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
Best Answer
I have tried to explain it using only words.
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.