume that $a. If $x_0=a$ then consider $\{a\}\times\left(\frac{y_0+b}{2},y_0+1\right)$. Clearly $(a,y_0)\in\{a\}\times\left(\frac{y_0+b}{2},y_0+1\right)$. Now, to see that $\{a\}\times\left(\frac{y_0+b}{2},y_0+1\right)$ we note that $(a,y_0)\in\{a\}\times\left(\frac{y_0+b}{2},y_0+1\right)\implies b<\frac{y_0+b}{2} and thus $(a,b)<(a,y_0)$ and since $a we automatically have that $(a,y)<(c,d)$. Now, if $x_0=c$ then choosing $\{c\}\times\left(y_0-1,\frac{y_0+d}{2}\right)$ and applying similar techniques works. Lastly, if $a then choosing $\{x_0\}\times\left(y_0-1,y_0+1\right)$ automatically works since $(x_0,y_0)\in\{x_0\}\times(y_0-1,y_0+1)$ and if $(x_0,y)\in\{x_0\}\times(y_0-1,y_0+1)$ we have that $x_0 and $x_0. Regardless, we’ve found a basic open set in $\mathbb{R}_d\times\mathbb{R}$ which contains $(x_0,y_0)$ and which is contained in $\left((a,b),(c,d)\right)$. It follows that the each topology is finer than the others, or that the topologies are equal. The order topology on $\mathbb{R}\times\mathbb{R}$ is finer than that of the usual topology on $\mathbb{R}^2$. Just note that if $(x_0,y_0)\in\mathbb{R}^2$ and $B_\delta((x_0,y_0))$ is an open ball centered at it that $\{x_0\}\times(y_0-\delta,y_0+\delta)$ is a basic open set in $\mathbb{R}\times\mathbb{R}$ containing $(x_0,y_0$ and which is contained inside the ball. The converse is not true. For example consider the origin $(0,0)$ and the basic open neighborhood $\{0\}\times(-1,1)$. Any open ball centered at zero will contain points not on that line from where the conclusion follows. $\blacksquare$ 10. Problem: Let $I=[0,1]$. Compare the product topology on $I\times I$, the dictionary order topology on $I\times I$, and the topology $I\times I$ inherits as a subspace of $\mathbb{R}\times\mathbb{R}$ in the dictionary order topology. Proof: Let $\mathfrak{J}_1,\mathfrak{J}_2,\mathfrak{J}_3$ denote the topologies in the order they were mentioned. We first claim that $\mathfrak{J}_1\subsetneq\mathfrak{J}_2$. To see this let $(x_0,y_0)\in I\times I$ be arbitrary and note that $\left\{(a,b)\times(c,d):a generates $\mathfrak{J}_1$. Then, if $(x_0,y_0)\in(a,b)\times(c,d)$ we have that $(x_0,y_0)\in\left((x_0,b),(x_0,d)\right)\subseteq(a,b)\times(c,d)$. To see why the inclusion’s strict we note as in the last example that for example $(\frac{1}{2},\frac{1}{2})\in\left((\frac{1}{2},0),(\frac{1}{2},1)\right)$ and any open square must intersect points of not in that interval. Clearly then $\mathfrak{J}_3\subsetneq\mathfrak{J}_2$ as was shown in the book. $\blacksquare$ May 21, 2010 ## Munkres Chapter 2 Section 1 This begins a substantial effort to complete all of (except the first chapter) the problems in James Munkre’s Topology I have chosen to start at chapter 2 considering the first chapter is nothing but prerequisites. Thus, without a minute to spare let us being. 1. Problem: Let $X$ be a topological space, let $A$ be a subset of $X$. Suppose that for each $x\in A$ there is an open set $U_x$ containing $x$ such that $U\subseteq A$. Prove that $A$ is open. Proof: We claim that $\displaystyle A=\bigcup_{x\in A}U_x=\overset{\text{def.