--Bhaskaracharya 05:41, 16 October 2013 (EDT)== Statement == If positive real numbers,then
Further if then
Every bounded sequence of reals contains a convergent subsequence.
Lemma 1: A bounded increasing sequence of reals converges to its least upper bound.
Proof: Suppose is a bounded increasing sequence of reals, so . Let . We want to show that . Assume . By the definition of convergence, we need to produce some such that for all , . Since is the least upper bound of the sequence, there exists an such that , and since is increasing, we know that for all , . Therefore for all , . This shows that converges to . This completes the proof of Lemma 1.
We can prove analogously that a bounded decreasing sequence of reals converges to its greatest lower bound.
Lemma 2: Every sequence of reals has a monotone subsequence.
Proof: Given a sequence of reals, we consider two cases:
- Case 1: Every infinite subsequence of contains a term that is strictly smaller than infinitely many later terms in the subsequence.
In this case, we build a strictly increasing subsequence .
Note that we can apply the assumption of Case 1 to itself. Thus there exists some such that there exist infinitely many such that . Let , and consider the subsequence where , and is the th term in after that is greater than . This exists by the assumption.
Now we can apply the assumption of Case 1 to . Thus there exists some such that there exist infinitely many such that . Since is a subsequence of , there exists an such that . We then let , and let be a subsequence of such that , and is the th term in after that is gerater tha . Note that is a subsequence of .
We can continue in this fashion to construct a strictly increasing subsequence of .
- Case 2: Case 1 fails.
In other words, there exists a subsequence of such that every term in is at least as large as some later term in . In this case, we can build a decreasing subsequence .
Consider such a sequence . Let . By the assumption of Case 2, there exists some . Let . By the assumption again, there exists some . Let . We can continue in this fashion to build a decreasing subsequence of . Since is a subsequence of , we have that is a decreasing subsequence of . This completes the proof of Lemma 2.
The Bolzano-Weierstrass Theorem follows immediately: every bounded sequence of reals contains some monotone subsequence by Lemma 2, which is in turn bounded. This subsequence is convergent by Lemma 1, which completes the proof.
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