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The proof I know of the Bolzano-Weierstrass Theorem (i.e. that every bounded sequence has a point of accumulation, and hence a convergent subsequence) proceeds by constructing an increasing sequence between the two bounds, not necessarily a subsequence of the original sequence. It goes on to show that there exists a subsequence of the original sequence that converges to the same limit as the increasing sequence.

Now the fact that every convergent sequence has either an increasing subsequence or decreasing subsequence (or both) follows from the Theorem that every convergent sequence is a Cauchy sequence.

It follows that every bounded sequence has either an increasing or decreasing (or both) subsequence.

My question is can this be turned on its head and used to prove the Bolzano-Weierstrass Theorem in the first place.

i.e. can we prove that every bounded sequence has either an increasing or decreasing (or both) subsequence (without first proving the Bolzano-Weierstrass Theorem) and hence actually prove the Bolzano-Weierstrass Theorem that every bounded sequence has a convergent subsequence since every bounded increasing or decreasing sequence converges.

Thank you.

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Yes, you can. You can refer to Theorem 3.4.7 on Page 80.

Given a sequence $(x_n)$, we can define $x_m$ is a peak if $x_m\ge x_n$ for all $n\ge m$. Then we separate the discussion into two cases, that is, $(x_n)$ has infinite many peaks (hence a decreasing subsequence) or $(x_n)$ has finite many peaks (hence an increasing subsequence).

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