Bolzano–Weierstrass theorem

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In real analysis, the Bolzano–Weierstrass theorem is a fundamental result about convergence in a finite-dimensional Euclidean space Rn. The theorem states that each bounded sequence in Rn has a convergent subsequence. An equivalent formulation is that a subset of Rn is sequentially compact if and only if it is closed and bounded.



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 { xn } in R has a monotone subsequence.

Proof: Let us call a positive integer n a "peak of the sequence", if m > n implies  xn > xmi.e., if  xn is greater than every subsequent term in the sequence. Suppose first that the sequence has infinitely many peaks, n1 < n2 < n3 < … < nj < …. Then the subsequence   \{x_{n_j}\}  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 n1 = N + 1. Then n1 is not a peak, since n1 > N, which implies the existence of an n2 > n1 with  x_{n_2} \geq x_{n_1}.  Again, n2 > N is not a peak, hence there is n3 > n2 with x_{n_3} \geq x_{n_2}.  Repeating this process leads to an infinite non-decreasing subsequence  x_{n_1} \leq x_{n_2} \leq x_{n_3} \leq \ldots, as desired. \square

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 Rn, 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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