# Proving that a compact Hausdorff space is a Baire space.

This is exercise 27.5 in Munkres:

Let $$X$$ be a compact Hausdorff space; Let $$\{A_n\}$$ be a countable collection of closed sets of $$X$$. Show that if each set $$A_n$$ has empty interior in $$X$$, then the union $$\bigcup A_n$$ has empty interior in $$X$$. [Hint: Imitate the proof of Theorem 27.7.]

Here is Theorem 27.7:

Let $$X$$ be a nonempty compact Hausdorff space. If $$X$$ has no isolated points, then $$X$$ is uncountable.

The proof of Theorem 27.7 boils down to this:

1. Show that given any nonempty open set $$U$$ of $$X$$ and any $$x\in X$$, there exists a nonempty open set $$V$$ contained in $$U$$ such that $$x\notin\overline V$$.
2. Given $$f:\mathbb Z_+\to X$$, use 1. to construct a sequence of points $$x_n$$ all distinct from $$x$$. It follows that $$f$$ is not surjective, and hence $$X$$ is uncountable.

I don't see any relation between the exercise and the proof of the theorem. The theorem constructs a sequence of points, while the exercise requires me to show that a set has empty interior. Any hints on how to relate these two would be greatly appreciated (I'd like to complete the proof myself).

• To prove the exercise, you wish to show that given any non-empty open set $U$, $U$ contains a point outside $\bigcup A_n$. In the same manner as the proof of theorem 27.7, you wish to construct a decreasing sequence of open sets which avoid $A_n$. – lc2r43 Nov 17 '19 at 5:44

Suppose that $$O \subseteq \bigcup_n A_n$$ is non-empty open.

$$A_1$$ has empty interior and is closed, so there is some $$O_1$$ non-empty open with $$\overline{O_1} \subseteq O$$ such that $$O_1 \cap A_1 = \emptyset$$.

$$A_1 \cup A_2$$ also has empty interior and is still closed, so there is some non-empty open $$O_2$$ such that $$O_2 \subseteq \overline{O_2} \subseteq O_1$$ and $$O_2 \cap (A_1 \cup A_2)=\emptyset$$.

Continue finding $$O_n$$ this way and note that $$\bigcap_n \overline{O_n}$$ is non-empty by compactness and see what contradiction you find.

With a minor adaptation at the start, this also works for locally compact Hausdorff spaces.

The common theme with the older proof is the decreasing open sets intersecting to achieve a "countable goal" when finite goals are achievable.

• Thanks, but where exactly did you use Hausdorffness? – Math1000 Nov 17 '19 at 23:52
• I think I see it: Since $\bigcap_{n=1}^\infty \overline O_n$ and $\bigcup_{n=1}^\infty A_n$ are disjoint, there exist points $x\in\bigcap_{n=1}^\infty \overline O_n$ and $y\in\bigcup_{n=1}^\infty A_n$ with disjoint neighborhoods $U$ and $V$. But this means that $V\subset \bigcup_{n=1}^\infty A_n$, contradicting the assumption that $\bigcup_{n=1}^\infty A_n$ has empty interior. Is that correct? – Math1000 Nov 18 '19 at 0:13
• @Math1000 I use it when I apply regularity in the shrinking (closures) of open sets. The contradiction lies in the fact that $x\in U$ but $x\notin \bigcup_n A_n$ despite the initial assumptions of non-empty interior (which gave us $O$ in the first place). – Henno Brandsma Nov 18 '19 at 5:11
• I didn't see a $U$ in your post. Are you referring to the $U$ in my comment? – Math1000 Nov 18 '19 at 18:10
• And as for "regularity," $X$ is regular because it is compact Hausdorff, correct? Sorry, I have not yet studied regular spaces. – Math1000 Nov 18 '19 at 18:13