# Cantor’s Intersection 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. ∎

Title Cantor’s Intersection Theorem CantorsIntersectionTheorem 2013-03-22 15:12:35 2013-03-22 15:12:35 paolini (1187) paolini (1187) 4 paolini (1187) Theorem msc 54E45