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I have $(X,\tau)$ a topological space, $Y \subseteq X$, $\mathcal{O} = \{Y \cap U \mid U \in \tau \}$, and I want to prove that $\mathcal{O}$ is a topology on $Y$. All too easy and good, but something caught my mind here.

Let $\mathscr{C} \subset \mathcal{O}$, and let's prove that $\bigcup \mathscr{C} \in \mathcal{O}$. For all $\Omega \in \mathscr{C}$, we can write $\Omega = U_\Omega \cap Y$, with $U_\Omega \in \tau$. Clearly: $$\bigcup \mathscr{C} = \bigcup_{\Omega \in \mathscr{C}}\Omega = \bigcup_{\Omega \in \mathscr{C}}(U_\Omega \cap Y) = \left(\bigcup_{\Omega \in \mathscr{C}}U_\Omega \right) \cap Y.$$ We do not know immediately that $\bigcup_{\Omega \in \mathscr{C}}U_\Omega \in \tau$, because $\mathscr{C}$ is not necessarily a subcollection of $\tau$. So I would need to rewrite $\bigcup_{\Omega \in \mathscr{C}}U_\Omega$, reindexing it. The only thing that comes to mind is writing: $$\bigcup_{\Omega \in \mathscr{C}}U_\Omega = \bigcup \{ U_\Omega \in \tau \mid \Omega \in \mathscr{C} \}.$$

My understanding is that $\{ U_\Omega \in \tau \mid \Omega \in \mathscr{C} \}$ yes, is indeed a subcollection of $\tau$.

Questions: Is this last equality correct? Am I being paranoid? If not, is there an easier way to go about this?

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You are indeed correct, and no, you are not being paranoid for checking this.

What needs addressing, however, is that the indexing set is not really what such a union is about. For example, consider:

$$\bigcup_{U\in\tau} \{\tau\}\notin \tau$$

where the result is not in $\tau$ even though the indexing set is trivially a subset of $\tau$. Conversely, I presume you wouldn't object to a construction like:

$$\bigcup_{n \in \Bbb N} U_n \in \tau$$

even though in most cases we won't have $\Bbb N \subseteq \tau$.

What should instead be verified is that all the elements considered in the union (which is the set $\{U_\Omega \in \tau \mid \Omega\in \mathscr C\}$) are in $\tau$. You did this correctly in your example; it effectively amounts to the rewriting of the union.

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  • $\begingroup$ Very helpful! I understood, thanks. $\endgroup$ – Ivo Terek Mar 20 '15 at 21:25

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