State and prove bolzano-weierstrass theorem
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First we prove the theorem when {\displaystyle n=1} n=1, in which case the ordering on {\displaystyle \mathbb {R} } \mathbb {R} can be put to good use. Indeed, we have the following result.
Lemma: Every infinite sequence {\displaystyle (x_{n})} (x_{n}) in {\displaystyle \mathbb {R} } \mathbb {R} has a monotone subsequence.
Proof: Let us call a positive integer {\displaystyle n} n a "peak of the sequence" if {\displaystyle n<m} n<m implies {\displaystyle x_{n}>x_{m}} {\displaystyle x_{n}>x_{m}} i.e., if {\displaystyle x_{n}} x_{n} is greater than every subsequent term {\displaystyle x_{m}} x_{m} in the sequence. Suppose first that the sequence has infinitely many peaks, {\displaystyle n_{1}<n_{2}<n_{3}<\dots <n_{j}<\dots } {\displaystyle n_{1}<n_{2}<n_{3}<\dots <n_{j}<\dots }. Then the subsequence {\displaystyle (x_{n_{j}})} {\displaystyle (x_{n_{j}})} corresponding to these peaks is monotonically decreasing. So suppose now that there are only finitely many peaks, let {\displaystyle N} N be the last peak and {\displaystyle n_{1}=N+1} {\displaystyle n_{1}=N+1}. Then {\displaystyle n_{1}} n_{1} is not a peak, since {\displaystyle N<n_{1}} {\displaystyle N<n_{1}}, which implies the existence of {\displaystyle n_{2}} n_{2} with {\displaystyle n_{1}<n_{2}} {\displaystyle n_{1}<n_{2}} and {\displaystyle x_{n_{1}}\leq x_{n_{2}}} {\displaystyle x_{n_{1}}\leq x_{n_{2}}}. Again, {\displaystyle n_{2}>N} {\displaystyle n_{2}>N} is not a peak, hence there is an {\displaystyle n_{3}} n_{3} where {\displaystyle n_{2}<n_{3}} {\displaystyle n_{2}<n_{3}} with {\displaystyle x_{n_{2}}\leq x_{n_{3}}} {\displaystyle x_{n_{2}}\leq x_{n_{3}}}. Repeating this process leads to an infinite non-decreasing subsequence {\displaystyle x_{n_{1}}\leq x_{n_{2}}\leq x_{n_{3}}\leq \ldots } x_{n_{1}}\leq x_{n_{2}}\leq x_{n_{3}}\leq \ldots , as desired
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