Abstract Nonsense

Crushing one theorem at a time

Munkres Chapter 2 Section 17


Theorem 17.11

Problem: Prove that every simply ordered set X with the order topology is Hausdorff. The product of two Hausdorff spaces is Hausdorff. A subspace of a Hausdorff space is Hausdorff.

Proof: Let X have the order topology and let a,b\in X be distinct and assume WLOG that a<b. If there does not exists a c\in X such that a<c<b then (-\infty,b),(a,\infty) are disjoint neighborhoods of a,b respectively. They are disjoint for to suppose that x\in(-\infty,b)\cap(a,\infty) would imply that a<x<b contradictory to our assumption. If there is some a<c<b then (-\infty,c),(c,\infty) are disjoint neighborhoods of a,b respectively.

Let X,Y be Hausdorff and (x,y),(x',y')\in X\times Y be distinct. Since (x,y)\ne(x',y') are distinct it follows we may assume WLOG that x\ne x'. But, since X is Hausdorff there exists disjoint neighborhoods U,V of x,y. So, consider U\times Y,V\times Y these are clearly neighborhoods of (x,y),(x',y') in X\times Y and to assume (u,v)\in U\times Y\cap V\times Y would imply \pi_1((u,v))-u\in\pi_1(U\times Y\cap V\times Y)\subseteq U\cap V.

Lastly, suppose that X is Hausdorff and let Y be a subspace of X. If x,y\in Y are distinct there exists disjoint neighborhoods U,V of them in X. Thus, U\cap Y,V\cap Y are disjoint neighborhoods of them in Y. \blacksquare

1.

Problem: Let \mathcal{C} be a collection of subsets of the set X. Suppose that \varnothing,X\in\mathcal{C}, and that the finite union and arbitrary intersection of elements of \mathcal{C} are in \mathcal{C}. Prove that the collection \mathfrak{J}=\left\{X-C:C\in\mathcal{C}\right\} is a topology on X.

Proof: Clearly X=X-\varnothing and \varnothing=X-X are in \mathfrak{J}. Now, suppose that \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}\subseteq\mathfrak{J} then \displaystyle X-\bigcup_{\alpha\in\mathcal{A}}U_\alpha=\bigcap_{\alpha\in\mathcal{A}}\left(X-U_\alpha\right) and since \left\{X-U_\alpha\right\}_{\alpha\in\mathcal{A}}\subseteq\mathcal{C} and \mathcal{C} is closed under arbitrary intersection it follows that \displaystyle \bigcap_{\alpha\in\mathcal{A}}\left(X-U_\alpha\right)=X-\bigcup_{\alpha\in\mathcal{A}}U_\alpha\in\mathcal{C} and thus \displaystyle \bigcup_{\alpha\in\mathcal{A}}U_\alpha\in\mathcal{C}. Lastly, suppose that \left\{U_1,\cdots,U_n\right\}\subseteq\mathcal{C} then \displaystyle X-\left(U_1\cap\cdots\cap U_n\right)=\left(X-U_1\right)\cup\cdots\cup\left(X-U_n\right) and since \left\{X-U_1,\cdots,X-U_n\right\}\subseteq\mathcal{C} and \mathcal{C} is closed under finite union it follows that X-\left(U_1\cap\cdots\cap U_n\right)\in\mathcal{C} and thus U_1\cap\cdots\cap U_n\in\mathfrak{J}. \blacksquare

2.

Problem: Show that if A is a closed in Y and Y is closed in X that A is closed in X.

Proof: Since A is closed in Y and Y is a subspace of X we have that A=Y\cap G for some closed set G\subseteq X but since Y is closed it follows that A is the intersection of two closed sets in X and thus closed in X. \blacksquare

3.

Problem: Show that if A is closed in X and B is closed in Y then A\times B is closed in X\times Y

Proof: This follows immediately from question 9. \blacksquare

4.

Problem: Show that if U is open in X and A is closed in X, then U-A is open in X and A-U is closed in X.

Proof: This follows immediately from the fact that U-A=U\cap\left(X-A\right) and A-U=A\cap\left(X-U\right). \blacksquare

5.

Problem: Let X be an ordered set in the order topology. Show that \overline{(a,b)}\subseteq[a,b]. Under what conditions does equality hold?

Proof: Let x\in(-\infty,a)\cup(b,\infty) then (-\infty,a)\cup(b,\infty) is a neighborhood of x disjoint from (a,b) and thus x\notin\overline{(a,b)}. Equality will hold when a,b are limit points of (a,b) or said otherwise whenever a<c<b we have that there are some $lated d,e$ such that a<d<c<e<b. \blacksquare

6.

