# Abstract Nonsense

## Munkres Chapter 2 Section 1

This begins a substantial effort to complete all of (except the first chapter) the problems in James Munkre’s Topology I have chosen to start at chapter 2 considering the first chapter is nothing but prerequisites. Thus, without a minute to spare let us being.

1.

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

Proof: We claim that $\displaystyle A=\bigcup_{x\in A}U_x=\overset{\text{def.}}{=}\Omega$ but this is obvious since for each $x\in A$ we have that $x\in U_x\subseteq\Omega$. Conversely, since each $U_x\subseteq A$ we have that the union of all of them is contained in $A$, namely $\Omega\subseteq A$. Thus, $A$ is the union of open sets and thus open. $\blacksquare$

2.

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

Solution: This is simple.

3.

Problem: Show that given a set $X$ if we denote $\mathfrak{J}$ to be cocountable topology (a set is open if it’s complement is countable or the full space) that $\left(X,\mathfrak{J}\right)$ is a topological space. Is it still a topological space if we let $\mathfrak{J}=\left\{U\in\mathcal{P}(X):X-U\text{ is infinite, empty, or all of }X\right\}$?

Proof: Clearly for the first part $\varnothing,X\in\mathfrak{J}$. Now, if $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ is a collection of open sets then we note that $X-U_\alpha$ is finite and thus $\displaystyle X-\bigcup_{\alpha\in\mathcal{A}}U_\alpha=\bigcap_{\alpha\in\mathcal{A}}\left(X-U_\alpha\right)\subseteq U_{\alpha_0}$ for any $\alpha_0$. Thus, $\displaystyle X-\bigcup_{\alpha\in\mathcal{A}}U_\alpha$ is finite and thus in $\mathfrak{J}$. Now, if $U_1,\cdots,U_n\in\mathfrak{J}$ we have that $X-(U_1\cap\cdots\cap U_n)=(X-U_1)\cup\cdots\cup (X-U_n)$ and thus $X-(U_1\cap\cdots\cap U_n)$ is the finite union of finite sets and thus finite, so $U_1\cap\cdots\cap U_n\in\mathfrak{J}$.

If we redefine the topology as described it is not necessarily a topology. For example, give $\mathbb{N}$ that topology and note that $\left\{\{n\}\right\}_{n\in\mathbb{N}-\{1\}}$ is a collection of elements of $\mathfrak{J}$ but $\displaystyle \bigcup_{n\in\mathbb{N}-\{1\}}\{n\}=\mathbb{N}-\{1\}$ whose complement is neither infinite, empty, or the full space. Thus, this isn’t a topology.

4.

Problem:

a) If $\left\{\mathfrak{J}_\alpha\right\}_{\alpha\in\mathcal{A}}$ is a family of topologies on $X$, show that $\displaystyle \bigcap_{\alpha\in\mathcal{A}}\mathfrak{J}_\alpha$ is a topology on $X$. Is $\displaystyle \bigcup_{\alpha\in\mathcal{A}}\mathfrak{J}_\alpha$?

b) Let $\left\{\mathfrak{J}\right\}_\alpha$ be a family of topologies on $X$. Show that there is unique topology on $X$ containing all the collections $\mathfrak{J}_\alpha$, and a unique largest topology contained in all of the $\mathfrak{J}_\alpha$.

c) If $X=\{a,b,c\}$, let $\mathfrak{J}_1=\{\varnothing,X,\{a\},\{a,b\}\}$ and $\mathfrak{J}_2=\left\{\varnothing,X,\{a\},\{b,c\}\right\}$. Find the smallest topology containing $\mathfrak{J}_1,\mathfrak{J}_2$ and the larges topology contained in $\mathfrak{J}_1,\mathfrak{J}_2$.

