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Let $\mathcal{T}_\alpha$ be topologies on space X for some $\alpha$ acting as arbitrary index. Define $V_\alpha \mathcal{T}_\alpha$ as topology with $\bigcup_\alpha\mathcal{T}_\alpha$ as its subbase. Prove that $V_\alpha \mathcal{T}_\alpha$ is the weakest topology that is also stronger than every $\mathcal{T}_\alpha$.

Proof:

  1. $V_\alpha \mathcal{T}_\alpha$ is stronger than every $\mathcal{T}_\alpha$

To prove that it is sufficient to show that for a fixed $\alpha$ we have $T \in \mathcal{T}_\alpha \implies T \in V_\alpha \mathcal{T}_\alpha$. Knowing that subbase of $V_\alpha \mathcal{T}_\alpha$ is $\bigcup_\alpha\mathcal{T}_\alpha$ and both $X \in \bigcup_\alpha\mathcal{T}_\alpha$ and $T \in \bigcup_\alpha\mathcal{T}_\alpha$ for every $T \in \mathcal{T}_\alpha$, then we may say $T = X \cap T$. Now, $X \cap T$ is a finite intersection so it is in $V_\alpha \mathcal{T}_\alpha$ by the definition of subbase. We thus conclude $V_\alpha \mathcal{T}_\alpha$ is stronger than $\mathcal{T}_\alpha$ for every $\alpha$.

  1. $V_\alpha \mathcal{T}_\alpha$ is the weakest topology with that property

Here I got stuck and appreciate any help.

Also, is the proof number 1 correct?

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I'll call your $V_\alpha \mathcal{T}_\alpha$ topology just $\mathcal{T}$, with subbase $\mathcal{S} := \bigcup_\alpha \mathcal{T}_\alpha$

For each $\alpha$ we have $\mathcal{T}_\alpha \subseteq \mathcal{T}$ because $\mathcal{T}_\alpha \subseteq \mathcal{S} \subseteq \mathcal{T}$. This takes care of 1.

Now suppose $\mathcal{T'}$ is any topology on $X$ that is stronger than every $\mathcal{T}_\alpha$. We have to show that $\mathcal{T} \subseteq \mathcal{T}'$ to show $\mathcal{T}$ is weakest. This is quite easy: let $O \in \mathcal{T}$ and let $x \in O$. Because the finite intersections of members of $\mathcal{S}$ is a base for $\mathcal{T}$, we have finitely many $S_1, S_2, \ldots, S_n \in \mathcal{S}$ such that $x \in (S_1 \cap S_2 \cap \ldots \cap S_n) \subseteq O$. By definition of $\mathcal{T}$ all $S_i$ are members of $\bigcup_\alpha \mathcal{T}_\alpha$ and so members of $\mathcal{T}'$ (because $\mathcal{T}'$ contains all $\mathcal{T}_\alpha$ by assumption!) and so, as topologies are closed under finite intersections, $(S_1 \cap S_2 \cap \ldots \cap S_n) \in \mathcal{T}'$, so $x$ is an interior point of $O$ w.r.t. $\mathcal{T}'$ and as $x \in O$ was arbitrary, $O \in \mathcal{T}'$, which is what we had to show.

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Show that if $\tau$ is stronger than any $\mathcal{T}_\alpha$, then $\tau$ contains $\bigcup_\alpha\mathcal{T}_\alpha$. The basis generated by $\bigcup_\alpha\mathcal{T}_\alpha$ would then be contained in $\tau$, and then by the definition of a basis, every open set in $V_\alpha \mathcal{T}_\alpha$ would be in $\tau$.

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