# What is the justification for these two steps in proof of Baire's theorem?

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 $$U_0 - \bigcup_{i = 1}^{\infty} A_i$$.

• 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. Commented Jan 1, 2022 at 22:17
• 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? Commented Jan 1, 2022 at 22:18
• Did they not prove the lemma that states precisely what I said above? It is a standard result. Commented Jan 1, 2022 at 23:03
• Have they shown that a locally compact Hausdorff space is regular in your text? Commented Jan 1, 2022 at 23:20
• 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). Commented Jan 2, 2022 at 10:14

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).