Bolzano-Weierstrass Property

Theorem 4.21 (Bolzano-Weierstrass Property) A set of real numbers \mathrm{E} is closed and bounded if and only if every sequence of points chosen from the set has a subsequence that converges to a point that belongs to \mathrm{E}.

Proof Suppose that \mathrm{E} is both closed and bounded and let \mathrm{\{x_n\}} be a sequence of points chosen from \mathrm{E}. Since \mathrm{E} is bounded this sequence \mathrm{\{x_n\}} must be bounded too. We apply the Bolzano-Weierstrass theorem for sequences (Theorem 2.40) to obtain a subsequence \mathrm{\{x_{n_k}\}} that converges. If \mathrm{x_{n_k}\rightarrow z} then since all the points of the subsequence \mathrm{\{x_{n_k}\}} belong to \mathrm{E} either the sequence is constant after some term or else \mathrm{z} is a point of accumulation of \mathrm{E}. In either case we see that \mathrm{z \in E}. This proves the theorem in one direction.

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