In real analysis, the Bolzano–Weierstrass theorem is a fundamental result about convergence in a finitedimensional Euclidean space R^{n}. The theorem states that each bounded sequence in R^{n} has a convergent subsequence. An equivalent formulation is that a subset of R^{n} is sequentially compact if and only if it is closed and bounded.
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Proof
First we prove the theorem when n = 1, in which case the ordering on R can be put to good use. Indeed we have the following result.
Lemma: Every sequence { x_{n} } in R has a monotone subsequence.
Proof: Let us call a positive integer n a "peak of the sequence", if m > n implies x_{ n} > x_{ m} i.e., if x_{n} is greater than every subsequent term in the sequence. Suppose first that the sequence has infinitely many peaks, n_{1} < n_{2} < n_{3} < … < n_{j} < …. Then the subsequence corresponding to peaks is monotonically decreasing, and we are done. So suppose now that there are only finitely many peaks, let N be the last peak and n_{1} = N + 1. Then n_{1} is not a peak, since n_{1} > N, which implies the existence of an n_{2} > n_{1} with Again, n_{2} > N is not a peak, hence there is n_{3} > n_{2} with Repeating this process leads to an infinite nondecreasing subsequence , as desired.
Now suppose we have a bounded sequence in R; by the Lemma there exists a monotone subsequence, necessarily bounded. But it follows from the Monotone convergence theorem that this subsequence must converge, and the proof is complete.
Finally, the general case can be easily reduced to the case of n = 1 as follows: given a bounded sequence in R^{n}, the sequence of first coordinates is a bounded real sequence, hence has a convergent subsequence. We can then extract a subsubsequence on which the second coordinates converge, and so on, until in the end we have passed from the original sequence to a subsequence n times — which is still a subsequence of the original sequence — on which each coordinate sequence converges, hence the subsequence itself is convergent.
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