# 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

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 $F_1$, $F_2$, $F_3$ 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.