# Abstract Nonsense

## Munkres Chapter 2 Section 17

Theorem 17.11

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

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

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

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

1.

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

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

2.

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

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

3.

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

Proof: This follows immediately from question 9. $\blacksquare$

4.

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

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

5.

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

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

6.

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

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

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

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

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

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

$\blacksquare$

7.

Problem: Criticize proof (see book).

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

8.

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

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

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

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

Proof:

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

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

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

9.

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

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

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

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

13.

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

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

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

14.

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

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

15.

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

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

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

16.

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

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

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

Proof:

a) As a reminder the topologies are

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

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

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

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

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

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

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

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

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

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

b)

Lemma: If $X$ is Hausdorff, then it is $T_1$

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

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

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

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

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

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

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

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

17.

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

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

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

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

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

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

The conclusion follows. $\blacksquare$

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

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

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

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

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

Proof:

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

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

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

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

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

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