Problem: Let A,B and \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}} denote subsets of X. Prove that if A\subseteq B then \overline{A}\subseteq\overline{B}, \overline{A\cup B}=\overline{A}\cup\overline{B}, and \displaystyle \overline{\bigcup_{\alpha\in\mathcal{A}}U_\alpha}\supseteq\bigcup_{\alpha\in\mathcal{A}}\overline{U_\alpha}. Give an example where this last inclusion is strict.

Proof: We choose to prove the second part first. Let x\notin\overline{A\cup B} then there is a neighborhood N of x such that N\cap (A\cup B)=(N\cap A)\cup (N\cap B)=\varnothing and thus x\notin \overline{A}\text{ and }x\notin\overline{B} and so x\notin\overline{A}\cup\overline{B}. Conversely, suppose that x\notin\overline{A}\cup\overline{B} then x\notin \overline{A} and x\notin\overline{B} and so there are neighborhoods N,G of x such that N\cap A=\varnothing,G\cap B=\varnothing clearly then N\cap G is a neighborhood of x such that (N\cap G)\cap(A\cup B)=\varnothing and thus x\notin\overline{A\cup B}.

Using this, if A\subseteq B then we have that \overline{B}=\overline{B\cup A}=\overline{B}\cup\overline{A} from where it follows that \overline{A}\subseteq\overline{B}.

Let \displaystyle x\notin\overline{\bigcup_{\alpha\in\mathcal{A}}U_\alpha} then there is a neighborhood N of x such that \displaystyle N\cap\bigcup_{\alpha\in\mathcal{A}}U_\alpha=\bigcup_{\alpha\in\mathcal{A}}\left(N\cap U_\alpha\right)=\varnothing and thus N\cap U_\alpha=\varnothing for every \alpha\in\mathcal{A} and thus x\notin\overline{U_\alpha} for every \alpha\in\mathcal{A} and so finally we may conclude that \displaystyle x\notin\bigcup_{\alpha\in\mathcal{A}}\overline{U_\alpha}.

To see when inclusion can be strict consider \mathbb{R} with the usual topology. Then,

\displaystyle \mathbb{R}=\overline{\bigcup{q\in\mathbb{Q}}\{q\}}\supsetneq\bigcup_{q\in\mathbb{Q}}\overline{\{q\}}=\bigcup_{q\in\mathbb{Q}}\{q\}=\mathbb{Q}

\blacksquare

7.

Problem: Criticize proof (see book).

Proof: There is no guarantee that the A_\alpha for which U intersected will b e the same A_\alpha that V will intersect if you pick another V. \blacksquare

8.

Problem: Let A,B and \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}} denote subsets of X. Determine whether the following equations hold, if any equality fails determine which inclusion holds.

a) \overline{A\cap B}=\overline{A}\cap\overline{B}

b) \displaystyle \overline{\bigcap_{\alpha\in\mathcal{A}}U_\alpha}=\bigcap_{\alpha\in\mathcal{A}}\overline{U_\alpha}

c) \overline{A-B}=\overline{A}-\overline{B}

Proof:

a) Equation does not hold. Consider that \overline{(-1,0)\cap(0,1)}=\overline{\varnothing}=\varnothing but \overline{(-1,0)}\cap\overline{(0,1)}=[-1,0]\cap[0,1]=\{0\}. The \subseteq inclusion always holds.

b) This follows from a) that equality needn’t hold. Once again the \subseteq inclusion is true.

c) This need be true either \overline{\mathbb{R}-\mathbb{Q}}=\mathbb{R}\ne\overline{\mathbb{R}}-\overline{\mathbb{Q}}=\varnothing. The \subseteq inclusion holds. \blacksquare

9.

Problem: Prove that if A\subseteq X,B\subseteq Y then \overline{A\times B}=\overline{A}\times\overline{B} in the product topology on X\times Y.

Proof: Let \displaystyle (x,y)\in\overline{A\times B}. Let U\times V be a basic open set in X\times Y which contains (x,y). Since x\in \overline{A},y\in \overline{B} we can choose some point x'\in U\cap A,y'\in V\cap B. Then, (x',y')\in U\times V\cap A\times B. It follows that \displaystyle (x,y)\in\overline{A\times B}

Conversely, let (x,y)\in\overline{A\times B}. Let U\subseteq X be any set such that x\in U. Since \pi_1^{-1}(U) is open in X\times Y it contains some point (x',y')\in A\times B. Then, x'\in U\cap A. It follows that x\in\overline{A}. A similar technique works for Y. \blacksquare

10., 11,. 12 Covered in theorem stated and proved at the beginning of the post.

13.

Problem: Prove that X is Hausdorff if and only if the diagonal \Delta=\left\{(x,x):x\in X\right\} is closed in X\times X with the product topology.

Proof: Suppose that X is Hausdorff then given x\ne y we may find disjoint neighborhoods U,V of them. So, (x,y)\in U\times V and U\times V\cap \Delta=\varnothing since U\cap V=\varnothing. It follows that \Delta is closed.