}}{=}\Omega$ but this is obvious since for each $x\in A$ we have that $x\in U_x\subseteq\Omega$. Conversely, since each $U_x\subseteq A$ we have that the union of all of them is contained in $A$, namely $\Omega\subseteq A$. Thus, $A$ is the union of open sets and thus open. $\blacksquare$ 2. Problem: Compare the nine topologies on $\{a,b,c\}$ given in example 1. Solution: This is simple. 3. Problem: Show that given a set $X$ if we denote $\mathfrak{J}$ to be cocountable topology (a set is open if it’s complement is countable or the full space) that $\left(X,\mathfrak{J}\right)$ is a topological space. Is it still a topological space if we let $\mathfrak{J}=\left\{U\in\mathcal{P}(X):X-U\text{ is infinite, empty, or all of }X\right\}$? Proof: Clearly for the first part $\varnothing,X\in\mathfrak{J}$. Now, if $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ is a collection of open sets then we note that $X-U_\alpha$ is finite and thus $\displaystyle X-\bigcup_{\alpha\in\mathcal{A}}U_\alpha=\bigcap_{\alpha\in\mathcal{A}}\left(X-U_\alpha\right)\subseteq U_{\alpha_0}$ for any $\alpha_0$. Thus, $\displaystyle X-\bigcup_{\alpha\in\mathcal{A}}U_\alpha$ is finite and thus in $\mathfrak{J}$. Now, if $U_1,\cdots,U_n\in\mathfrak{J}$ we have that $X-(U_1\cap\cdots\cap U_n)=(X-U_1)\cup\cdots\cup (X-U_n)$ and thus $X-(U_1\cap\cdots\cap U_n)$ is the finite union of finite sets and thus finite, so $U_1\cap\cdots\cap U_n\in\mathfrak{J}$. If we redefine the topology as described it is not necessarily a topology. For example, give $\mathbb{N}$ that topology and note that $\left\{\{n\}\right\}_{n\in\mathbb{N}-\{1\}}$ is a collection of elements of $\mathfrak{J}$ but $\displaystyle \bigcup_{n\in\mathbb{N}-\{1\}}\{n\}=\mathbb{N}-\{1\}$ whose complement is neither infinite, empty, or the full space. Thus, this isn’t a topology. 4. Problem: a) If $\left\{\mathfrak{J}_\alpha\right\}_{\alpha\in\mathcal{A}}$ is a family of topologies on $X$, show that $\displaystyle \bigcap_{\alpha\in\mathcal{A}}\mathfrak{J}_\alpha$ is a topology on $X$. Is $\displaystyle \bigcup_{\alpha\in\mathcal{A}}\mathfrak{J}_\alpha$? b) Let $\left\{\mathfrak{J}\right\}_\alpha$ be a family of topologies on $X$. Show that there is unique topology on $X$ containing all the collections $\mathfrak{J}_\alpha$, and a unique largest topology contained in all of the $\mathfrak{J}_\alpha$. c) If $X=\{a,b,c\}$, let $\mathfrak{J}_1=\{\varnothing,X,\{a\},\{a,b\}\}$ and $\mathfrak{J}_2=\left\{\varnothing,X,\{a\},\{b,c\}\right\}$. Find the smallest topology containing $\mathfrak{J}_1,\mathfrak{J}_2$ and the larges topology contained in $\mathfrak{J}_1,\mathfrak{J}_2$. Proof: a) Let $\displaystyle \Omega\overset{\text{def.