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Below is the statement and proof of Baire theorem from the textbook I'm using. I have some difficulties understanding it and I would appreciate if someone can explain it in more details. In particular I can't understand:

  1. Why there exists a relatively compact neighbourhood $U_1 \subseteq U_0$ of point $x'$, such that $\overline{U_1} \subseteq U_0 - A_1$. From local compactness (in Hausdorff space) I understand that there exists a relatively compact neighbourhood $U_1 \subseteq U_0$ of point $x'$, such that $U_1 \subseteq U_0 - A_1$. But why is the closure $\overline{U_1}$ included in $U_0 - A_1$ as well?
  2. Why is $U_1$ not contained in $A_2$? In other words, how do we know that $U_1$ is an open set?

Theorem (Baire)

Let $A_1, A_2, \ldots$ be a countable family of closed subsets with empty interior in a locally compact Hausdorff space $X$. Then $\bigcup_{i=1}^{\infty} A_i$ also has empty interior in $X$.

Proof:

Let $U_0$ be any open set in $X$. We will show that $U_0$ is not contained in $\bigcup_{i=1}^{\infty} A_i$, in other words, that $U_0 - \bigcup_{i=1}^{\infty} A_i \neq \emptyset$. Because $A_1$ has empty interior, there exists a point $x' \in U_0 - A_1$. Because of local compactness, there exists a relatively compact neighbourhood $U_1 \subseteq U_0$ of point $x'$, such that $\overline{U_1} \subseteq U_0 - A_1$. Similarily, because $U_1$ is not contained in $A_2$, there exists a relatively compact neighbourhood $U_2$, such that $\overline{U_2} \subseteq U_1 - A_2$. If we keep doing the same thing, we get a series of compact sets $\overline{U_1} \supset \overline{U_2} \supset \overline{U_3} \supset \ldots$, such that $\overline{U_i} \cap A_i = \emptyset$, for all $i$. Finally, $\bigcap_i \overline{U_i}$ is nonempty and by construction included in $U_0 - \bigcup_{i = 1}^{\infty} A_i$.

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    $\begingroup$ To me, "neighborhood" signifies open set, so $U_1$ is of course open. (It's hard to guess how your textbook defines all its vocabulary.) Local compactness gives you an open subset of the given open set $U$ whose closure is contained in $U$. And I think the easiest answer to your first question is to let take $U=U_0-A_1$ as the open set to which to apply the definition. $\endgroup$ Commented Jan 1, 2022 at 22:17
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    $\begingroup$ Yes, it defines neighborhood of x as open set that contains x. My textbook says that local compactness means there is a basis from relatively compact sets (and a relatively compact set is one whose closure is compact). Why exactly is the closure contained in U? $\endgroup$
    – Jesus
    Commented Jan 1, 2022 at 22:18
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    $\begingroup$ Did they not prove the lemma that states precisely what I said above? It is a standard result. $\endgroup$ Commented Jan 1, 2022 at 23:03
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    $\begingroup$ Have they shown that a locally compact Hausdorff space is regular in your text? $\endgroup$ Commented Jan 1, 2022 at 23:20
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    $\begingroup$ It is fairly rare that neighborhoods are defined to be open, usually we say "open neighborhood" to indicate that it is supposed to be open. In general a neighborhood (of a point) is any subspace containing an open neighborhood (of that point). $\endgroup$ Commented Jan 2, 2022 at 10:14

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A locally compact Hausdorff space is regular and so if $x \in O$ is a situation in such a space, where $O$ is open, we can find an open $O_1$ so that $$x \in O_1 \subseteq \overline{O_1} \subseteq O$$ by regularity. After that we can find a relatively compact $U$ so that $x \in U \subseteq O_1$ (because relatively compact sets form a base). It then follows that $\overline{U} \subseteq \overline{O_1} \subseteq O$: i.e. the closure of the relatively compact sets sits inside the original open set (and we can forget about the "auxiliary" $O_1$ again).

So the first part about why $U$ exists for $x' \in U_0-A_1$: it's the previous argument (lemma?) applied to the open set $O=U_0-A_1$ (which is open as an "open minus closed" set) and $x=x'$. The resulting $U$ is then called $U_1$ and is the next step in the recursion.

Next, $A_2$ also has empty interior. As $U_1$ is non-empty (as witnessed by $x'$ from before) it cannot be a subset of $A_2$. So $U_1 - A_2$ is open (open - closed again) and non-empty and "the show can continue" (we can proceed with the recursion).

Hope this answers your doubts.

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