Cantor's intersection theorem: Difference between revisions

Revision as of 22:22, 15 October 2019

Revision as of 14:25, 21 October 2019

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==Topological Statement==

'''Theorem.''' ''Let'' ''S'' ''be a [[Topological Space|topological space]]. A decreasing nested sequence of non-empty compact,'' ''closed subsets of S has a non-empty intersection. In other words, supposing'' <math>(C_k)</math> ''is a sequence of non-empty compact, closed subsets of'' ''S'' ''satisfying''

:<math>C_0 \supset C_1 \supset \cdots C_n \supset C_{n+1} \cdots, </math>

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'''Note:''' We may leave out the closedness condition 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, since 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>. 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==

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'''Theorem.''' ''Let'' <math>(C_k)</math> ''be a family of non-empty, closed, and bounded subsets of'' <math>\mathbf{R}</math> ''satisfying''

:<math>C_0 \supset C_1 \supset \cdots C_n \supset C_{n+1} \cdots. </math>

''Then,''

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:<math>x=\lim_{k\to \infty} x_k.</math>

For fixed ''k'', <math>x_j\in C_k</math> for all <math>j\geq k</math> and since <math>C_k</math> was closed and ''x'' is a [[limit point]], it follows that <math>x\in C_k</math>. Our choice of ''k'' was arbitrary, hence ''x'' belongs to ''<math>\bigcap_k C_k</math>'' and the proof is complete. ∎

== Variant in complete metric spaces ==

In a [[complete metric space]], the following variant of Cantor's intersection theorem holds.

'''Theorem.''' ''Suppose that X is a complete metric space, and <math>C_k</math> is a sequence'' ''of non-empty closed nested subsets of X whose [[diameter]]s tend to zero:''

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''for some x in X.''

''Proof (sketch).'' A proof goes as follows. Since the diameters tend to zero, the diameter of the intersection of the ''<math>C_k</math>'' is zero, so it is either empty or consists of a single point. So it is sufficient to show that it is not empty. Pick an element <math>x_k\in C_k</math> for each ''k''. Since the diameter of ''<math>C_k</math>'' tends to zero and the ''<math>C_k</math>'' are nested, the ''<math>x_k</math>'' form a Cauchy sequence. Since the metric space is complete this Cauchy sequence converges to some point ''x''. Since each ''<math>C_k</math>'' is closed, and ''x'' is a limit of a sequence in ''<math>C_k</math>'', ''x'' must lie in ''<math>C_k</math>''. This is true for every ''k'', and therefore the intersection of the ''<math>C_k</math>'' must contain ''x''. ∎

A converse to this theorem is also true: if ''X'' is a metric space with the property that the intersection of any nested family of non-empty closed subsets whose diameters tend to zero is non-empty, then ''X'' is a complete metric space. (To prove this, let ''<math>(x_k)</math>'' be a Cauchy sequence in ''X'', and let ''<math>C_k</math>'' be the closure of the tail of this sequence.)

== References ==