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→Proof: clarification. The open cover is it on the compact space C_0, not in S |
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Assume, by way of contradiction, that <math>\bigcap C_n=\emptyset</math>. For each n, 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>, thus <math>\bigcup U_n=C_0</math>.
Since <math>C_0</math>⊂<math>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_k</math> be the largest set of this cover; then <math>C_0</math>⊂<math>U_k</math>. But then <math>C_k=C_0\setminus U_k=\emptyset</math>, a contradiction.
==Statement for Real Numbers==
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