# Abstract Nonsense

## Munkres Chapter two Section 12 & 13: Topological Spaces and Bases

Point of post: This is the solutions to Munkres Chapter two Section 12 as the heading indicates.

1.

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

Proof: So, we know that for each $x\in A$ there exists some open $U_x$ such that $x\in U_x\subseteq$. Clearly

$\displaystyle A=\bigcup_{x\in A}U_x$

and so $A$ is the union of open sets, and thus open.

2.

Problem: Consider the nine topologies on the set $X=\{a,b,c\}$ indicated in Example 1 of section 12. Compare them.

Proof: This is incredibly laborious to type out, and really not that difficult, so we omit it.

3.

Problem: Show that the collection $\mathfrak{J}_c$ given in example 4 of section 12 is a topology on the set $X$. Is the collection

$\mathfrak{J}_\infty=\left\{U:X-U\text{ is infinite or empty or all of }X\right\}$

a topology on $X$?

Proof: This $\mathfrak{J}_c$ is just the cocountable topology. We can see that $\varnothing\in\mathfrak{J}_c$ since it’s complement is all of $X$. Also, $X\in\mathfrak{J}_c$ since $X-X=\varnothing$ which is finite. Also, we notice that if $\left\{U_{\alpha}\right\}_{\alpha\in\mathcal{A}}$ is a nonempty subset of $\mathfrak{J}_c$ that

$\displaystyle X-\bigcup_{\alpha\in\mathcal{A}}U_\alpha=\bigcap_{\alpha\in\mathcal{A}}\left(X-U_{\alpha}\right)\subseteq U_{\alpha_0},\text{ }\alpha_0\in\mathcal{A}$

and so the complement of $\displaystyle \bigcup_{\alpha\in\mathcal{A}}U_\alpha$ is a subset of a countable set and thus countable. Similarly, if $U_1,\cdots,U_n\in\mathfrak{J}_c$ then

$X-\left(U_1\cup\cdots\cup U_n\right)=\left(X-U_1\right)\cup\cdots\cup\left(X-U_n\right)$

and so the complement of $U_1\cap\cdots\cap U_n$ is the finite union of countable sets, and thus countable, so $U_1\cap\cdots\cap U_n\in\mathfrak{J}_c$. It follows that $\mathfrak{J}_c$ really is a topology on $X$

Now, $\mathfrak{J}_c$ is not a topology on any infinite set. To see this let $X$ be infinite and let $\varphi:X\to\mathbb{Z}$ be any surjection. Then, we can see that $\varphi^{-1}\left(\{0\}\cup\mathbb{N}\right)$ and $\varphi^{-1}\left(-\mathbb{N}\cup\{0\}\right)$ are in $\mathbb{J}_{\infty}$ but their intersection is $\{0\}$ which is not in $\mathfrak{J}_\infty$

4.

Problem:

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

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

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

Proof:

a) Clearly since $\varnothing,X\in\mathfrak{J}_{\alpha}$ for every $\alpha\in\mathcal{A}$  we have that $\varnothing,X\in\Omega$. Now, if $\left\{U_{\beta}\right\}_{\beta\in\mathcal{B}}\subseteq\Omega$ then $\left\{U_{\beta}\right\}_{\beta\in\mathcal{B}}\subseteq \mathfrak{J}_\alpha$ for every $\alpha\in\mathcal{A}$. But, since each $\mathfrak{J}_{\alpha}$ is a topology it follows that $\displaystyle \bigcup_{\beta\in\mathcal{B}}U_\beta\in\mathfrak{J}_\alpha$ for every $\alpha\in\mathcal{A}$ and so $\displaystyle \bigcup_{\beta\in\mathcal{B}}U_\beta\in\Omega$. Lastly, if $U_1,\cdots,U_n\in\Omega$ we have that $U_1,\cdots,U_n\in\mathfrak{J}_\alpha$ for every $\alpha\in\mathcal{A}$. But, since these are topologies this implies that $U_1\cap\cdots\cap U_n\in\mathfrak{J}_\alpha$ for every $\alpha\in\mathcal{A}$ and so $U_1\cap\cdots\cap U_n\in\Omega$

b) To show that there is a unique topology which contains all the $\mathcal{J}_\alpha$ we define

$\Gamma=\left\{\mathfrak{T}:\mathfrak{T}\text{ is a topology on }X\text{ and }\mathfrak{T}\supseteq\bigcup_{\alpha\in\mathcal{A}}\mathfrak{J}_\alpha\right\}$

We first notice that $\Gamma$ is not empty since $2^X\in\Gamma$. So, let

$\displaystyle \mathfrak{J}=\bigcap_{\mathfrak{T}\in\Gamma}\mathfrak{T}$

Clearly then $\mathfrak{J}$ is a topology which contains all the $\mathfrak{J}_\alpha$, and any other topology which contains them must contain $\mathfrak{J}$. More formally, $\mathfrak{J}$ is a topology on $X$ such that $\mathfrak{J}_\alpha\subseteq\mathfrak{J},\text{ }\forall\alpha\in\mathcal{A}$. Furthermore, given a topology $\mathfrak{K}$ which contains every $\mathfrak{J}_\alpha$ we see that $\mathfrak{K}\in\Gamma$ so that $\mathfrak{J}\subseteq\mathfrak{K}$. Also, if $\mathfrak{J}'$ were another smallest topology containing all the $\mathfrak{J}_\alpha$‘s that both the inclusions $\mathfrak{J}\subseteq\mathfrak{J}'$ and $\mathfrak{J}'\subseteq\mathfrak{J}$ and so $\mathfrak{J}=\mathfrak{J}'$, from where it follows that the smallest such topology is unique.

