Setup for question:

Let $A$ bet a set. Let $\tau_2$ and $\tau_1$ be two topologies on $A$.

Statements to prove that are equivalent:

(I) $\tau_2 \subseteq \tau_1$ (i.e. $\tau_1$ is finer than $\tau_2$);

(II) for every $a\in A$ and $U \in \tau_2$ with $a\in U$ there exists $V \in \tau_1$ such that $a \in V \subseteq U$; and

(III) for every $a\in A$ and $U \in \tau_2$ with $a\in U$ there is a finite set $F \subseteq \tau_1$ such that $a \in \bigcap F \subseteq U$.


(I) $\implies$ (II):

Suppose that $\tau_1$ is finer than $\tau_2$, i.e. $\tau_2 \subseteq \tau_1$. Let $a \in A$ and let $U \in \tau_2$. Then $U\in \tau_1$. If we set $U=V$ then $a\in V \subseteq U$.

(II) $\implies$ (I)

Suppose that for every $a\in A$ and $U \in \tau_2$ with $a\in U$ there exists $V \in \tau_1$ such that $a \in V \subseteq U$.

Let $B \in \tau_2$. For each point $a \in B$ there's a guaranteed existence of a set $S_a \in \tau_1$ such that $a \in S_a \subseteq B$. Indeed:

$$ B = \bigcup_{a \in B}S_a.$$

Since arbritary unions of $\tau_1$-open sets are $\tau_1$-open, the above equality shows that $B$ is $\tau_1$ open, i.e. $ B \in \tau_1$. Therefore, $\tau_2 \subseteq \tau_1$.

Help needed for (I) $\iff$ (III) OR (II) $\iff$ (III).

  • $\begingroup$ As for the proof of $(I) \iff (III)$, it is very straightforward if you prove $(II) \iff (III)$ [and then use $(I) \iff (II)$]. $\endgroup$ – Daniel Fischer Aug 15 '18 at 13:04
  • $\begingroup$ @DanielFischer Yes you're correct on both typos and they have been fixed. Any tips on how to start it at least? Also isn't it redudant to do $(I) \iff (III)$ if I do $(II) \iff (III)$ considering the fact that $(I) \iff (II)$ has already been proven? $\endgroup$ – James Bundster Aug 15 '18 at 13:05
  • $\begingroup$ Yes, $(I) \iff (III)$ is implied by $(I) \iff (II)$ and $(II) \iff (III)$. You added the "or $(II) \iff (III)$" while I was writing my comment, I hadn't seen that. The $(II) \implies (III)$ direction is immediate from the remark that singleton sets are finite. $\endgroup$ – Daniel Fischer Aug 15 '18 at 13:10
  • $\begingroup$ @DanielFischer Are you saying $V$ is a singleton set? If yes, why? $\endgroup$ – James Bundster Aug 15 '18 at 13:15
  • $\begingroup$ No, $V$ usually isn't a singleton set. But $\{V\}$ is. $\endgroup$ – Daniel Fischer Aug 15 '18 at 13:21

Your proof of $(I) \iff (II)$ is correct. To prove $(II) \implies (III)$ we can take $F = \{V\}$. Then $F$ is clearly a finite subset of $\tau_1$, and $$a \in \bigcap F = \bigcap \{V\} = V \subseteq U\,.$$

For the other direction: By definition of a topology, $V := \bigcap F$ is a member of $\tau_1$, and by assumption on $F$ it satisfies the condition in $(II)$.


We show:

$I)\Rightarrow II)\Rightarrow III)\Rightarrow I)$

$I)\Rightarrow II):$

Let $a\in A$ be arbitrary and $U\in\tau_2$ with $a\in U$. Since $\tau_2\subseteq \tau_1$ it is $U\in\tau_1$ and we can choose $V=U\ni a$.

$II)\Rightarrow III):$

Let $a\in A$ be arbitrary and $U\in\tau_2$ with $a\in U$. We have to show, that it exists a finite set $\mathcal{F}\subseteq\tau_1$, such that $a\in\bigcap_{F\in\mathcal{F}} F\subseteq U$.

We know from the assumption, that it exists $V\in\tau_1$ with $a\in V\subseteq U$. Choose $\mathcal{F}=\{V\}$, which is a finite set and $\bigcap_{F\in\{V\}} F=V\subseteq U$.

$III)\Rightarrow I):$

Let $U\in\tau_2$. We have to show, that $U\in\tau_1$.

Let $a\in U$ be arbitrary. Then exists a finite set $\mathcal{F}_a\subseteq\tau_1$ with $a\in\underbrace{\bigcap_{F\in\mathcal{F}_a}}_{\in\tau_1} F\subseteq U$.

[It is $\bigcap_{F\in\mathcal{F}_a} F\in\tau_1$ because of the definition of a topology. Which states, that the intersection of finite open sets in $\tau_1$ is again in $\tau_1$ and $\mathcal{F}_a$ just contains finite elements, so we take a finite interesection!]

Then is $\underbrace{\bigcup_{a\in U} \mathcal{F}_a}_{\in\tau_1}=U\in\tau_1$

[It is $\underbrace{\bigcup_{a\in U} \mathcal{F}_a}\in\tau_1$, because we know $\mathcal{F}_a\in\tau_1$ for every $a\in U$. From the definition of a topology we know, that an arbitrary union of sets in $\tau_1$ is again in $\tau_1$!]

Which concludes the proof.


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