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In my functional analysis textbook there's the statement:

This is often called the category theorem, for the following reason. Let $S$ be either a complete metric space or a locally compact Hausdorff space. If $\left\{ E_i \right\}$ is a countable collection of nowhere dense subsets of $S$, and if $V_i$ is the complement of $\bar{E}_i$, then each $V_i$ is dense in $S$, and the conclusion of the Baire's theorem is that $\bigcap V_i \neq \emptyset$.Hence $S \neq \bigcup E_i$.

Why is $V_i$ dense?

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  • $\begingroup$ Does this help : math.stackexchange.com/questions/829752/…? $\endgroup$ – Arnaud D. Sep 24 '18 at 12:38
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    $\begingroup$ That is because the complement of the interior is equal to the closure of the complement. It's relatively easy to prove from the definitions, but I'll see if it has been asked before $\endgroup$ – Arnaud D. Sep 24 '18 at 12:44
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    $\begingroup$ Are you asking why it's called the category theorem or why $V_i$ is dense? $\endgroup$ – Teddy38 Sep 24 '18 at 12:49
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    $\begingroup$ Here's a question about the fact I mentioned in my comment above; it's give for metric spaces but one of the answers is given for topological spaces. By the way, I've found another relevant question (again, one of the answers is topological). $\endgroup$ – Arnaud D. Sep 24 '18 at 12:56
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    $\begingroup$ As it was pointed out by @ArnaudD (+1 for that). use the fact that for any $B \subseteq X$, $\overline{B} = X \setminus (X \setminus B)^{\circ}$ and also that $B = X \setminus (X \setminus B)$. For the question in the title, the conclusion of the theorem could have been written simply as $S$ is second category in itself (mathworld.wolfram.com/SecondCategory.html). Informally, Bair categories tell you that the first category sets are "small" or "negligible", second category sets are "big". The terminology of first and second category sets is independent of category theory. $\endgroup$ – m_gnacik Sep 24 '18 at 13:19
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As requested by @PaulFrost.

Set $V_i = S \setminus \overline{E_i}$ for $i \in I$, where $I$ is a countable set of indices. Recall that, given a topological space $(X, \tau)$ and $B \subseteq X$ one have $$ \overline{B} = X \setminus (X \setminus B)^{\circ}.$$ You want to show that $\overline{V_i} = S.$ We have that $$\overline{V_i} = \overline{S \setminus \overline{E_i}} = S \setminus \left( S \setminus (S \setminus \overline{E_i})\right)^\circ = \mbox{you fill the gap} = S \setminus \emptyset = S.$$

To fill the gap use the fact that $E_i$ is nowhere dense in $S$ and note that $B = X \setminus (X \setminus B)$ for any $B \subseteq X$, where $(X, \tau)$ is a topological space.

For the question in the title, the conclusion of the theorem could have been written simply as $S$ is the second category in itself (link). Informally, Bair categories tell you that the first category sets are "small" or "negligible", second category sets are "big". The terminology of first and second category sets is independent of category theory.

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