}}{=}\bigcap_{\alpha\in\mathcal{A}}\mathfrak{J}_\alpha$ and let $\left\{U_\beta\right\}_{\beta\in\mathcal{B}}\subseteq\Omega$ be arbitrary. Then, by assumption we have that $\left\{U_\beta\right\}_{\beta\in\mathcal{B}}\subseteq\mathfrak{J}_\alpha$ for every $\alpha\in\mathcal{A}$ but since this was assumed to be a topology we have that for every $\alpha\in\mathcal{A}$ that $\displaystyle \bigcup_{\beta\in\mathcal{B}}U_\beta\subseteq\mathfrak{J}_\alpha$ and thus $\displaystyle \bigcup_{\beta\in\mathcal{B}}U_\beta\in\Omega$. Now, if $\{U_1,\cdots,U_n\}\subseteq\Omega$ we must have that $\{U_1,\cdots,U_n\}\subseteq\mathfrak{J}_\alpha$ for every $\alpha$ and thus $U_1\cap\cdots\cap U_n\in\mathfrak{J}_\alpha$ for every $\alpha\in\mathcal{A}$ and thus $U_1\cap\cdots\cap U_n\in\Omega$. Thus, noting that for every $\alpha\in\mathcal{A}$ we must have that $\varnothing,X\in\mathfrak{J}_\alpha$ the conclusion follows. No, the union of two topologies needn’t be a topology. Let $\mathfrak{J}_1,\mathfrak{J}_2$ be defined as in part c and note that $\mathfrak{J}_1\cup\mathfrak{J}_2=\left\{\varnothing,X,\{a\},\{a,b\},\{b,c\}\right\}$ but $\{a,b\}\cap\{b,c\}=\{c\}\notin\mathfrak{J}_1\cup\mathfrak{J}_2$ b) This follows immediately from part a. For the first part let $\Omega=\left\{\mathfrak{J}\in\text{Top }X:\mathfrak{J}_\alpha\subseteq\mathfrak{J},\text{ for all }\alpha\in\mathcal{A}\right\}$ (where $\text{Top }X$ is the set of all topologies) and let $\mathfrak{T}=\bigcap_{\mathfrak{J}\in\Omega}\mathfrak{J}$, this clearly satisfies the conditions. For the second one merely take $\displaystyle \bigcap_{\alpha\in\mathcal{A}}\mathfrak{J}_\alpha$ c) We (as pointed out very poignantly in the previous problem) just take the union of the two topologies. So, we claim that the desired smallest topology is in fact $\left\{X,\varnothing,\{a\},\{b\},\{a,b\},\{a,c\}\right\}$. But this is just computation and we leave it to the reader. For the second part just intersect the two topologies. 5. Problem: Show that if $\mathcal{A}$ is a base for a topology on $X$, then the topology generated by $\mathcal{A}$ equals the intersection of all topologies on $X$ which contain $\mathcal{A}$. Prove the same if $\mathcal{A}$ is a subbase Proof: Let $\Omega$ be the intersection of all topologies on $X$ which contain $\mathcal{A}$ and $\mathfrak{J}_g$ the topology generated by $\mathcal{A}$. Clearly $\Omega\subseteq\mathfrak{J}_g$ since $\mathfrak{J}_g$ is itself a topology on $X$ containing $\mathcal{A}$. Conversely, let $U\in\mathfrak{J}_g$ we show that $U\in\mathfrak{J}$ where $\mathfrak{J}$ is any topology on $X$ containing $X$. But, this is obvious since $\displaystyle U=\bigcup_{B\in\mathcal{B}}B$ for some $\mathcal{B}\subseteq\mathcal{A}$ and thus $U$ is the union of open sets in $\mathfrak{J}$ and thus in $\mathfrak{J}$. The conclusion follows. Next, let $\Omega,\mathfrak{J}_g$ be above except now $\mathcal{A}$ is a subbase. For the same reasons as above we have that $\Omega\subseteq\mathfrak{J}_g$. Conversely, for any topology $\mathfrak{J}$ containing $\mathcal{A}$ we have that if $U\in\mathfrak{J}_g$ then $\displaystyle U=\bigcup_{\beta\in\mathcal{B}}V_\beta$ where each $V_\beta$ is the finite union of elements of $\mathcal{A}$. But, by construction it follows that each $V_\beta$ is open (it is the finite intersection of open sets in $\mathfrak{J}$) and thus $U$ is the union of open sets in $\mathfrak{J}$ and thus $V\in\mathfrak{J}$. $\blacksquare$ 6. Problem: Show that the topologies on $\mathbb{R}_K$ and the Sorgenfrey line aren’t comparable Proof: See the last part of the next problem 7. Problem: Consider the following topologies on$\mathbb{R}$: $\mathfrak{J}_1=\text{usual topology}$ $\mathfrak{J}_2=\text{topology on }\mathbb{R}_K$ $\mathfrak{J}_3=\text{cofinite topology}$ $\mathfrak{J}_4=\text{topology having the set set of all }(a,b]\text{ as a base}$ $\mathfrak{J}_5=\text{the topology having all sets }(-\infty,a)\text{ as a base}$ For each determine which of the others contain it. Solution: $\mathfrak{J}_1\subseteq\mathfrak{J}_2$:Clearly we have that $\mathfrak{J}_1\subseteq\mathfrak{J}_2$ since the defining open base for $\mathfrak{J}_1$ is contained entirely in $\mathfrak{J}_2$. $\mathfrak{J}_1\supseteq\mathfrak{J}_3$: But, $\mathfrak{J}_1\not\subseteq\mathfrak{J}_3$ since $(0,1)\in\mathfrak{J}_1$ but $\mathbb{R}-(0,1)\simeq\mathbb{R}$ and thus not in $\mathfrak{J}_3$. Now, to prove the inclusion indicated we know that for each open set $U$ in the cofinite topology we have that $U=\mathbb{R}-\{x_1,\cdots,x_n\}$ and so if we assume WLOG that $x_1<\cdots then $\displaystyle U=\bigcup_{ax_n}(x_n,b)$ And thus $U$ is open in the usual topology. $\mathfrak{J}_1\subseteq\mathfrak{J}_4$: But, $\mathfrak{J}_1\subseteq\mathfrak{J}_4$. To see this it suffices to show that $(a,b)$ is open in $\mathfrak{J}_4$ since this is a base for $\mathfrak{J}_1$. But, to see this we must merely note that $\displaystyle (a,b)=\bigcup_{c. $\mathfrak{J}_1\supseteq\mathfrak{J}_5$: Lastly, $\mathfrak{J}_1\supseteq\mathfrak{J}_5$. To see this we must merely note that $\displaystyle (-\infty,b)=\bigcup_{a. For $\mathfrak{J}_2$ the result is obvious except possibly how it relates to $\mathfrak{J}_3$. But, in fact they aren’t comparable. To see this we first show that $(0,1]$ is not open in $\mathbb{R}_K$. To see this we show it can’t be written as the union of sets of the form $(a,b)$ and $(c,d)-K$ but it clearly suffices to do this for the latter sets. Now, to see that $(0,1]$ can’t be written as the union of sets of the form $(a,b)$ we recall from basic real number topology that $(0,1]$ is not open ($1$ is not an interior point) and thus it can’t be written as the union of intervals or it would be open in the usual topology on $\mathbb{R}$. Also, consider $(-1,1)-K$. 8. Problem: Show that the countable collection $\mathfrak{B}=\left\{(a,b):a is a base for the usual topology on $\mathbb{R}$ Proof: This follows from the density of $\mathbb{Q}$. It suffices to show that given any $x\in\mathbb{R}$ and any $(a,b)\supseteq\{x\}$ that there is some element $(p,q)\in\mathfrak{B}$ such that $x\in(p,q)\subseteq(a,b)$. But, from basic analysis we know there is some rational number $q$ such that $a and similarly there is some $p\in\mathbb{Q}$ such that $x. Thus, $x\in(p,q)\subseteq(a,b)$. The conclusion follows. $\blacksquare$. 9. Problem: Show that the collection $\mathcal{C}=\left\{[a,b):a generates a different topology from the one on $\mathbb{R}_\ell$ (the lower limit topology)$.

