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 $n$?

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).

Why would it imply the entire space has empty interior?

@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.

Thank you for your patience

You are very welcome!

Hi, sorry... from your last comment. If $X_n$ is closed with empty interior, then $X_n = \partial X_n$. Therefore if $U$ is open in $X$ we have $U \subset X - X_n^\circ$ = $X - (X_n - \partial X_n) = X_n^c \cup \partial X_n$. I can't reach the conclusion that $X_n^c$ is dense.

Use that $A$ is dense in $X$ if $cl(A)=X$ together with the relation $cl(A)=X\setminus(int(X\setminus A))$

@user8469759 Did it work out?

Not really, sorry.

What don't you understand about it?

Is $A$ supposed to be some $X_n^c$ as a starter?

Yes, tthe idea is that you apply my two facts with $A= X_n^c$. Then you get $cl(X_n^c) = int(X_n)^c = \emptyset^c = X$ so $X_n^c$ is dense in $X$.

Hi, are you available for a cht?

Yes, I have some time now.

hi!

I do apologize

first of all, the way I was trying to prove this was the following

if $X_n$ is closed and has empty interior and I wanted to show that $X_n^c$ is dense in $X$ then I need to show that for any $U$ open in $X$ the intersection $U \cap X_n^c$ is not the empty set.

But why don't you just try to use the identity that I said in the comment?

The relation between closure and interior

ok I didn't know that relationship

But maybe we can try to prove directly without using that relationship.

I guess I can prove that identity at some point as exercise

Do you agree it suffices to show the following: If A is a subset of a topological space X such that A is closed and has empty interior, then cl(X\A) = X?

I think so, isn't this a way to justify why Baire's theorem is called Category theorem

Let's prove this directly: If x is an element of X, then we show that x \in cl(X\A) by showing that every neighborhood V of x intersects X\A. Suppose to the contrary that there is a neighborhood V of x that does NOT intersect X\A, that is V \cap X\A = \emptyset. Then clearly V is a subset of A containing x, so A has an interior point.

it's a basic fact. If $A$ has empty interior then the X\A is dense

Yes, indeed I did not use that A was closed but we need that in our case so that the complement is OPEN and dense.

before you carry on

actually never mind, please carry on

I'm not sure this solves your problem?

I think so... I think I got confused for something silly.

I need to re-visit your answer with this conversation on hand

Sure, I'll remain online for a while