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?

  • 1
    $\begingroup$ You have to show what? $\endgroup$ – Jakobian May 19 at 14:55
  • $\begingroup$ 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 $n\in\Bbb N$). $\endgroup$ – ryan221b May 19 at 14:56
  • $\begingroup$ 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. $\endgroup$ – Jakobian May 19 at 14:57
  • $\begingroup$ By showing that $\sup A$ exists, I prove the aforementioned assertion...? $\endgroup$ – Analysis May 19 at 14:57
  • $\begingroup$ I want to avoid Limits @Jakobian $\endgroup$ – 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 $[a_n,b_n]$.

  • $\begingroup$ And thus $\bigcap_{n\in\mathbb{N}}[a_n,b_n]$ can not be empty. $\endgroup$ – Analysis May 19 at 15:04
  • $\begingroup$ Yes! Intersection is non empty $\endgroup$ – Chinnapparaj R May 19 at 15:05

Assume that the intersection is empty.

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.


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