Proof:

a) Let $\displaystyle \Omega\overset{\text{def.}}{=}\bigcap_{\alpha\in\mathcal{A}}\mathfrak{J}_\alpha$ and let $\left\{U_\beta\right\}_{\beta\in\mathcal{B}}\subseteq\Omega$ be arbitrary. Then, by assumption we have that $\left\{U_\beta\right\}_{\beta\in\mathcal{B}}\subseteq\mathfrak{J}_\alpha$ for every $\alpha\in\mathcal{A}$ but since this was assumed to be a topology we have that for every $\alpha\in\mathcal{A}$ that $\displaystyle \bigcup_{\beta\in\mathcal{B}}U_\beta\subseteq\mathfrak{J}_\alpha$ and thus $\displaystyle \bigcup_{\beta\in\mathcal{B}}U_\beta\in\Omega$. Now, if $\{U_1,\cdots,U_n\}\subseteq\Omega$ we must have that $\{U_1,\cdots,U_n\}\subseteq\mathfrak{J}_\alpha$ for every $\alpha$ and thus $U_1\cap\cdots\cap U_n\in\mathfrak{J}_\alpha$ for every $\alpha\in\mathcal{A}$ and thus $U_1\cap\cdots\cap U_n\in\Omega$. Thus, noting that for every $\alpha\in\mathcal{A}$ we must have that $\varnothing,X\in\mathfrak{J}_\alpha$ the conclusion follows.

No, the union of two topologies needn’t be a topology. Let $\mathfrak{J}_1,\mathfrak{J}_2$ be defined as in part c and note that $\mathfrak{J}_1\cup\mathfrak{J}_2=\left\{\varnothing,X,\{a\},\{a,b\},\{b,c\}\right\}$ but $\{a,b\}\cap\{b,c\}=\{c\}\notin\mathfrak{J}_1\cup\mathfrak{J}_2$

b) This follows immediately from part a. For the first part let $\Omega=\left\{\mathfrak{J}\in\text{Top }X:\mathfrak{J}_\alpha\subseteq\mathfrak{J},\text{ for all }\alpha\in\mathcal{A}\right\}$ (where $\text{Top }X$ is the set of all topologies) and let $\mathfrak{T}=\bigcap_{\mathfrak{J}\in\Omega}\mathfrak{J}$, this clearly satisfies the conditions. For the second one merely take $\displaystyle \bigcap_{\alpha\in\mathcal{A}}\mathfrak{J}_\alpha$

c) We (as pointed out very poignantly in the previous problem) just take the union of the two topologies. So, we claim that the desired smallest topology is in fact $\left\{X,\varnothing,\{a\},\{b\},\{a,b\},\{a,c\}\right\}$. But this is just computation and we leave it to the reader. For the second part just intersect the two topologies.

5.

Problem: Show that if $\mathcal{A}$ is a base for a topology on $X$, then the topology generated by $\mathcal{A}$ equals the intersection of all topologies on $X$ which contain $\mathcal{A}$. Prove the same if $\mathcal{A}$ is a subbase

Proof: Let $\Omega$ be the intersection of all topologies on $X$ which contain $\mathcal{A}$ and $\mathfrak{J}_g$ the topology generated by $\mathcal{A}$. Clearly $\Omega\subseteq\mathfrak{J}_g$ since $\mathfrak{J}_g$ is itself a topology on $X$ containing $\mathcal{A}$. Conversely, let $U\in\mathfrak{J}_g$ we show that $U\in\mathfrak{J}$ where $\mathfrak{J}$ is any topology on $X$ containing $X$. But, this is obvious since $\displaystyle U=\bigcup_{B\in\mathcal{B}}B$ for some $\mathcal{B}\subseteq\mathcal{A}$ and thus $U$ is the union of open sets in $\mathfrak{J}$ and thus in $\mathfrak{J}$. The conclusion follows.

Next, let $\Omega,\mathfrak{J}_g$ be above except now $\mathcal{A}$ is a subbase. For the same reasons as above we have that $\Omega\subseteq\mathfrak{J}_g$. Conversely, for any topology $\mathfrak{J}$ containing $\mathcal{A}$ we have that if $U\in\mathfrak{J}_g$ then $\displaystyle U=\bigcup_{\beta\in\mathcal{B}}V_\beta$ where each $V_\beta$ is the finite union of elements of $\mathcal{A}$. But, by construction it follows that each $V_\beta$ is open (it is the finite intersection of open sets in $\mathfrak{J}$)  and thus $U$ is the union of open sets in $\mathfrak{J}$ and thus $V\in\mathfrak{J}$. $\blacksquare$

6.

