Proof
(related to Theorem: Theorem of Bolzano-Weierstrass)
- Let X be a metric space.
- Let A\subset X be a compact subset.
- Let (x_n)_{n\in\mathbb N} be a sequence of points x_n\in A.
- We will prove by contradiction that (x_n)_{n\in\mathbb N} contains a subsequence (x_{n_k})_{k\in\mathbb N}, which converges against some point a\in A.
- Assume, there is no such subsequence.
- Then for every x\in A we can choose arbitrary small \epsilon_x > 0 such that the open ball B(x,\epsilon_x) contains only finitely many sequence members. (Otherwise we could construct a subsequence convergent against x, since we can choose \epsilon_x arbitrarily small).
- Note that (B(x,\epsilon_x))_{x\in A} is an open cover of A, i.e. A\subset \bigcup_{x\in A} B(x,\epsilon_x).
- Since A is compact, there are finitely many points x_1,\ldots,x_n\in A forming with their open balls a finite open subcover of A, formally
A\subset \bigcup_{k=1}^n B(x_k,\epsilon_{x_k}).
- But then, the sequence (x_n)_{n\in\mathbb N} would contain only finitely many points, which is a contradiction.
- Therefore, the assumption that there is no convergent subsequence (x_{n_k})_{k\in\mathbb N} of (x_n)_{n\in\mathbb N} is wrong.
∎
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References
Bibliography
- Forster Otto: "Analysis 2, Differentialrechnung im \mathbb R^n, Gewöhnliche Differentialgleichungen", Vieweg Studium, 1984