Help with problem of basis of topology

Question

Again, I'm still in trouble with the problems of bases related.

The problem is the following.

Let $(X, \tau_1), (X, \tau_2)$ be topological spaces, and $B_1, B_2$ bases for $\tau_1 \text{and } \tau_2$ respectively. Prove that $\tau_2 \subset \tau_1$ if and only if $\forall ( b_2 \in B_2 \text{ and } x \in b_2 )$ there exists a $b_1 \in B_1$ such that $x \in b_1 \subset b_2$.

Proof. I have done one way.

If $\forall ( b_2 \in B_2 \text{ and } x \in b_2 )$ there exists a $b_1 \in B_1$ such that $x \in b_1 \subset b_2$ then $\tau_2 \subset \tau_1$.

Let $O \in \tau_2$, because $B_2$ is a base for it, we have that there exists $b_2 \in B_2$ such that $b_2 \subset O$, but by the hypothesis If such set exists then there has to be $b_1 \in B_1$ such that $b_1 \subset b_2 \subset O$, moreover, it implies that there is an open set $O$ that $b_1 \subset O$.

Now, since $B_1$ is a basis for $\tau_1$ It follows that $\forall (O' \in \tau_1 )$ there exists $b_1 \in B_1 $ s.t. $b_1 \subset O'$ So because of this and the previous result:

It is necessary that $O_{\tau_2} \in \tau_1 $, hence, $\tau_2 \subset \tau_1$.

I know that in the statements there are points which should be handled carefully, but since what I'm trying to prove does not involve the points x explicitly, I think they aren't necessary. Or should I include them?

Is this correct?

Now to prove the other way.

If $\tau_2 \subset \tau_1$ then $\forall ( b_2 \in B_2 \text{ and } x \in b_2 )$ there exists a $b_1 \in B_1$ such that $x \in b_1 \subset b_2$

Let $b_2 \in B_2, x\in b_2$ But because $B_2$ is a basis for $\tau_2$ there is an open set $O \in \tau_2$ such that $b_2 \subset O$, now using th hypothesis, since this is true for a open set in $\tau_2$, it should be true for an open set in $\tau_1$ ,so $\color{red}{ x \in b_2 \subset O_{\tau_1} }$ In this case I end up with $b_2 \subset O_{\tau_1}$ and $b_1 \subset O_{\tau_1}$, but I don't see how to imply that $b_1 \subset b_2$

I still don't know how this is helpful, tho.

Now what I have found is that because $B_1$ es a basis for $\tau_1$, the, $\forall ( O_{\tau_1} , x \in O_{\tau_1} ), \exists b_1 \in B_1 $ s.t. $x \in b_1 \subset O_{\tau_1}$. To prove what is desired I have to options whether $\color{blue} { O_{\tau_1} \subset b_2} $ or $\color{blue} {O_{\tau_1} = b_2}$ .

Because in general, somehow this way is saying that, $B_1$ is a basis for $B_2$, To try this I was thinking of using the topology induced by the basis $\tau_{B_2} = \{ A \subset X: \exists b_2 \in B_2, A=\bigcup b_2\}$ , basically to prove that $b_1 \in \tau_{B_2}$ but I end up with the same problem because if this is true, $b_1 = \bigcup b_2$ and $b_2 \subset \bigcup b_2$.

I've come to a blind point.

Any help is appreciated.

Same, I'm sorry for any mistake of spelling or grammar.

Answer 1

So we want to show that $\tau_2 \subseteq \tau_1$ iff: $$\forall B_2 \in \mathcal{B}_2: \forall x \in B_2: \exists B_1 \in \mathcal{B}_1 : x \in B_1 \subseteq B_2 \text{ (1) }$$

What does (1) say, really? It says exactly that every $B_2 \in \mathcal{B}_2$ is $\tau_1$-open. This holds as for any base $\mathcal{B}$ of any topology $\tau$ we can characterise the open sets by the base: $$O \in \tau \text{ iff } \forall x \in O: \exists B \in \mathcal{B} : x \in B \subseteq O \text{ (2) }$$

And we can do this for every $x \in B_2$ w.r.t. $\mathcal{B}_1$, so $B_2 \in \tau_1$.

So if (1) holds, $\mathcal{B}_2 \subseteq \tau_1$ and so $\tau_2 \subseteq \tau_1$ as open sets in $\tau_2$ are unions of sets from $\mathcal{B}_2$, so unions of sets from $\tau_1$, hence in $\tau_1$.

And if $\tau_2 \subseteq \tau_1$, then it's clear that $\mathcal{B}_2 \subseteq \tau_1$ (as $\mathcal{B}_2 \subseteq \tau_2$ by definition) and this is just what (1) says.

Answer 2

(I). LEMMA. Let $T$ be a topology on $X$ and let $S\subset X$. Then $S\in T$ iff for every $p\in S$ there exists $V(p)\in T$ such that $p\in V(p)\subset S.$

One direction is trivial: If $S\in T$ then let $V(p)=S$ for each $p\in S.$

For the other direction: For $p\in S$ let $W(p)=\{V\in T:p\in V\subset S\}.$ Let $V(p)=\cup W(p).$ We have $W(p)\subset T$ so $V(p)\in T.$

So $\cup \{V(p): p\in S\}\in T.$

And each $V)p)\subset$ S so $\cup \{V(p):p\in S\}\subset S.$

And we have $S=\cup \{\{p\}:p\in S\}\subset \cup \{V(p):p\in S\}\subset S.$ Therefore $S=\cup \{V(p):p\in S\}\in T.$...QED.

(The use of $W(p)$ is to avoid the Axiom of Choice (AC). If we use AC we could simply choose $V(p)$ to be some member of $W(p).$)

(II). In your Q, suppose whenever $p\in b_2\in B_2$ there exists $b_1\in B_1$ with $p\in b_1\subset b_2.$

Then whenever $p\in S\in \tau_2,$ there exists $b_2\in B_2$ such that $p\in b_2\subset S$ (....because $B_2$ is a basis for $\tau_2.$..) and there exists $b_1\in B_1$ such that $p\in b_1\subset b_2\subset S.$

Now $b_1\in \tau_1$ so whenever $p\in S\in \tau_2$ there exists $b_1\in \tau_1$ such that $p\in b_1\subset S.$ That is, every $S\in \tau_2$ satisfies the conditions of the Lemma (...with $T=\tau_1$ and $V(p)=b_1$...) so by the Lemma we have $$S\in\tau_2\implies S\in \tau_1.$$

Remark: I think an easier proof of the non-trivial direction of the Lemma is: Let $S^*=\{t\in T:t\subset S\}$. Now the hypothesis implies that every $p\in S$ belongs to a member of $S^*,$ so $S\subset \cup S^*.$ But every member of $S^*$ is a subset of $S$ so $\cup S^*\subset S.$ Hence $S=\cup S^*\in T.$