# Baire's Theorem with locally compact Hausdorff space

Statement of the theorem:

If $$S$$ is either

a) a complete metric space, or

b) a locally compact Hausdorff space,

Then the intersection of every countable collection of dense open subsets of $$S$$ is dense in $$S$$.

The idea of the proof is to show that every open set $$B$$ intersects the countable union of given dense open subsets. Specifically if $$\left\{ V_i \right\}_{i \in \mathbb{N}}$$ is such a collection and $$B$$ is an arbitrary open set the following recursion is then defined

$$\begin{array}{l} B_0 = B \\ \bar{B}_{n} \subset V_n \cap B_{n-1} \end{array}$$

Later we define

$$K = \bigcap_{n=1}^{\infty} \bar{B}_n$$

The author at this point states that $$K$$ isn't empty by compactness. I cannot really understand why.

From wikipedia:

Let $$X$$ be a topological space. Most commonly $$X$$ is called locally compact, if every point $$x$$ of $$X$$ has a compact neighbourhood, i.e., there exists an open set $$U$$ and a compact set $$K$$, such that $${\displaystyle x\in U\subseteq K}$$

I guess this is the definition that we're trying to apply, but I can't figure how exactly we apply it.

• There are some crucial facts missing here. With what you have quoted it is not possible to show that $K$ is not empty. Please look at the entire proof. – Kavi Rama Murthy Jul 14 '20 at 12:36
• The only bit I missed is that, according to the Rudin, $\bar{B}_n$ can be chosen to be compact is this what I'm missing? – user8469759 Jul 14 '20 at 12:45
• Yes, you missed the most important assumption. – Kavi Rama Murthy Jul 14 '20 at 13:02
• Another case that works for Baire category theorem: locally countably compact regular space. – GEdgar Jul 14 '20 at 13:15

$$(\overline {B_n})$$ is decreasing sequence of nonempty compact sets and hence their intersection is not empty: if it is empty then complements of $$\overline {B_n}, n=2,3,,$$ cover the compact set $$\overline {B_1}$$. Hence there is a finite sub-cover. But this means $$\cap_{n=1}^{N} \overline {B_n} (=\overline {B_N})$$ is empty for some $$N$$, a contradiction.