Using the exact same logic, the smallest topology contained in all of the $\mathfrak{J}_{\alpha}$‘s is $\displaystyle \mathfrak{J}=\bigcap_{\alpha\in\mathcal{A}}\mathfrak{J}_\alpha$.

c) One can verify that $\mathfrak{J}_1\cup\mathfrak{J}_2\cup\{b\}$ is the smallest topology containing both and $\mathfrak{J}_1\cap\mathfrak{J}_2$ is a topology, and thus the smallest contained in both.

5.

Problem: Show that if $\mathfrak{B}$ is a basis for a topology on $X$, then the topology generated by $\mathfrak{B}$ equals the intersection of all topologies on $X$ that contain $\mathfrak{B}$. Show the same is true if $\mathfrak{B}$ is a subbasis.

Proof:Let $T(X)$ be the set of all topologies on $X$ and let

$\displaystyle \mathfrak{J}=\bigcap_{\mathfrak{I}\in T(X):\mathfrak{I}\supseteq\mathfrak{B}}\mathfrak{I}$

and $\mathfrak{T}$ the topology generated by $\mathfrak{B}$. Clearly since $\mathfrak{T}$ is a topology on $X$ which contains $\mathfrak{B}$ we have that $\mathfrak{T}\supseteq\mathfrak{J}$. Conversely, let $U\in\mathfrak{T}$, then $\displaystyle U=\bigcup_{\beta\in\mathcal{B}}S_\beta$ where $\left\{S_\beta\right\}_{\beta\in\mathcal{B}}\subseteq\mathfrak{B}$. But, for any topology $T$ containing $\mathfrak{B}$ we must have that $\displaystyle \bigcup_{\beta\in\mathcal{B}}S_\beta\in T$ (since it’s  closed under arbitrary unions). Thus, $\displaystyle \bigcup_{\beta\in\mathcal{B}}S_\beta$ is contained in all topologies which are supersets of $\mathfrak{B}$, namely it is in their intersection $\mathfrak{J}$. It follows that $\mathfrak{J}=\mathfrak{T}$ as desired.

Now, let $\mathfrak{B}$ be a subbasis and

$\displaystyle \mathfrak{J}=\bigcap_{\mathfrak{I}\in T(X):\mathfrak{I}\supseteq\mathfrak{B}}\mathfrak{I}$

and $\mathfrak{T}$ the topology generated by $\mathfrak{B}$.

Clearly $\mathfrak{T}$ being a topology on $X$ containing $\mathfrak{B}$ guarantees that $\mathfrak{T}\supseteq\mathfrak{J}$. Conversely, if $U\notin\mathfrak{J}$, then $U\notin T$ for some topology $T$ on $X$ which contains $\mathfrak{B}$. But, this then means that $T$ cannot be an arbitrary union of finite intersections of elements of $\mathfrak{B}$, since any topology $T$ containing $\mathfrak{B}$ must contain them. Thus, $U\notin\mathfrak{T}$. It follows that $\mathfrak{T}=\mathfrak{J}$ as required.

6.

Problem: Show that the topologies $\mathbb{R}_\ell$ and $\mathbb{R}_K$ aren’t comparable.

Proof: We claim that $(-1,1)-K\in\mathbb{R}_K$ but $(-1,1)-K\notin\mathbb{R}_\ell$. The fact that $(-1,1)-K\in\mathbb{R}_K$ follows from the definition of the basis. Now, suppose there was a basis element $[\alpha,\beta)\in\mathbb{R}_\ell$ such that $0\in[\alpha,\beta)\subseteq (-1,1)-K$. Clearly $\beta>0$ otherwise $0\notin[\alpha,\beta)$. But, by the Archimedean principle there exists some $n\in\mathbb{N}$ such that $\displaystyle \frac{1}{n}<\beta$, thus $\displaystyle \frac{1}{n}\in[\alpha,\beta)$ but this contradicts that $[\alpha,\beta)\subseteq (-1,1)-K$. It follows that no such $[\alpha,\beta)$ exists.

We now claim that $[0,1)\in\mathbb{R}_\ell$ but $[0,1)\notin\mathbb{R}_K$. To see this suppose that $(\alpha,\beta)-K$ is a basic element for $\mathbb{R}_K$ such that $0\in (\alpha,\beta)-K\subseteq[0,1)$. Now, clearly $\alpha<0$ otherwise $0\notin(a,b)-K$. But, by the Archimedean principle we may choose $\displaystyle n\in\mathbb{N}$ such that $\displaystyle \alpha<\frac{-1}{n}<0$ and so $\displaystyle \frac{-1}{n}\in(\alpha,\beta)$ and thus $\displaystyle \frac{-1}{n}\in(\alpha,\beta)-K$, but this contradicts that $(\alpha,\beta)-K\subseteq[0,1)$

7.

