Show that there exist $x\in X$ such that $x\notin\bigcup_{n=1}^{\infty}{A_n}$. Let $X$ a compact Hausdorff topological space. Let $\{A_n\}_{n\in\mathbb{Z}^{+}}$ a collection of countable closed, such that the interior of $A_{n}$ is empty, for all $n\in\mathbb{Z}^{+}$. Show that, there exist  $x\in X$ such that $x\notin\bigcup_{n=1}^{\infty}{A_n}$.
So, I know this union has empty interior, Prob. 5, Sec. 27 in Munkres. But, if the interior of the union is empty, maybe there exist a point in the boundary of $\bigcup_{n=1}^{\infty}{A_n}$. 
Can give me some hint, or any idea to prove this, pls. Thanks! 
 A: Any Compact Hausdorff space is a Baire space.
And any non-empty open set in a baire space is of the second category, that is it cannot be written as countable union of nowhere dense sets.
The sets $A_n$ are all nowhere dense as they are closed and has an empty interior, and since $X$ is open in itself so $X\neq \bigcup_{n=1}^{\infty}{A_n}$, hence there exists $x\in X$ such that $x\notin\bigcup_{n=1}^{\infty}{A_n}$.
A: Elaborating on Robert Israel's comment:  let $x\in X$ be such that $x \not\in \cup_{n=1}^{\infty} A_n$. Then there exists an open ball $B(x,r)$ of some positive radius $r>0$ centred at $x$ (because $X$ is a compact Hausdorff space) and so we claim that $B(x,r)\not\subset \left( \cup_{n=1}^{\infty}A_n \right)$.  Suppose the ball were contained in the union: then by the definition of interior the ball would be contained in that (because it's an open set) -- but we know that $\cup_{n=1}^{\infty}A_n$ has empty interior.
Since $B(x,r)\not\subset \left( \cup_{n=1}^{\infty}A_n \right)$ there exists some $p\in B(x,r)\in X$ with $p\not\in \cup_{n=1}^{\infty}A_n$, which is what you wanted to show.
There is an extra point for completeness: if $X=\cup_{n=1}^{\infty}A_n$ then our assumption we can find an $x$ not in the union doesn't hold -- but then as Arpan1729 says $X$ would not be compact Hausdorff at that point, so we don't need to worry.
A: $X \setminus A_n$ is dense as $\overline{X \setminus A_n} = X\setminus \operatorname{Int}(A_n) = X$, and open as the $A_n$ are closed. 
So $$\bigcap_n (X \setminus A_n) = X \setminus (\bigcup_n A_n)$$ is dense by Baire's theorem, so $\bigcup_n A_n \neq X$. (the empty set is not dense as $X \neq \emptyset$; we can even say $\operatorname{Int}(\bigcup_n A_n) = \emptyset$, or this interior would be a non-empty open set disjoint from the dense set.
