# Union of topologies as subbase of a topology is the weakest topology that is also stronger than each of summed topologies

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?

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.
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 $$\tau$$.