# Let $\mathbb R$ be a countable union of closed sets, then at least one has nonempty interior

Let $\mathbb R=\bigcup_{n=1}^\infty F_n$, I know that a way of proving this is suppose that all the closed sets have empty interior, and recursively construct closed intervals $I_n$ of length less than $\frac{1}{n}$ and $I_1\supset I_2\supset I_3\supset ...$ and $I_n\cap F_n=\emptyset$. But I am not sure how to construct this set and how to finish the proof.

• Do you know Baire Category Theorem? – Arpit Kansal Nov 11 '15 at 5:38
• No this is a real analysis course so we have not learned much about topology – icicle Nov 11 '15 at 5:39
• First of all, can you find an interval $I_1$ so that $I_1\cap F_1 = \emptyset$? You need to do that inductively. – user99914 Nov 11 '15 at 5:56
• I once wanted to put a question about the Baire category theorem on a topology qualifying exam, but I was told that I can't do that because this theorem counts as real analysis. – Andreas Blass Nov 11 '15 at 17:23

## 1 Answer

Did you learn and do you remember this proof, due to Cantor, that an interval $[a,b]$ cannot be countable?

Let $r_1$, $r_2$, $r_3$ be a sequence of points that (possibly) may include every point of $[a,b]$. Choose an interval $[a_1,b_1]$ that does not include the point $r_1$. Choose a subinterval of that $[a_2,b_2]$ that does not include the point $r_2$. Just keep doing that and use the fact that there must be some point $z\in [a,b]$ that belongs to each of the intervals. But then $z$ is not equal to $r_i$ for any $i$ so this construction is impossible.

Let's steal this idea from Cantor and apply it here. [That's how you do mathematical research: steal ideas and twist them around to suit your own purposes.]

Let $F_1$, $F_2$, $F_3$ be a sequence of closed sets with empty interiors that (possibly) may include every point of $[a,b]$. Choose an interval $[a_1,b_1]$ that is not included in the set $F_1$. Choose a subinterval of that $[a_2,b_2]$ that is not included in the set $F_2$. Just keep doing that ... construction is impossible.

At each stage you aren't so much "constructing" as claiming the existence of such an interval. This is exactly your idea and exactly Mr. Ma's hint. Had you remembered Cantor's proof I think you wouldn't be agonizing over the idea quite so much.