Conversely, suppose \Delta is closed in X\times X and x,y\in X are distinct. Then, (x,y)\notin\Delta and so there exists a basic open neighborhood U\times V of (x,y) such that U\times V\cap \Delta=\varnothing and so U,V are neighborhoods of x,y in X which are disjoint. For, to suppose they were not disjoint would to assume that z\in U\cap V\implies (z,z)\in U\times V contradicting the assumption that U\times V\cap\Delta=\varnothing. \blacksquare

14.

Problem: In the cofinite topology on \mathbb{R} to what point or points does the sequence \left\{\frac{1}{n}\right\}_{n\in\mathbb{N}} converge to?

Proof: It converges to every point of \mathbb{R}. To see this let x\in\mathbb{R} be arbitrary and let U be any neighborhood of it. Then, \mathbb{R}-U is finite and in particular \left(\mathbb{R}-U\right)\cap K is finite. If it’s empty we’re done, so assume not and let \frac{1}{n_0}=\min\left(\left(\mathbb{R}-U\right)\cap K\right) then clearly for all n\in\mathbb{N} such that n>n_0 we have that \frac{1}{n}<\frac{1}{n_0} and thus \frac{1}{n} is in U. The conclusion follows. \blacksquare

15.

Problem: Prove that the T_1 axiom is equivalent to the condition that for each pair of points x,y\in X there are neighborhoods of each which doesn’t contain the other.

Proof: Suppose that X is T_1 then given distinct x,y\in X the sets X-\{y\},X-\{x\} are obviously neighborhoods of x,y respectively which don’t contain the other.

Conversely, suppose the opposite is true and let y\in X-\{x\} then there is a neighborhood U of it such that x\notin U\implies U\subseteq X-\{x\} and thus X-\{x\} is open and \{x\} is therefore closed. \blacksquare

16.

Problem: Consider the five topologies on \mathbb{R} given in exercise 7 of section 13 (my section 2).

a) Deterime the closure of K under each of these topologies.

b) Which of these topologies are Hausdorff? Which are T_1

Proof:


a) As a reminder the topologies are

\mathfrak{J}_1=\text{usual topology}

\mathfrak{J}_2=\text{topology on }\mathbb{R}_K

\mathfrak{J}_3=\text{cofinite topology}

\mathfrak{J}_4=\text{upper limit topology}

\mathfrak{J}_5=\text{left ray topology}

For the first one we easily see that \overline{K}=K\cup\{0\}.

For the second one we can see that \overline{K}=K. To see this note that \displaystyle \bigcup_{a<b}(a,b)-K=\mathbb{R}-K is open by definition and thus K being the complement of it is closed. Thus, \overline{K}=K

For the third one \overline{K}=\mathbb{R}. To see this note that we in a sense already proved this in 14, but for any x\in\mathbb{R} and any neighborhood N of it we have that N=\mathbb{R}-\{x_1,\cdots,x_n\} and thus if \displaystyle \frac{1}{n_0}=\min\left(\{x_1,\cdots,x_n\}\cap K\right) (assuming it’s non-empty) we see that n>n_0\implies \frac{1}{n}\notin\{x_1,\cdots,x_n\}\implies \frac{1}{n}\in N. Thus, given any point of \mathbb{R} and any neighborhood N of it we have that N\cap K is infinite, and thus clearly x\in\overline{K}. The conclusion follows from that.

For the fourth topology we note that \displaystyle \mathbb{R}-K=(-\infty,0]\cup\bigcup_{n\in\mathbb{N}}\left(\frac{1}{n+1},\frac{1}{n}\right)\cup(1,\infty) which is open in \mathbb{R} with the upper limit topology. Remember that \displaystyle (a,b)=\bigcup_{a<c<b}(a,c]

For the last one \overline{K}=[0,\infty). Clearly K\subseteq[0,\infty) and since [0,\infty) is closed and \overline{K} is the intersection of all closed supersets of K it follows that \overline{K}\subseteq[0,\infty). Now, suppose that x\notin[0,\infty) then x\in(-\infty,0) which is a neighborhood of x which doesn’t intersect K and thus x\notin\overline{K}. So, [0,\infty)\subseteq\overline{K}. The conclusion follows.

b)

Lemma: If X is Hausdorff, then it is T_1

Proof: Let x\in X and y\in X-\{x\} by assumption there exists disjoint neighborhoods U,V of x,y respectively and so clearly y\in V\subseteq X-\{x\} and thus X-\{x\} is open and so \{x\} is closed. \blacksquare

\mathfrak{J}_1: Clearly the first topology is Hausdorff since it is an order topology and we proved in the initial post that all order topologies are Hausdorff.