Problem: Show that the topologies on $\mathbb{R}_K$ and the Sorgenfrey line aren’t comparable

Proof: See the last part of the next problem

7.

Problem: Consider the following topologies on$\mathbb{R}$:

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

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

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

$\mathfrak{J}_4=\text{topology having the set set of all }(a,b]\text{ as a base}$

$\mathfrak{J}_5=\text{the topology having all sets }(-\infty,a)\text{ as a base}$

For each determine which of the others contain it.

Solution:

$\mathfrak{J}_1\subseteq\mathfrak{J}_2$:Clearly we have that $\mathfrak{J}_1\subseteq\mathfrak{J}_2$ since the defining open base for $\mathfrak{J}_1$ is contained entirely in $\mathfrak{J}_2$.

$\mathfrak{J}_1\supseteq\mathfrak{J}_3$: But, $\mathfrak{J}_1\not\subseteq\mathfrak{J}_3$ since $(0,1)\in\mathfrak{J}_1$ but $\mathbb{R}-(0,1)\simeq\mathbb{R}$ and thus not in $\mathfrak{J}_3$. Now, to prove the inclusion indicated we know that for each open set $U$ in the cofinite topology we have that $U=\mathbb{R}-\{x_1,\cdots,x_n\}$ and so if we assume WLOG that $x_1<\cdots then

$\displaystyle U=\bigcup_{ax_n}(x_n,b)$

And thus $U$ is open in the usual topology.

$\mathfrak{J}_1\subseteq\mathfrak{J}_4$: But, $\mathfrak{J}_1\subseteq\mathfrak{J}_4$. To see this it suffices to show that $(a,b)$ is open in $\mathfrak{J}_4$ since this is a base for $\mathfrak{J}_1$. But, to see this we must merely note that $\displaystyle (a,b)=\bigcup_{c.

$\mathfrak{J}_1\supseteq\mathfrak{J}_5$: Lastly, $\mathfrak{J}_1\supseteq\mathfrak{J}_5$. To see this we must merely note that $\displaystyle (-\infty,b)=\bigcup_{a.

For $\mathfrak{J}_2$ the result is obvious except possibly how it relates to $\mathfrak{J}_3$. But, in fact they aren’t comparable. To see this we first show that $(0,1]$ is not open in $\mathbb{R}_K$. To see this we show it can’t be written as the union of sets of the form $(a,b)$ and $(c,d)-K$ but it clearly suffices to do this for the latter sets. Now, to see that $(0,1]$ can’t be written as the union of sets of the form $(a,b)$ we recall from basic real number topology that $(0,1]$ is not open ($1$ is not an interior point) and thus it can’t be written as the union of intervals or it would be open in the usual topology on $\mathbb{R}$. Also, consider $(-1,1)-K$.

8.

Problem: Show that the countable collection $\mathfrak{B}=\left\{(a,b):a is a base for the usual topology on $\mathbb{R}$

Proof: This follows from the density of $\mathbb{Q}$. It suffices to show that given any $x\in\mathbb{R}$ and any $(a,b)\supseteq\{x\}$ that there is some element $(p,q)\in\mathfrak{B}$ such that $x\in(p,q)\subseteq(a,b)$. But, from basic analysis we know there is some rational number $q$ such that $a and similarly there is some $p\in\mathbb{Q}$ such that $x. Thus, $x\in(p,q)\subseteq(a,b)$. The conclusion follows. $\blacksquare$.

9.

Problem: Show that the collection $\mathcal{C}=\left\{[a,b):a generates a different topology from the one on $\mathbb{R}_\ell$ (the lower limit topology)\$.

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