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Every bounded [[sequence]] of reals contains a convergent subsequence. | Every bounded [[sequence]] of reals contains a convergent subsequence. | ||

## Latest revision as of 17:48, 1 February 2021

Every bounded sequence of reals contains a convergent subsequence.

## Proof

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.

## See also

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