# $\mathbb{R}$ is a countable union of closed sets, then at least one of them has nonempty interior

Formally stated, Let $S_1, S_2, \cdots$ be closed sets and assume that $\bigcup_{j} S_j = \mathbb{R}$. Prove that at least one of the sets $S_j$ has nonempty interior.

After doing some searching before I asked this question, I found it to be a special case of Baire Category Theorem. (ZF) If $\mathbb{R}^k$ is a countable union of closed sets, then at least one has a nonempty interior.

However this is my first real analysis class, so we haven't learned anything about category sets/nowhere dense sets yet. This is left as a starred textbook problem of the section perfect set, so I am wondering if it can be solved with more fundamental techniques.

The hint from the textbook is to use an idea from the proof that perfect sets are uncountable. (The textbook I am using is Real Analysis and Foundation, Third Edition, by Steven Krantz)

• I guess your title should say "nonempty" rather than "empty"? – MPW Oct 31 '15 at 15:47
• This isn't "measure theory." This is about nowhere dense sets, 1st category sets, 2nd category sets, residual sets, and of course the Baire category theorem. Find out about these. – B. S. Thomson Oct 31 '15 at 15:48
• In the "related" section, this turns up: math.stackexchange.com/questions/88632/… — I guess that should be helpful. – celtschk Oct 31 '15 at 15:55

Let $F_n$ be closed sets with empty interior ($n=1,2,3,\dots$). Construct recursively closed intervals $J_n$ of positive length ${} \le 1/n$ with $J_1 \supset J_2 \supset J_3 \supset \dots$ and $J_n \cap F_n = \varnothing$. Then $\bigcap J_n$ is nonempty, but disjoint from $\bigcup F_n$.
• This is nice, but you should use the word compactness when you say that $\cap J_n$ is non-empty. – Zach Stone Oct 31 '15 at 17:07
• I am sorry I am afraid I didn't get it after staring at it for a long time. Can you possibly give out more hint/explanation as for how to construct recursively intervals $J_n$ of positive length $\leq \frac{1}{n}$ with $J_1 \subset J_2 \subset J_3 \subset \cdots$ and $J_n \cap F_n = \emptyset$? – LKSR Oct 31 '15 at 22:35
If none would have empty interior then they would have been nowhere dense thus $\mathbb R$ becomes a set in first category which is false since $\mathbb R$ is complete.