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

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

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

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

$\mathfrak{J}_4=\text{ the upper limit topology, the topology with sets of the form }(\alpha,\beta]\text{ as a basis}$

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

Determine, for each of these topology, which of the others contain it.

Proof:We know from the book that $\mathfrak{J}_1\subseteq\mathfrak{J}_2$.

Next, we claim that $\mathfrak{J}_1\supseteq\mathfrak{J}_3$. To see this let $U\in\mathfrak{J}_3$, then $X-U=\{x_1,\cdots,x_n\}$ where we may assume WLOG that $x_1<\cdots. So then, $U=(-\infty,x_1)\cup(x_1,x_2)\cup\cdots\cup(x_n,\infty)$ which is the finite union of open sets in $\mathbb{R}$, thus $U\in\mathfrak{J}_1$

We claim that $\mathfrak{J}_1\subseteq\mathfrak{J}_4$. To see this we must merely note that

$\displaystyle (\alpha,\beta)=\bigcup_{\gamma<\beta}(\alpha,\gamma]$

so that every basic set, and thus every set, in $\mathfrak{J}_1$ may be written as the union of basic open sets in $\mathfrak{J}_4$ from where the conclusion follows.

It’s also clear that $\mathfrak{J}_1\supseteq\mathfrak{J}_5$ since

$(-\infty,\alpha)=\bigcup_{\beta<\alpha}(\beta,\alpha)$

and so every element of $\mathfrak{J}_5$ is the union of open sets in $\mathfrak{J}_1$ and so $\mathfrak{J}_1\supseteq\mathfrak{J}_5$

Now, clearly $\mathfrak{J}_3\supseteq\mathfrak{J}_2\supseteq\mathfrak{J_1}$ and $\mathfrak{J}_3\supseteq\mathfrak{J}_1\supseteq\mathfrak{J}_5$ and so it suffices to check how it relates to $\mathfrak{J}_4$. We claim that they aren’t comparable. To see this we note that $(0,1]$ is open in the upper limit topology, but any set of the form $(a,b)-K$ must have $b>1$ and so it will entrap points not in $(0,1]$ that aren’t removed by $K$. Also, using a very similar argument one can show that $(-1,1)-K$ is not open in the upper limit topology.

The last inclusion we must check is how $\mathfrak{J}_3$ relates to $\mathfrak{J}_5$ and we claim they aren’t comparable. Clearly $(-\infty,0)\notin\mathfrak{J}_2$ sine it’s complement is infinite but it’s in $\mathfrak{J}_5$. Also, $(-\infty,0)\cup(0,\infty)\in\mathfrak{J}_3$ but it’s not in $\mathfrak{J}_5$ since any basic set containing $1$, must contain $0$

8.

a) Apply lemma 13.2 to show that the countable collection $\mathfrak{B}=\left\{(a,b):a  is a basis that generates the standard topology on $\mathbb{R}$.

b) Show that the collection $\mathfrak{C}=\left\{[a,b):a generates a topology different than $\mathbb{R}_\ell$

Proof:

a) Clearly since $\mathfrak{B}$ is contained is the usual basis for $\mathbb{R}$ the topology generated by it is coarser than the usual one. Conversely, let $(a,b)$ be any basic open set in the usual topology on $\mathbb{R}$ and $x\in(a,b)$ arbitrary. We know from the structure of the real numbers that there is some $a'\in\mathbb{Q}$ such that $a and some $b\in\mathbb{Q}$ such that $x. Thus, $(a',b')\in\mathfrak{B}$ and $x\in(a',b')\subseteq(a,b)$ from where it follows that the topology generated by $\mathfrak{B}$ is finer than the usual topology. Thus, it follows that the usual topology and the one generated by $\mathfrak{B}$ are equal.

b) We claim that $[\sqrt{2},3)$ is not in the topology generated by $\mathfrak{C}$. To see this suppose that $[\sqrt{2},3)$ could be written as the union of sets of the form $[\alpha,\beta),\text{ }\alpha,\beta\in\mathbb{Q}$, say $\displaystyle \bigcup_{\gamma\in\mathcal{C}}[\alpha_\gamma,\beta_\gamma)$. We see that since $[\alpha_\gamma,\beta_\gamma)\subseteq[\sqrt{2},3)$ that $\alpha_\gamma\geqslant\sqrt{2}$ and since it’s rational we may conclude that $\alpha_\gamma>\sqrt{2}$. But, this means that for every $\gamma\in\mathcal{C}$ we have that $\sqrt{2}\notin[\alpha_\gamma,\beta_\gamma)$ and thus $\displaystyle \sqrt{2}\notin\bigcup_{\gamma\in\mathcal{C}}[\alpha_\gamma,\beta_\gamma)$ contradictory to our assumption.