# Cantor’s Intersection Theorem

###### Theorem 1.

Let ${K}_{\mathrm{1}}\mathrm{\supset}{K}_{\mathrm{2}}\mathrm{\supset}{K}_{\mathrm{3}}\mathrm{\supset}\mathrm{\dots}\mathrm{\supset}{K}_{n}\mathrm{\supset}\mathrm{\dots}$ be a sequence of non-empty, compact subsets of a metric space $X$. Then the intersection ${\mathrm{\bigcap}}_{i}{K}_{i}$ is not empty.

###### Proof.

Choose a point ${x}_{i}\in {K}_{i}$ for every $i=1,2,\mathrm{\dots}$
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.
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