Abstract Nonsense

Crushing one theorem at a time

Halmos Section 36: Inverses (Pt. III)

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

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December 4, 2010 Posted by | Fun Problems, Halmos, Linear Algebra | , , , , , | Leave a comment

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.

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November 22, 2010 Posted by | Fun Problems, Halmos, Linear Algebra | , , , | 3 Comments

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.

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November 10, 2010 Posted by | Fun Problems, Halmos, Linear Algebra | , , , , , , , , | Leave a comment

Halmos Section 26 and 27: Permutations and Cycles cont.

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

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November 10, 2010 Posted by | Fun Problems, Halmos, Linear Algebra, Uncategorized | , , , , , , | Leave a comment

Munkres Chapter one Section two: Functions


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


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.


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


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.


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


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)


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?


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


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}


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


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.


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 Posted by | Fun Problems, Munkres, Topology | , , , , , | 7 Comments

Halmos Chaper One, Section 1: Fields


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


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



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)?


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




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


2) Addition is associative- This is also clear since


which is equal to


which finally is equal to


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


for every \alpha\in F.

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


which equals


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


which equals


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.


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?


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)<m-1 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


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<a,b<p. 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


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.


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?


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})



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

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


a) No.

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



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?


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 Posted by | Fun Problems, Halmos, Munkres, Topology | , , , , , , | 4 Comments

Munkres Chapter 2 Section 18


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))



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


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}'


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


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


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


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



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?


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.


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.


a) Let x_0\notin\Omega then f(x_0)>g(x_0). Suppose first that there is no g(x_0)<\xi<f(x_0) 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<f(x_0) 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)<f(x_0)\leqslant f(z) and thus z\notin\Omega.

Now, suppose that there is some \xi such that g(x_0)<\xi<f(x_0) 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


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.$


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


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


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.


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


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)


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 Posted by | Fun Problems, Munkres, Topology, Uncategorized | , , , , , | 4 Comments

Munkres Chapter 2 Section 2


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


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



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



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


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





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


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


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


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<b,c<d,\text{ }a,b,c,d\in\mathbb{Q}\right\} is an open base for \mathbb{R}^2

Proof: This follows from the fact that \left\{(a,b):a<b,\text{ }a,b\in\mathbb{Q}\right\} is an open base for \mathbb{R} and \mathbb{R}^2 has the product topology. \blacksquare


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<c<b we have that a^2<c^2<b^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


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<b,\text{ }a,b\in\mathbb{R}\right\} and \left\{(a,b):a<b,\text{ }a,b\in\mathbb{R}\right\} are open bases for \mathbb{R}_\ell and \mathbb{R} we know that \mathfrak{B}=\left\{(a,b)\times[c,d):a<b,c<d,\text{ }a,b,c,d\in\mathbb{R}\right\} 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<d,\text{ }c,d\in\mathbb{R}\right\}\cup\{\varnothing\}.  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<d\text{ }c,d\in\mathbb{R}\right\}\cup\{\varnothing\}

Conversely, let U\in\left\{\{x_0\}\times[c,d):c<d,\text{ }c,d\in\mathbb{R}\right\}\cup\{\varnothing\}. 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<b,\text{ }a,b\in\mathbb{R}\right\}

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


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}<y<b 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<b. 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}<y_0 and thus (a,b)<(a,y_0) and since a<c 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<x_0<c 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<a\implies (a,b)<(x_0,y) and x_0<c\implies (x_0,y)<(c,d). 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


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<b,c<d\text{ }a,b,c,d\in I\right\} 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 Posted by | Fun Problems, Topology | , , , , , | Leave a comment

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.


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


Problem: Compare the nine topologies on \{a,b,c\} given in example 1.

Solution: This is simple.


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.



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.


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.


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


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


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.


\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<x_n then

\displaystyle U=\bigcup_{a<x_1}(a,x_1)\cup(x_1,x_2)\cup\cdots\cup(x_{n-1},x_n)\bigcup_{b>x_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<b}(a,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<b}(a,b).

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.


Problem: Show that the countable collection \mathfrak{B}=\left\{(a,b):a<b,\text{ }a,b\in\mathbb{Q}\right\} 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<q<x and similarly there is some p\in\mathbb{Q} such that x<p<b. Thus, x\in(p,q)\subseteq(a,b). The conclusion follows. \blacksquare.


Problem: Show that the collection \mathcal{C}=\left\{[a,b):a<b,\text{ }a,b\in\mathbb{R}\right\} 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 Posted by | Fun Problems, Munkres, Topology | , , , , , , | Leave a comment

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


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


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


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.


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 Posted by | Analysis, Fun Problems, Topology, Uncategorized | , , , , , , , , | Leave a comment