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.

  • $\begingroup$ 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. $\endgroup$ – Kavi Rama Murthy Jul 14 '20 at 12:36
  • $\begingroup$ 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? $\endgroup$ – user8469759 Jul 14 '20 at 12:45
  • $\begingroup$ Yes, you missed the most important assumption. $\endgroup$ – Kavi Rama Murthy Jul 14 '20 at 13:02
  • $\begingroup$ Another case that works for Baire category theorem: locally countably compact regular space. $\endgroup$ – 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.


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