# If $\mathbb{R^k}= \cup^{\infty} F_n$ where $F_n$ is closed, then at least one $F_n$ has non empty interior.

Question: If $$\mathbb{R^k}= \cup^{\infty} F_n$$ where $$F_n$$ is closed, then at least one $$F_n$$ has non empty interior.

closed definition: a set E is closed if every limit point of E is a point of E.

Proof: Assume that $$\mathbb{R^k}= \cup^{\infty} F_n$$ where each $$F_n$$ is closed and has nonempty interior. Let $$N_O$$ be a ball of finite radius around a point $$x_1 \in F_1$$ so that $$\bar{N_O}$$ is compact. Assume $$N _ {i-1}$$ is open and does not comtain any points of $$F _ 1,...,F _ {i-1}$$. this set must contain a point $$x _ i$$ not in $$F _ i$$, otherwise it would belong to the interior of $$F _ i$$. $$x _ i$$ must be contained in a neighborhood $$N _ i \subset N$$ that does mot intersect $$F_i$$ as $$x_i$$ otherwise would be a limit point of $$F_i$$ and therefore belong to $$F_i$$. We can choose $$N_i$$ such that $$\bar{N_{i-1}}$$,and we observe that it does not comtain any points of $$F_1,..F_i$$.

Since each $$\bar{N_i}$$ is compact, and that $$\bar{N_{i+1}} \subset \bar{N_i}$$, so that by the corollary (if $${K_n}$$ is a sequence of nonempty compact sets such that $$K_{n+1} \subset K_n$$, then $$\cap_{1}^{\infty} K_n$$ is not empty), $$I=\cap_i \bar{N_i}$$ is nonempty. By construction, if $$x \in I$$, then $$x \notin F_i$$ for any i. This implies $$x \notin \cup F_i= \mathbb{R^k}$$, a contradiction.

I know this is a proof by contradiction, but I don't get the overall idea of the proof. Can someone help me out with this? Thanks

Let $$X$$ be a complete metric space. Write $$X$$ as countable union of closed sets. If possible, let each of these closed sets have empty interior. Choose a point $$x\in X$$, and consider a relatively compact nbd of $$x$$ which does not intersect with the $$n$$-th closed set appears in the union. Find another smaller relatively compact nbd of $$x$$, whose closure is contained in previously chosen nbd of $$x$$, so that it doesn't intersect with $$(n+1)$$-th closed set appears in the union.
Iterating this process, we have a decreasing sequence of relatively compact nbds of $$x$$ such that closure of $$(n+1)$$-th nbd is contained in $$n$$-th nbd. So, taking closure of these nbds we again have another decreasing sequence compact subsets, so its intersection is non-empty. Take a point in this non-empty intersection, then this point belongs to $$X$$ but not in the any closed set in the countable union, so a contradiction.