# Proof that $\bigcap_{n\in\mathbb{N}}[a_n,b_n]$ is a non-emtpy set

Let $$a_n,b_n \in \mathbb{R}$$, for $$n\in \mathbb N$$ with $$a_n \leq a_{n+1} \leq b_{n+1} \leq b_n$$.

Proof that $$\bigcap_{n\in\mathbb{N}}[a_n,b_n]$$ is a non-emtpy set.

My attempt:

Observe $$A:=\{a_n : n\in \mathbb{N}\}$$. I now have to to show $$\sup A$$. But how do I do that?

• You have to show what? – Jakobian May 19 at 14:55
• Try to guess: what point might you find in the intersection of all those intervals? (Hint: consider the simpler case of intervals $[-1/n, 1/n]$ for $n\in\Bbb N$). – ryan221b May 19 at 14:56
• For any $k$ we have $a_k\leq \lim_{n\to\infty} a_n \leq \lim_{n\to\infty} b_n\leq b_k$ and the limits exist because $a_n$ and $b_n$ are monotone. – Jakobian May 19 at 14:57
• By showing that $\sup A$ exists, I prove the aforementioned assertion...? – Analysis May 19 at 14:57
• I want to avoid Limits @Jakobian – Analysis May 19 at 14:57

Every $$b_n$$ is an upper bound for $$A$$. So supremum of $$A$$ exist, call it as $$x$$. Thus, $$a_n \leq x$$ for all $$n$$ and note that every $$b_n$$ is an upper bound and $$x$$ is the supremum, so $$x \leq b_n$$ for all $$n$$ . Hence $$x$$ belongs to every $$[a_n,b_n]$$.
• And thus $\bigcap_{n\in\mathbb{N}}[a_n,b_n]$ can not be empty. – Analysis May 19 at 15:04
Then the open sets $$[a_n,b_n]^{\complement}$$ cover the closed and bounded set $$[a_1,b_1]$$ and there is no finite subcover.
This however contradicts that $$[a_1,b_1]$$ is compact, so the assumption must be wrong.