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By Cantor's intersection theorem I know that a sequence of nonempty compact sets which are nested has nonempty intersection. But how can I use that to prove that arbitrary intersection of compact sets is compact?

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  • $\begingroup$ If you're just talking about compact sets of the complex plane, just use the Heine-Borel theorem. $\endgroup$ – Vik78 Oct 24 '16 at 22:31
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In an finite dimensional metric space (such as the complex numbers), all compact sets are closed and bounded. Then, all of your compact sets are closed and therefore, their intersection is a closed set. Then, because the intersection is closed and contained in any of your compact sets, it is a compact set (This property can be used because metric spaces are, in particular, Hausdorff spaces).

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  • $\begingroup$ But how do you prove that the intersection of bounded sets is a bounded set? $\endgroup$ – shot22 Sep 15 '17 at 9:55
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    $\begingroup$ Supose we have $\{K_i\}_i a colection of compact sets. This means that for every $i$ there exists some $R_i>0$ such that $K_i$ is bounded: $K_i\subseteq B(0,R_i)$. Now let $K$ be the intersection of the $K_i$. By property of intersections $K\subseteq K_i$ for all $i$, this implies, $K\subseteq B(0, R_i)$ for all $i$, i.e. $K$ is bounded. $\endgroup$ – N. Bitar Sep 15 '17 at 14:20

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