How to prove that the union of the interiors of the members of a covering of countable closed subsets is dense in $\mathbb{R}^n$? If a sequence of closed subsets $\{F_k\}$ of $\mathbb{R}^n$ constitute a cover of $\mathbb{R}^n$ then the union of
their interiors is dense in $\mathbb{R}^n$.
Let $x\in \mathbb{R}^n$, then$$x\in\bigcup_{k\in K}F_k$$ for $K$ in $\mathbb{N}$; if $$x\notin\overline{\bigcup_{k\in K}int(F_k)}$$ then there exist $r>0:B_r(x)$ is disjoint from $int(F_k)\;\forall k\in K$; which means $B_r(x)$ should be covered only by boundary type points. But I can tell (but not able to state mathematical) that there is no way a countable family of boundary point sets can cover an open subset of $\mathbb{R}^n$. How can I do?
 A: For this you definitely want the Baire category theorem. To show the basic idea clearly, I’ll prove first that some $F_k$ has non-empty interior.
Let $V_k=\Bbb R^n\setminus F_k$ for $k\in\Bbb N$. Each $V_k$ is open, and $\bigcap_{k\in\Bbb N}V_k=\varnothing$, so the Baire category theorem says that there is some $k\in\Bbb N$ such that $V_k$ is not dense in $\Bbb R^n$, and therefore $\operatorname{int}F_k\ne\varnothing$. (The Baire category theorem certainly applies to $\Bbb R^n$: the space is both complete metric space and locally compact Hausdorff!)
To get the desired result, you want to relativize this argument to an arbitrary non-empty open set in $\Bbb R^n$. That is, you want to show that if $U$ is a non-empty open set in $\Bbb R^n$, there is a $k\in\Bbb N$ such that $U\cap \operatorname{int}F_k\ne\varnothing$. This is easy: $U$ is a locally compact Hausdorff space, so the Baire category theorem applies to it, and you can just repeat the above argument within $U$.
It follows immediately that $\bigcup_{k\in\Bbb N}\operatorname{int}F_k$ is dense in $\Bbb R^n$.
