A while back I posted a long proof of the Bolzano-Weierstrass theorem -- also known as the "sequential compactness theorem" -- which basically says every sequence that's bounded has a subsequence within it that converges. Here's a much shorter and simpler version of it.

First we'll prove a lemma that shows for any sequence we can always find a monotone subsequence -- that is, a subsequence that's always increasing or decreasing.

*Lemma.* Every sequence has a monotone subsequence.

*Proof.* Let be a sequence. Define a "peak" of as an element such that for all . That is is a peak if forever after that point going forward, there is no other element of the sequence that is greater than . Intuitively, think of shining a flashlight from the right onto the "mountain range" of a sequence's plotted elements. If the light hits an element, that element is a peak.

If has infinitely many peaks, then collect those peaks into a subsequence . This is a monotone decreasing subsequence, as required.

If has finitely many peaks, take *n* to be the position of the last peak. Then we know is not a peak. So there exists an such that . Call this point . We also know that is not a peak either. So there also exists an such that . Call this point .

Continuing, we can create a subsequence that is monotone increasing. In either case -- if our sequence has infinite or finitely many peaks -- we can always find a monotone subsequence, as required.

Now that we've proved the above lemma, the proof of the Bolzano-Weierstrass theorem follows easily.

*Theorem (Bolzano-Weierstrass).*Every bounded sequence has a convergent subsequence.

*Proof.* By the previous lemma, every sequence has a monotone subsequence. Call this . Since is bounded by assumption, then the subsequence is also bounded. So by the monotone convergence theorem, since is monotone and bounded, it converges. So every bounded sequence has a convergent subsequence, completing the proof.

Posted by Andrew on Tuesday July 20, 2010 | Feedback?