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Cantor's Intersection Theorem (Theorem)
Theorem 1   Let $ K_1\supset K_2\supset K_3 \supset \ldots \supset K_n\supset \ldots$ be a sequence of non-empty, compact subsets of a metric space $ X$. Then the intersection $ \bigcap_i K_i$ is not empty.
Proof. Choose a point $ x_i\in K_i$ for every $ i=1,2,\ldots$ Since $ x_i \in K_i \subset K_1$ is a sequence in a compact set, by Bolzano-Weierstrass Theorem, there exists a subsequence $ x_{i_j}$ which converges to a point $ x\in K_1$. Notice, however, that for a fixed index $ n$, the sequence $ x_{i_j}$ lies in $ K_n$ for all $ j$ sufficiently large (namely for all $ j$ such that $ i_j>n$). So one has $ x\in K_n$. Since this is true for every $ n$, the result follows. $ \qedsymbol$



"Cantor's Intersection Theorem" is owned by paolini.
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Cross-references: index, fixed, converges, subsequence, Bolzano-Weierstrass theorem, compact set, point, intersection, metric space, compact subsets, sequence

This is version 1 of Cantor's Intersection Theorem, born on 2005-04-27.
Object id is 6970, canonical name is CantorsIntersectionTheorem.
Accessed 2197 times total.

Classification:
AMS MSC: 54E45 (General topology :: Spaces with richer structures :: Compact metric spaces)
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