Proof: Clearly $[\sqrt{2},3)$ is open in $\mathbb{R}_\ell$ but we show that it can’t be written as the union of elements of $\mathcal{C}$. So, suppose that $\displaystyle \bigcup_{\alpha\in\mathcal{A}}[a_\alpha,b_\alpha)=[\sqrt{2},3)$ where $\left\{[a_\alpha,b_\alpha)\right\}_{\alpha\in\mathcal{A}}\subseteq\mathcal{C}$. Then, there exists some $[a_\alpha,b_\alpha)$ such that $\sqrt{2}\in[a_\alpha,b_\alpha)$. Now, since $[a_\alpha,b_\alpha)\subseteq[\sqrt{2},b_\alpha)$ we must have that $a_\alpha\geqslant \sqrt{2}$ but $a_\alpha\neq\sqrt{2}$ and thus $a_\alpha>\sqrt{2}$ and thus $\sqrt{2}\notin[a_\alpha,b_\alpha)$. Contradiction. $\blacksquare$

May 20, 2010

## Just For Fun(Rudin’s Topology Section) Part IV

28.

Problem: Prove that ever closed set in a separable metric space $\mathcal{M}$  is the union of a (possibly empty) perfect set and a set which is a set which is countable.

Proof: It is easy to see that the above proof (proof 27.) holds in any space which is second countable, in particular for separable metric spaces. Thus, if $E\subseteq\mathcal{M}$ is closed we surely have that $E\subseteq \mathfrak{C}\cup\left(\left(\mathbb{R}-\mathfrak{C}\right)\cap E\right)$. Thus, $E=\left(E\cap\mathfrak{C}\right)\cup\left(\left(\mathbb{R}-\mathfrak{C}\right)\cap E\right)$ but since $\mathfrak{C}\subseteq D(E)$ and $D(E)\subseteq E$ we have then that $E=\mathfrak{C}\cup\left(\left(\mathbb{R}-\mathfrak{C}\right)\cap E\right)$ which is the union of a perfect and countable set respectively. $\blacksquare$

29.

Problem: Prove that every open set $U\subseteq\mathbb{R}$ may be written as the disjoint union of countably many open intervals.

Proof: We need to prove a quick lemma

Lemma: Let $X$ be a topological space and let $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ be a class of connected subspace of $X$ such that $U_{\alpha_1}\cap U_{\alpha_2}\ne\varnothing,\text{ }\alpha_1,\alpha_2\in\mathcal{A}$ then $\displaystyle \Lambda=\bigcup_{\alpha\in\mathcal{A}}$ is a connected subspace of $X$.

Proof: Suppose that $\left(E\cap \Lambda\right)\cup\left(G\cap \Lambda\right)=\Lambda$ is a separation of $\Lambda$. We may assume WLOG that $U_{\alpha_0}\cap E\ne\varnothing$ for some $\alpha_0\in\mathcal{A}$. So, now we see that $U_{\alpha_0}\subseteq E\cap \Lambda$ otherwise $E\cap U_{\alpha_0},G\cap U_{\alpha_0}$ would be non-empty disjoint subsets of $U_{\alpha_0}$ whose union is $U_{\alpha_0}$ contradicting that $U_{\alpha_0}$ is connected. Thus, it easily follows that for any  $U_\alpha$ we have that $U_\alpha\cap U_{\alpha_0}\ne\varnothing$ so that $U_{\alpha}\cap E\ne\varnothing$ and thus by a similar reasoning we see that $U_\alpha\subseteq E\cap\Lambda$. Thus, since $\alpha$ was arbitrary it follows that $\Lambda\subseteq E\cap\Lambda$ contradicting that $G\cap\Lambda\ne\varnothing$. $\blacksquare$

So, now for each $x\in U$ define

$\mathcal{C}(x)=\left\{V\subseteq U: V\text{ is open in }U,\text{ }V\text{ is connected },\text{ and }x\in V\right\}$

And let $\displaystyle C(x)=\bigcup_{V\in\mathcal{C}(x)}V$ and finally we prove that

$\Omega=\left\{C(x):x\in U\right\}$

is a countable class of disjoint open intervals whose union is $U$. The fact that each $C(x)$ is an open interval follows since it is the union of open connected spaces with non-empty intersection (each element of $\mathcal{C}(x)$ contains $x$) it is also an open connected subspace of $\mathbb{R}$ (note that each element of $\mathcal{C}(x)$ is open in $U$ but since $U$ is open it is also open in $\mathbb{R}$. But, it was proven in the book the only connected subspace of $\mathbb{R}$ are intervals and thus $C(x)$ is an interval for each $C(x)\in\Omega$.

