# The union of all the open sets in a family of topologies

I'm starting studying topology for the first time and my teacher just wrote this. I just don't understand the last line:

Let $\{\tau_\alpha\}$ be a family of topologies on X. [...] To say that a collection of subsets of X contains all the collections $\tau_\alpha (\alpha\in A)$ is equivalent to saying that this collection contains $\bigcup_{\alpha\in A} \tau_\alpha$. We can see $X\in \bigcup_{\alpha\in A}\tau_\alpha$, then $$\bigcup_{H\in\bigcup_{\alpha\in A}\tau_\alpha} H = X$$

Why is he using the H set?

Is it not $\bigcup_{\alpha\in A}\tau_\alpha = X$?

Thanks!

• the union without the H is a union of sets of subsets of X, while the union over H is a union of subsets of X. Commented Jul 24, 2014 at 20:39
• In the future, you should try and find a more descriptive title, a title that refers to the mathematics of your problem. Commented Jul 24, 2014 at 20:43

The $\tau_{\alpha}$'s are topologies, so collections of subsets of $X$. The union of topologies, in the sense above, is a topology then, and not a subset of $X$. So, $\bigcup_{\alpha \in A} \tau_{\alpha} = X$ does not make sense, because on one side of the equality you have a topology and on the other you have a subset of $X$.
However, $\bigcup_{H \in \cup_{\alpha \in A}} H = X$ makes sense, because on the left we have a union of subsets of $X$, where the index runs over all subsets of $X$ which are in the collection $\cup_{\alpha \in A} \tau_{\alpha}$. Of course, on the right hand side we have a subset of $X$.
$\tau_\alpha$ is not a subset of $X$, it is a collection of subsets of $X$. The last line is stating that the union of all the elements $H$ of all the topologies $\tau_\alpha$ is $X$. Each $H$ is a subset of $X$. Also since each $\tau_\alpha$ must contain the set $X$, this union is $X$ trivially.