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?


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.

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

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