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Alsosaid1987 (talk | contribs) openness of U_n noted |
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''Note'': We may leave out the closedness condition in situations where every compact subset of <math>S</math> is closed, for example when <math>S</math> is [[Hausdorff space|Hausdorff]].
=== Proof ===
Assume, by way of contradiction, that <math>\bigcap C_n=\emptyset</math>. For each <math>n</math>, let <math>U_n=C_0\setminus C_n</math>. Since <math>\bigcup U_n=C_0\setminus\bigcap C_n</math> and <math>\bigcap C_n=\emptyset</math>, we have <math>\bigcup U_n=C_0</math>. Note that, since the <math>C_n</math> are closed, the <math>U_n</math>, their set complements in <math>C_0</math>, are open relative to <math>C_0</math>.
Since <math>C_0\subset S</math> is compact and <math>(U_n)</math> is an open cover (on <math>C_0</math>) of <math>C_0</math>, we can extract a finite cover <math>\{U_{n_1}, U_{n_2}, \ldots, U_{n_m}\}</math>. Let <math>U_M</math> be the set such that <math>U_{n_i}\subset U_M</math> for <math>i=1,2,\ldots, m</math>, which exists because <math>U_0\subset U_1\subset\cdots\subset U_k\subset U_{k+1}\cdots</math>, by the ordering hypothesis on the collection <math> (C_n).</math> Consequently, <math>C_0=\bigcup U_{n_i}\subset U_M</math>. But then <math>C_M=C_0\setminus U_M=\emptyset</math>, a contradiction.
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