Closed and boundary subsets Let $X$ be a nonempty compact space and let $F_1, F_2, ...$be its closed and boundary subsets. Prove that $\bigcup_{n=1}^{\infty} F_n \neq X$ 
I have no idea how to do it. My only plan would be to use somehow the fact that $int F_i=\emptyset$. How to prove it?
 A: This is false in general. Consider $X=\mathbb N$, equipped with the cofinite topology, i.e. $F\subseteq\mathbb N$ is closed if it is either finite or $F=\mathbb N$. This is a compact space, and $$X=\bigcup_{n=1}^\infty\{n\}.$$ Note that $\{n\}$ is closed, since it is finite, and its interior is empty, since any non-empty open set is infinite.
It is true if $X$ is Hausdorff, by the Baire category theorem (see BCT2), which states that for locally compact Hausdorff spaces, the intersection of a countable family of open dense sets $U_n,n\in\mathbb N,$ is a dense set. In your case, take $U_n=X\setminus F_n$. These are open and dense, since $F_n$ are closed and have empty interior. So, by the theorem $$\bigcap_{n=1}^\infty U_n=X\setminus\bigcup_{n=1}^\infty F_n$$ is dense in $X$. In particular, it is non-empty, so $$X\neq\bigcup_{n=1}^\infty F_n.$$ I doubt an easier proof is possible. In fact, I think your statement implies the Baire category theorem for locally compact Hausdorff spaces with a little work. (Use the fact that an open neighborhood of any point in a locally compact Hausdorff space contains a smaller open neighborhood, whose closure is compact and contained in the original neighborhood.)
