Cantor's intersection theorem: Difference between revisions

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Topological Statement: fixing convoluted notation of a rather simple proof.
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'''Cantor's{{short intersection theorem''' refers to two closely related theorems in [[general topology]] and [[real analysis]], named after [[Georg Cantor]], about intersections ofdescription|On decreasing nested [[sequence|sequences]] of non-empty compact sets.}}
'''Cantor's intersection theorem''',<ref>{{Cite web |last=Weisstein |first=Eric W. |title=Cantor's Intersection Theorem |url=https://mathworld.wolfram.com/CantorsIntersectionTheorem.html |access-date=2025-06-20 |website=mathworld.wolfram.com |language=en}}</ref> also called '''Cantor's nested intervals theorem''',<ref>{{Cite book |last=Segura |first=Julio |url=https://www.google.com.br/books/edition/An_Eponymous_Dictionary_of_Economics/Z6Oy4L-6LSwC |title=An Eponymous Dictionary of Economics: A Guide to Laws and Theorems Named After Economists |last2=Braun |first2=Carlos Rodríguez |date=2004-01-01 |publisher=Edward Elgar Publishing |isbn=978-1-84542-360-5 |pages=38 |language=en}}</ref><ref>{{Cite book |last=Denlinger |first=Charles G. |url=https://www.google.com.br/books/edition/Elements_of_Real_Analysis/CeTkVSXlj4cC |title=Elements of Real Analysis |date=2010-05-08 |publisher=Jones & Bartlett Publishers |isbn=978-1-4496-5993-6 |pages=103 |language=en}}</ref> refers to two closely related theorems in [[general topology]] and [[real analysis]], named after [[Georg Cantor]], about intersections of decreasing nested [[sequence]]s of non-empty compact sets.
 
==Topological Statementstatement==
'''Theorem.''' ''Let <math> S </math> 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 (''C''<submath>''(C_k)_{k'' \geq 0}</submath>) is a sequence of non-empty compact, closed subsets of <math>S</math> satisfying''
 
:<math>C_0 \supset C_1 \supset \cdots C_k\supset C_n \supset C_{kn+1} \supset \cdots, </math>
 
''it follows that''
 
:<math>\left(\bigcap_{k = 0}^\infty C_k\right) \neq \emptyset. </math>
 
''Note'':The Weclosedness condition may leavebe out the closedness conditionomitted 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>\bigcap C_n=\emptyset</math>. For each <math>n</math>, 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>, we have <math>\bigcup U_n=C_0</math>.
 
Since'''Proof.''' Assume, by way of contradiction, that <math>C_0{\subsettextstyle S</math>\bigcap_{k is= compact and0}^\infty <math>(U_n)C_k}=\emptyset</math>. isFor an open cover (oneach <math>C_0k</math>), oflet <math>U_k=C_0\setminus C_k</math>,. we can extract a finite coverSince <math>{\textstyle \bigcup_{U_{n_1k = 0},^\infty U_{n_2U_k},=C_0\setminus {\ldots,textstyle U_\bigcap_{n_mk = 0}^\infty C_k}</math>. Letand <math>U_M{\textstyle \bigcap_{k = 0}^\infty C_k}=\emptyset</math>, be the set such thatwe have <math>U_{n_i\textstyle \bigcup_{k = 0}^\subsetinfty U_MU_k}=C_0</math>. forSince the <math>i=1,2,\ldots, mC_k</math>, whichare closed existsrelative becauseto <math>U_0\subset U_1\subset\cdots\subset U_k\subset U_{k+1}\cdotsS</math> and therefore, byalso theclosed orderingrelative hypothesisto on<math>C_0</math>, the collection <math> (C_n).U_k</math>, their Consequently,set complements in <math>C_0=\bigcup U_{n_i}\subset U_M</math>., Butare thenopen relative to <math>C_M=C_0\setminus U_M=\emptyset</math>, a contradiction.
 
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.|∎]]
==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 '''R'''. It states that a decreasing nested sequence (''C''<sub>''k''</sub>) of non-empty, closed and bounded subsets of '''R''' has a non-empty intersection.
==Statement for Realreal Numbersnumbers==
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>\mathbb{R'''}</math>. It states that a decreasing nested sequence (''C''<submath>''(C_k)_{k'' \geq 0}</submath>) of non-empty, closed and bounded subsets of '''<math>\mathbb{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>''(C_k)_{k'' \geq 0}</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 ''<math>n''</math>-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.
 
