Cantor's intersection theorem: Difference between revisions

Revision as of 20:40, 3 March 2021

Revision as of 20:41, 3 March 2021

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The closedness condition may be omitted 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>{\textstyle \bigcap_{k = 0}^\infty C_k}=\emptyset</math>. For each <math>k</math>, let <math>U_k=C_0\setminus C_k</math>. Since <math>{\textstyle \bigcup_{k = 0}^\infty U_k}=C_0\setminus\left( {\textstyle \bigcap_{k = 0}^\infty C_k}\right)</math> and <math>{\textstyle \bigcap_{k = 0}^\infty C_k}=\emptyset</math>, we have <math>{\textstyle \bigcup_{k = 0}^\infty U_k}=C_0</math>. Since the <math>C_k</math> are closed relative to <math>S</math> and therefore, also closed relative to <math>C_0</math>, the <math>U_k</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_k \vert k \geq 0\}</math> is an open cover (on <math>C_0</math>) of <math>C_0</math>, a finite cover <math>\{U_{k_1}, U_{k_2}, \ldots, U_{k_m}\}</math> can be extracted. Let <math>M=\max_{1\leq i\leq m} {k_i}</math>. Then <math>{\textstyle \bigcup_{i = 1}^m U_{k_i}}=U_M</math> because <math>U_1\subset U_2\subset\cdots\subset U_n\subset U_{n+1}\cdots</math>, by the nesting hypothesis for the collection <math>(C_k)_{k \geq 0}</math>. Consequently, <math>C_0={\textstyle \bigcup_{i = 1}^m U_{k_i}} = U_M</math>. But then <math>C_M=C_0\setminus U_M=\emptyset</math>, a contradiction. [[Q.E.D.|∎]]