Now, to see that they are disjoint we show that if $C(x)\cap C(y)\ne\varnothing$ then $C(x)=C(y)$ from where the conclusion will follow. So, to see this we first note that if $C(x)\cap C(y)$ is non-empty then $C(x)\cup C(y)$ is an open connected subspace of $U$ containing both $x$ and $y$ and thus $C(x)\cup C(y)\subseteq C(x)$ and $C(x)\cup C(y)\subseteq C(y)$ but this implies that $C(y)\subseteq C(x)$ and $C(x)\subseteq C(y)$ respectively from where the conclusion follows.

Now, to see that $\Omega$ is countable we notice that for each $\Omega$ we have that $C(x)\cap\mathbb{Q}\ne\varnothing$ and so if we let $F\left(C(x)\right)$ be a fixed but arbitrary $q\in C(x)\cap\mathbb{Q}$ then

$F:\Omega\to\mathbb{Q}:C(x)\mapsto F\left(C(x)\right)$

is an injection since the elements of $\Omega$ are pairwise disjoint. The fact that $\Omega$ is countable follows immediately.

Thus, $\Omega$ is a countable collection of open intervals and

$\displaystyle U=\coprod_{C(x)\in\Omega}C(x)$

Thus, the conclusion follows. $\blacksquare$

30.

Problem: Prove that if $\displaystyle \mathbb{R}^n=\bigcup_{n\in\mathbb{N}}F_n$ where each $F_n$ is a closed subset of $\mathbb{R}^n$ then at least one $F_n$ has non-empty interior.

Proof:

Lemma: Let $\mathcal{M}$ be a complete metric space and $\left\{K_n\right\}_{n\in\mathbb{N}}$ a descending sequence of non-empty closed subsets of $\mathcal{M}$ such that $\text{diam }K_n\to 0$. Then, $\displaystyle \bigcap_{n\in\mathbb{N}}K_n$ contains one point.

Proof: Clearly $\displaystyle \bigcap_{n\in\mathbb{N}}K_n$ does not contain more than one point since $\text{diam }K_n\to 0$. So, now for each $n\in\mathbb{N}$ choose some $x_n\in K_n$ and let $P=\left\{x_n:n\in\mathbb{N}\right\}$. Now, if $P$ were somehow finite a similar argument to that used in the first case of problem 26. is applicable, so assume that $P$ is infinite. But, since $\text{diam }K_n\to 0$ it is evident that $\left\{x_n\right\}_{n\in\mathbb{N}}$ is a Cauchy sequence and thus by assumption it converges to some point $x\in\mathcal{M}$. We claim that $\displaystyle x\in\bigcap_{n\in\mathbb{N}}K_n$. To see this we note similarly to problem 26. that since $P$ is infinite it is easy to see that $x$ is a limit point of $P$ and so if $x\notin K_{n_0}$ for some $n_0\in\mathbb{N}$ then $\mathcal{M}-K_{n_0}$ is a neighborhood of $x$ containing only finitely many points of $P$ which clearly contradicts that it is a limit point. The conclusion follows. $\blacksquare$

So, now suppose each $F_n$ had empty interior (i.e. nowhere dense) . Then, since $\mathcal{M}$ is open and $F_1$ is nowhere dense there exists an open ball $B_1$ of radius less than one such that $B_1\cap F_1=\varnothing$. Let $E_1$  be the concentric closed ball of $B_1$ whose radius is half that of $B_1$. Since $F_2$ is nowhere dense $E_1^{\circ}$ contains an open ball $B_2$ of radius less than one-half which is disjoint from $F_2$.  Let $E_2$ be the concentric closed ball of $B_2$ whose radius is one-half that of $B_2$. Since $F_3$ is nowhere dense we have that $E_2^{\circ}$ contains an open ball $B_3$ of radius less than one-fourth which is disjoint from $F_3$. Let $E_3$ be the concentric closed ball of $B_3$ whose radius is half that of $E_3$. Continuing in this we get a descending sequence of non-empty closed subsets $\left\{E_n\right\}$ for which $\text{diam }E_n\to 0$. Thus, by our lemma we have that there is some $\displaystyle x\in\bigcap_{n\in\mathbb{N}}E_n$. This point is clearly not in any of the $F_n$‘s from where the conclusion follows. $\blacksquare$

May 14, 2010