'''Theorem.''' ''Let'' <math>(C_k)_{k \geq 0}</math> ''be a sequence of non-empty, closed, and bounded subsets of'' <math>\mathbb{R}</math> ''satisfying''
=== Proof ===
 
:<math>C_0 \supset C_1 \supset \cdots C_n \supset C_{n+1} \cdots. </math>
Each closed bounded non-empty subset ''C''<sub>''k''</sub> of '''R''' admits a minimal element ''x''<sub>''k''</sub>. Since for each ''k'', we have
 
:<math>x_{k+1} \in C_{k+1} \subseteq C_k</math>,
''Then,''
 
:<math>\bigcap_{nk =1 0}^\infty C_nC_k =\neq \{x\}emptyset. </math>
:
 
''Proof.'' Each nonempty, closed, and bounded non-empty subset ''C''<submath>''k''C_k\subset\mathbb{R}</submath> of '''R''' admits a minimal element ''x''<submath>''k''x_k</submath>. Since for each ''<math>k''</math>, we have
 
:<math>x_{k+1} \in C_{k+1} \subseteqsubset C_k</math>,
it follows that
:<math>x_k \le x_{k+1}</math>,
 
so (''x''<submath>''(x_k)_{k'' \geq 0}</submath>) is an increasing sequence contained in the bounded set ''C''<submath>0C_0</submath>. The [[Monotonemonotone convergence theorem]] for bounded sequences of real numbers now guarantees the existence of a [[Limit of a sequence|limit point]]
 
:<math>x=\lim_{k\to \infty} x_k.</math>
 
For fixed ''k'', we have that ''x''<sub>''j''</sub>∈''C''<sub>''k''</sub> for all ''j''≥''k'' and since ''C''<sub>''k''</sub> was closed, it follows that ''x''∈''C''<sub>''k''</sub>.
For fixed <math>k</math>, <math>x_j\in C_k</math> for all <math>j\geq k</math>, and since <math>C_k</math> is closed and <math>x</math> is a limit point, it follows that <math>x\in C_k</math>. Our choice of ''<math>k''</math> wasis arbitrary, hence ''<math>x''</math> belongs to the<math>{\textstyle intersection\bigcap_{k of= all0}^\infty ''C''<sub>''k''C_k}</submath> 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 ''<math>X''</math> is a complete metric space, and ''C''<submath>''n''(C_k)_{k \geq 1}</submath> is a sequence of non-empty closed nested subsets of ''<math>X''</math> whose [[diameter]]s tend to zero:''
 
:<math>\lim_{nk\to\infty} \operatorname{diam}(C_nC_k) = 0,</math>
where diam(''C''<sub>''n''</sub>) is defined by
 
:<math>\operatorname{diam}(C_n) = \sup\{d(x,y) | x,y\in C_n\}.</math>
''where <math>\operatorname{diam}(C_k)</math> is defined by''
Then the intersection of the ''C''<sub>''n''</sub> contains exactly one point:
 
:<math>\bigcap_{n=1}^\infty C_n = \{x\}</math>
:<math>\operatorname{diam}(C_nC_k) = \sup\{d(x,y) |\mid x,y\in C_nC_k\}.</math>
for some ''x'' in ''X''.
 
''Then the intersection of the ''C''<submath>''n''C_k</submath> contains exactly one point:''
 
:<math>\bigcap_{k=1}^\infty C_k = \{x\}</math>
 
''for some ''<math>x'' \in ''X</math>.''.
 
A''Proof proof goes as follows(sketch). '' 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''<submath>''n''x_k\in C_k</submath> offor each ''C''<submath>''n''k</submath> for each ''n''. 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 ''<math>x''</math>. Since each ''C''<submath>''n''C_k</submath> is closed, and ''<math>x''</math> is a limit of a sequence in ''C''<submath>''n''C_k</submath>, ''<math>x''</math> must lie in ''C''<submath>''n''C_k</submath>. This is true for every ''n''<math>k</math>, and therefore the intersection of the ''C''<submath>''n''C_k</submath> must contain ''<math>x''</math>.
 
A converse to this theorem is also true: if ''<math>X''</math> 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 ''<math>X''</math> is a complete metric space. (To prove this, let ''x''<submath>''n''(x_k)_{k \geq 1}</submath> be a Cauchy sequence in ''<math>X''</math>, and let ''C''<submath>''n''C_k</submath> be the closure of the tail <math>(x_j)_{j \geq k}</math> of this sequence.)
=== Proof ===
 
== See also ==
A proof goes as follows. Since the diameters tend to zero, the diameter of the intersection of the ''C''<sub>''n''</sub> 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''</sub> of ''C''<sub>''n''</sub> for each ''n''. Since the diameter of ''C''<sub>''n''</sub> tends to zero and the ''C''<sub>''n''</sub> are nested, the ''x''<sub>''n''</sub> form a Cauchy sequence. Since the metric space is complete this Cauchy sequence converges to some point ''x''. Since each ''C''<sub>''n''</sub> is closed, and ''x'' is a limit of a sequence in ''C''<sub>''n''</sub>, ''x'' must lie in ''C''<sub>''n''</sub>. This is true for every ''n'', and therefore the intersection of the ''C''<sub>''n''</sub> must contain ''x''.
 
* [[Kuratowski's intersection theorem]]
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''<sub>''n''</sub> be a Cauchy sequence in ''X'', and let ''C''<sub>''n''</sub> be the closure of the tail of this sequence.)
* [[Helly's theorem]] - another theorem on intersection of sets.
 
== References ==
{{Reflist}}
* {{MathWorld | urlname=CantorsIntersectionTheorem | title=Cantor's Intersection Theorem}}
* Jonathan Lewin. An interactive introduction to mathematical analysis. Cambridge University Press. {{ISBN|0-521-01718-1}}. Section 7.8.