Cantor's intersection theorem: Difference between revisions

Revision as of 07:53, 9 October 2020

Revision as of 07:55, 9 October 2020

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The closedness condition may be omitted in situations where every compact subset of ''S'' is closed, for example when ''S'' is [[Hausdorff space|Hausdorff]].

'''Proof.''' Assume, by way of contradiction, that <math>\bigcap C_k=\emptyset</math>. For each ''k'', let <math>U_k=C_0\setminus C_k</math>. Since <math>\bigcup U_k=C_0\setminus\left(\bigcap C_k\right)</math> and <math>\bigcap C_k=\emptyset</math>, we have <math>\bigcup U_k=C_0</math>. Note that, sinceSince the <math>C_k</math> are closed relative to ''S'' 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)</math> is an open cover (on <math>C_0</math>) of <math>C_0</math>, we can extract 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>\bigcup 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).</math> Consequently, <math>C_0=\bigcup U_{k_i} = U_M</math>. But then <math>C_M=C_0\setminus U_M=\emptyset</math>, a contradiction. [[Q.E.D.|∎]]

==Statement for Real Numbers==