Cantor's intersection theorem: Difference between revisions

Revision as of 05:36, 15 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 <math>S</math> 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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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==

The theorem in real analysis draws the same conclusion for [[closed set|closed]] and [[bounded set|bounded]] subsets of the set of [[real number]]s '''<math>\mathbf{R'''}</math>. It states that a decreasing nested sequence (''C''<submath>''k''(C_k)</submath>) of non-empty, closed and bounded subsets of '''<math>\mathbf{R'''}</math> has a non-empty intersection.

This version follows from the general topological statement in light of the [[Heine&ndash;Borel theorem]], which states that sets of real numbers are compact if and only if they are closed and bounded. However, it is typically used as a lemma in proving said theorem, and therefore warrants a separate proof.

As an example, if ''C''<submath>''k''</sub> C_k=&nbsp;[0,&nbsp;1/''k'']</math>, the intersection over {''C''<submath>''k''(C_k)</submath>} is&nbsp;<math>\{0\}</math>. On the other hand, both the sequence of open bounded sets ''C''<submath>''k''</sub> C_k=&nbsp;(0,&nbsp;1/''k'')</math> and the sequence of unbounded closed sets ''C''<submath>''k''</sub> C_k=&nbsp;[''k'',&nbsp;∞\infty)</math> have empty intersection. All these sequences are properly nested.

This version of the theorem generalizes to '''R'''<supmath>''\mathbf{R}^n''</supmath>, the set of ''n''-element vectors of real numbers, but does not generalize to arbitrary [[metric space]]s. For example, in the space of [[rational number]]s, the sets

: <math>C_k = [\sqrt{2}, \sqrt{2}+1/k] = (\sqrt{2}, \sqrt{2}+1/k)</math>

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are closed and bounded, but their intersection is empty.

Note that this contradicts neither the topological statement, as the sets ''C''<submath>''k''C_k</submath> are not compact, nor the variant below, as the rational numbers are not complete with respect to the usual metric.

A simple corollary of the theorem is that the [[Cantor set]] is nonempty, since it is defined as the intersection of a decreasing nested sequence of sets, each of which is defined as the union of a finite number of closed intervals; hence each of these sets is non-empty, closed, and bounded. In fact, the Cantor set contains uncountably many points.

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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. Suppose that ''X'' is a complete metric space, and ''C''_''n'' is a sequence of non-empty closed nested subsets of ''X'' whose [[diameter]]s tend to zero:

:<math>\lim_{n\to\infty} \operatorname{diam}(C_n) = 0</math>

where diam(''C''_''n'') is defined by

:<math>\operatorname{diam}(C_n) = \sup\{d(x,y) | x,y\in C_n\}.</math>

Then the intersection of the ''C''_''n'' contains exactly one point:

:<math>\bigcap_{n=1}^\infty C_n = \{x\}</math>

for some ''x'' in ''X''.

'''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:''

=== Proof ===

:<math>\lim_{nk\to\infty} \operatorname{diam}(C_nC_k) = 0,</math>

''where <math>\operatorname{diam}(C_k)</math> is defined by''

:<math>\operatorname{diam}(C_nC_k) = \sup\{d(x,y) | x,y\in C_nC_k\}.</math>

''Then the intersection of the ''C ''<submath>''n''C_k</submath> contains exactly one point:''

:<math>\bigcap_{nk=1}^\infty C_nC_k = \{x\}</math>

''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 ''C''<submath>''n''C_k</submath>'' 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 ''x''<sub>''n''</submath>x_k\in of ''C''<sub>''n''C_k</submath> for each ''nk''. Since the diameter of ''C''<submath>''n''C_k</submath>'' tends to zero and the ''C''<submath>''n''C_k</submath>'' are nested, the ''x''<submath>''n''x_k</submath>'' form a Cauchy sequence. Since the metric space is complete this Cauchy sequence converges to some point ''x''. Since each ''C''<submath>''n''C_k</submath>'' is closed, and ''x'' is a limit of a sequence in ''C''<submath>''n''C_k</submath>'', ''x'' must lie in ''C''<submath>''n''C_k</submath>''. This is true for every ''nk'', and therefore the intersection of the ''C''<submath>''n''C_k</submath>'' 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 ''x''<submath>''n''(x_k)</submath>'' be a Cauchy sequence in ''X'', and let ''C''<submath>''n''C_k</submath>'' be the closure of the tail of this sequence.)

== References ==