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, $K \cap nE \neq \emptyset$.

  • $\begingroup$ 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 $n$? $\endgroup$ – user8469759 Jul 17 '20 at 16:19
  • $\begingroup$ Baire category theorem implies that a compact Hausdorff space is not the countable union of closed sets with empty interior (or Baire would imply that the entire space has empty interior which is absurd). $\endgroup$ – QuantumSpace Jul 17 '20 at 17:17
  • $\begingroup$ Why would it imply the entire space has empty interior? $\endgroup$ – user8469759 Jul 17 '20 at 18:55
  • $\begingroup$ @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. $\endgroup$ – QuantumSpace Jul 19 '20 at 10:59
  • $\begingroup$ Thank you for your patience $\endgroup$ – user8469759 Jul 19 '20 at 11:19

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