# Theorem 2.9 Rudin functional analysis - Inferring exists $n$ such that $K \cap nE \neq \emptyset$

Follow up to this question.

I realized that question, which I've asked, explains "why" we can apply Baire's Theorem to $$K$$. It doesn't address however why $$\exists n$$ such that $$K \cap nE \neq \emptyset$$, so this question it's just a check (as I'm reviewing my knowledge of Functional Analysis).

According to Baire's theorem such a $$K$$ is of second category, which means it is not a countable union of nowhere dense (so it's of second category) or equivalently the countable intersection of open dense in $$K$$ is not empty.

However I'm not able to reach the conclusion I want (or maybe I'm just not convinced). I guess I can pick a collection of open dense in $$K$$ therefore (collection is $$\left\{ V_i \right\}$$)

$$K = \overline{\bigcap V_i} = \bigcup K \cap nE$$

the bit that is confusing me is when I say "I can pick", can I actually pick such a collection as a consequence of Baire's theorem?

Note that $$K = \bigcup_{n=1}^\infty K \cap nE$$
and for all $$n \geq 1$$ we have that $$K \cap nE$$ is closed in $$K$$ (since $$E$$ is closed in $$X$$). By the Baire category theorem (applied to the compact Hausdorff space $$K$$), there is $$n \geq 1$$ such that $$K \cap nE$$ has non-empty interior. In particular, $$K \cap nE \neq \emptyset$$.
• How exactly are you applying Baire's theorem here? That's bit I struggle to get. Why does Baire's theorem imply there's such an $n$? – user8469759 Jul 17 '20 at 16:19
• @user8469759 Suppose that $X$ is a Baire space (a space in which the countable intersection of open dense subsets is again dense, for example locally compact Hausdorff spaces are such spaces). Suppose that $X= \cup_{n=1}^\infty X_n$ where $X_n$ where all $n \geq 1$ are closed sets. Then there must be $n \geq 1$ such that $X_n$ has non-empty interior. To see this, suppose to the contrary that for all $n \geq 1$ the interior of $X_n$ is empty. Then we have $\emptyset = \bigcap_{n=1}^\infty X_n^c$ and $X_n^c$ is open and dense for all $n \geq 1$, contradicting that $X$ is a Baire space. – QuantumSpace Jul 19 '20 at 10:59