## Two other proofs of the Bolzano-Weierstrass Theorem

We prove the result:
• If is a sequence of real numbers, then has a monotone subsequence.
Then the Bolzano-Weierstrass Theorem follows immediately, since if is bounded, so is any subsequence, so there is a monotone bounded subsequence, which we know has a limit: its in the increasing case and its in the decreasing case.

To prove the result, let be a given sequence.
Let so is a subsequence of for each .
Note that the first term of is and we have when .
• We say that is a peak if and only if exists and is .
Let be the set of all peaks.
• If is a finite (or empty) set, let be the smallest positive integer larger than any .
• Then is not a peak, so some exists, with , such that .
• Then is not a peak, so some exists, with , such that .
• Then is not a peak, so some exists, with , such that .
This process may be continued forever, with the result that we construct a subsequence of , such that , for each . So the sequence is a strictly increasing subsequence of and we are done.
• If is an infinite set, we list them in order , where , for each . Then put . Then if , we have , since . So we have, since is a peak, . So the sequence is a monotonic decreasing subsequence of and we are done.
So in all cases we produce a monotonic subsequence of and the result is proved. For the third proof, we use interval subdivision. Let be a bounded sequence, so for some real numbers and with .
• Bisect : , where and .
Put .
Then define the sets and . Then , so at least one of the sets and is infinite.
• If is infinite, put , with and and .
Also put and .
• If is finite or empty, put , with and and .
Also put and .
• Now bisect : , where and .
Then define the sets and . Then , which is an infinite set, so at least one of the sets and is infinite.
• If is infinite, put , with and and .
Also put and .
• If is finite or empty, put , with and and .
Also put and .
We may continue this process for ever, producing a nested sequence of intervals , for , such that each interval contains an infinite number of elements of . At the same time, we produce a subsequence of , such that , for all .
Note that the length of , for each , which has limit zero, as goes to infinity. So by the theory of nested intervals, we have , where .
So by squeeze applied to the inequality , for all , we see that converges with limit and we are done: the sequence has a convergent subsequence, proving Bolzano-Weierstrass again.