Cantor's intersection theorem: Difference between revisions

Revision as of 20:41, 3 March 2021

Revision as of 20:48, 3 March 2021

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==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>\mathbb{R}</math>. It states that a decreasing nested sequence <math>(C_k)_{k \geq 0}</math> 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 <math>C_k=[0,1/k]</math>, the intersection over <math>(C_k)_{k \geq 0}</math> is&nbsp;<math>\{0\}</math>. On the other hand, both the sequence of open bounded sets <math>C_k=(0,1/k)</math> and the sequence of unbounded closed sets <math>C_k=[k,\infty)</math> have empty intersection. All these sequences are properly nested.

This version of the theorem generalizes to <math>\mathbf{R}^n</math>, 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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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 familysequence of non-empty, closed, and bounded subsets of'' <math>\mathbb{R}</math> ''satisfying''

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

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''Then,''

:<math>\left(\bigcap_{k = 0}^\infty C_k\right) \neq \emptyset. </math>

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''Proof.'' Each nonempty, closed, and bounded subset <math>C_k\subset\mathbb{R}</math> admits a minimal element <math>x_k</math>. 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 <math>(x_k)_{k \geq 0}</math> is an increasing sequence contained in the bounded set <math>C_0</math>. The [[monotone convergence theorem]] for bounded sequences of real numbers now guarantees the existence of a limit point

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

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> wasis 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 ''<math>{\textstyle \bigcap_{k = 0}^\bigcap_kinfty C_k}</math>'' and the proof is complete. ∎

== Variant in complete metric spaces ==