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.∎