Lemma: Let X be a set and \mathfrak{J},\mathfrak{J}' two topologies on X such that \mathfrak{J}' is finer than \mathfrak{J}. Then, if X is Hausdorff with the \mathfrak{J} topology it is Hausdorff with the \mathfrak{J}' topology.

Proof: Clearly if x,y\in X are distinct we may find disjoint neighborhoods U,V\in\mathfrak{J} of them in the topology given by \mathfrak{J} and thus the same neighborhoods work in consideration of the topology given by \mathfrak{J}'. \blacksquare

\mathfrak{J}: From this lemma it follows that \mathbb{R}_K having a finer topology than \mathbb{R} with the usual topology is Hausdorff

\mathfrak{J}_3: The cofinite topology is T_1 but not Hausdorff. To see that it’s T_1 it suffices to prove that \{x\} is closed for any x\in\mathbb{R}. But, this is trivial since \mathbb{R}-\{x\} being the complement of a finite set is open, thus \{x\} closed. But, it is not Hausdorff since it cannot support two disjoint, non-empty open sets. To see this suppose that U is open in \mathbb{R} with the cofinite topology then \mathbb{R}-U is finite and since a disjoint set V would have to be a subset of \mathbb{R}-U it follows that V is finite and thus it’s complement not finite. Thus, V is not open.

\mathfrak{J}_4: Once again this topology finer than that of \mathbb{R} with the usual topology since \displaystyle (a,b)=\bigcup_{c<b}(a,c]

\mathfrak{J}_5: This isn’t even T_1. To see this we must merely note that if U is any set containing 1 we must have that there is some basic open set (-\infty,\alpha),\text{ }\alpha>1 such that 1\in(-\infty,\alpha)\subseteq U, but this means that 0\in U. So, there does not exist a neighborhood of 1 which does not contain 0.

17.

Problem: Consider the lower limit topology on \mathbb{R} and the topology given by the basis \mathcal{C}=\left\{[a,b):a<b,\text{ }a,b\in\mathbb{Q}\right\}. Determine the closure of the intervals A=(0,\sqrt{2}) and B=(\sqrt{2},3) in these two topologies.

Proof: We first prove more generally that if (a,b)\subseteq\mathbb{R}_\ell then \overline{(a,b)}=[a,b). To see this we first note that [a,b) is open since \mathbb{R}-[a,b)=(-\infty,a)\cup[b,\infty) which we claim is open. To see this we note that \displaystyle (-\infty,a)\cup[b,\infty)=\bigcup_{x<a}[x,a)\cup\bigcup_{y>b}[b,y). So, since \overline{(a,b)} is the intersection of all closed supersets of (a,b) and [a,b)\supseteq(a,b) is closed we see that \overline{(a,b)}\subseteq[a,b). So, we finish the argument by showing that \overline{(a,b)}\ne(a,b). To see this we show that a is a limit point for (a,b) from where the conclusion will follow. So, let N be any neighborhood of a, then we may find some basic open neighborhood [\alpha,\beta) such that a\in[\alpha,\beta)\subseteq N, but clearly [\alpha,\beta)\cap(a,b) contains infinitely many points from where it follows that a is a limit point of (a,b). From this we may conclude for \mathbb{R}_\ell that \overline{(0,\sqrt{2})}=[0,\sqrt{2}) and \overline{(\sqrt{2},3)}=[\sqrt{2},3)

We now claim that in the topology generated by \mathcal{C} that \overline{(0,\sqrt{2})}=[0,\sqrt{2}] and \overline{(\sqrt{2},3)}=[\sqrt{2},3). More generall, let us prove that in this topology

\overline{(a,b)}=\begin{cases}[a,b]\quad\text{if}\quad b\notin\mathbb{Q}\\ [a,b)\quad\text{if}\quad b\in\mathbb{Q}\end{cases}

Clearly if \alpha<a then choosing some rational number q<\alpha and some rational number \alpha<p<a then \alpha\in[q,p) and [q,p)\cap(a,b)=\varnothing so that \alpha\notin\overline{(a,b)}. Also, if \alpha>b then choosing some p\in\mathbb{Q} such that b<p<\alpha we see that \alpha\in[p,\alpha+1) and [p,\alpha+1)\cap(a,b)=\varnothing. Thus, the only possibilities for \overline{(a,b)} are (a,b),[a,b),[a,b]. But, just as before if N is any neighborhood a we may find some open basic neighborhood $latex [p,q)$ such that a\in[p,q)\subseteq N but clearly [p,q)\cap(a,b)\ne\varnothing and thus since N was arbitrary it follows that a\in\overline{(a,b)}.

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

The conclusion follows. \blacksquare

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

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

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

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

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

Proof:

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

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

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

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

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

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

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May 22, 2010 - Posted by | Fun Problems, Munkres, Topology, Uncategorized | , , , ,

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