A Simpler Proof of the Bolzano-Weierstrass Theorem
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 _{n\in \mathbb{N}} "\inline (a_n)_{n\in \mathbb{N}}")
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 _{n \in \mathbb{N}} "\inline (a_n)_{n \in \mathbb{N}}")
has infinitely many peaks, then collect those peaks into a subsequence _{k \in \mathbb{N}} "\inline (a_{n_{k}})_{k \in \mathbb{N}}")
. 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 _{k \in \mathbb{N}} "\inline (a_{n_k})_{k \in \mathbb{N}}")
. 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?
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