Theorem. Let ${\displaystyle S}$  be a topological space. A decreasing nested sequence of non-empty compact, closed subsets of ${\displaystyle S}$  has a non-empty intersection. In other words, supposing ${\displaystyle (C_{k})_{k\geq 0}}$  is a sequence of non-empty compact, closed subsets of S satisfying

${\displaystyle C_{0}\supset C_{1}\supset \cdots \supset C_{n}\supset C_{n+1}\supset \cdots ,}$ 

it follows that

${\displaystyle \bigcap _{k=0}^{\infty }C_{k}\neq \emptyset .}$ 

The closedness condition may be omitted in situations where every compact subset of ${\displaystyle S}$  is closed, for example when ${\displaystyle S}$  is Hausdorff.

Proof. Assume, by way of contradiction, that ${\displaystyle {\textstyle \bigcap _{k=0}^{\infty }C_{k}}=\emptyset }$ . For each ${\displaystyle k}$ , let ${\displaystyle U_{k}=C_{0}\setminus C_{k}}$ . Since ${\displaystyle {\textstyle \bigcup _{k=0}^{\infty }U_{k}}=C_{0}\setminus {\textstyle \bigcap _{k=0}^{\infty }C_{k}}}$  and ${\displaystyle {\textstyle \bigcap _{k=0}^{\infty }C_{k}}=\emptyset }$ , we have ${\displaystyle {\textstyle \bigcup _{k=0}^{\infty }U_{k}}=C_{0}}$ . Since the ${\displaystyle C_{k}}$  are closed relative to ${\displaystyle S}$  and therefore, also closed relative to ${\displaystyle C_{0}}$ , the ${\displaystyle U_{k}}$ , their set complements in ${\displaystyle C_{0}}$ , are open relative to ${\displaystyle C_{0}}$ .

Since ${\displaystyle C_{0}\subset S}$  is compact and ${\displaystyle \{U_{k}\vert k\geq 0\}}$  is an open cover (on ${\displaystyle C_{0}}$ ) of ${\displaystyle C_{0}}$ , a finite cover ${\displaystyle \{U_{k_{1}},U_{k_{2}},\ldots ,U_{k_{m}}\}}$  can be extracted. Let ${\displaystyle M=\max _{1\leq i\leq m}{k_{i}}}$ . Then ${\displaystyle {\textstyle \bigcup _{i=1}^{m}U_{k_{i}}}=U_{M}}$  because ${\displaystyle U_{1}\subset U_{2}\subset \cdots \subset U_{n}\subset U_{n+1}\cdots }$ , by the nesting hypothesis for the collection ${\displaystyle (C_{k})_{k\geq 0}}$ . Consequently, ${\displaystyle C_{0}={\textstyle \bigcup _{i=1}^{m}U_{k_{i}}}=U_{M}}$ . But then ${\displaystyle C_{M}=C_{0}\setminus U_{M}=\emptyset }$ , a contradiction. ∎

The theorem in real analysis draws the same conclusion for closed and bounded subsets of the set of real numbers ${\displaystyle \mathbb {R} }$ . It states that a decreasing nested sequence ${\displaystyle (C_{k})_{k\geq 0}}$  of non-empty, closed and bounded subsets of ${\displaystyle \mathbb {R} }$  has a non-empty intersection.

This version follows from the general topological statement in light of the Heine–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 ${\displaystyle C_{k}=[0,1/k]}$ , the intersection over ${\displaystyle (C_{k})_{k\geq 0}}$  is ${\displaystyle \{0\}}$ . On the other hand, both the sequence of open bounded sets ${\displaystyle C_{k}=(0,1/k)}$  and the sequence of unbounded closed sets ${\displaystyle C_{k}=[k,\infty )}$  have empty intersection. All these sequences are properly nested.

This version of the theorem generalizes to ${\displaystyle \mathbf {R} ^{n}}$ , the set of ${\displaystyle n}$ -element vectors of real numbers, but does not generalize to arbitrary metric spaces. For example, in the space of rational numbers, the sets

${\displaystyle C_{k}=[{\sqrt {2}},{\sqrt {2}}+1/k]=({\sqrt {2}},{\sqrt {2}}+1/k)}$ 

are closed and bounded, but their intersection is empty.

Note that this contradicts neither the topological statement, as the sets ${\displaystyle C_{k}}$  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 ${\displaystyle (C_{k})_{k\geq 0}}$  be a sequence of non-empty, closed, and bounded subsets of ${\displaystyle \mathbb {R} }$  satisfying

${\displaystyle C_{0}\supset C_{1}\supset \cdots C_{n}\supset C_{n+1}\cdots .}$ 

Then,

${\displaystyle \bigcap _{k=0}^{\infty }C_{k}\neq \emptyset .}$ 

Proof. Each nonempty, closed, and bounded subset ${\displaystyle C_{k}\subset \mathbb {R} }$  admits a minimal element ${\displaystyle x_{k}}$ . Since for each ${\displaystyle k}$ , we have

${\displaystyle x_{k+1}\in C_{k+1}\subset C_{k}}$ ,

it follows that

${\displaystyle x_{k}\leq x_{k+1}}$ ,

so ${\displaystyle (x_{k})_{k\geq 0}}$  is an increasing sequence contained in the bounded set ${\displaystyle C_{0}}$ . The monotone convergence theorem for bounded sequences of real numbers now guarantees the existence of a limit point

${\displaystyle x=\lim _{k\to \infty }x_{k}.}$ 

For fixed ${\displaystyle k}$ , ${\displaystyle x_{j}\in C_{k}}$  for all ${\displaystyle j\geq k}$ , and since ${\displaystyle C_{k}}$  is closed and ${\displaystyle x}$  is a limit point, it follows that ${\displaystyle x\in C_{k}}$ . Our choice of ${\displaystyle k}$  is arbitrary, hence ${\displaystyle x}$  belongs to ${\displaystyle {\textstyle \bigcap _{k=0}^{\infty }C_{k}}}$  and the proof is complete. ∎

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

Theorem. Suppose that ${\displaystyle X}$  is a complete metric space, and ${\displaystyle (C_{k})_{k\geq 1}}$  is a sequence of non-empty closed nested subsets of ${\displaystyle X}$  whose diameters tend to zero:

${\displaystyle \lim _{k\to \infty }\operatorname {diam} (C_{k})=0,}$ 

where ${\displaystyle \operatorname {diam} (C_{k})}$  is defined by

${\displaystyle \operatorname {diam} (C_{k})=\sup\{d(x,y)\mid x,y\in C_{k}\}.}$ 

Then the intersection of the ${\displaystyle C_{k}}$  contains exactly one point:

${\displaystyle \bigcap _{k=1}^{\infty }C_{k}=\{x\}}$ 

for some ${\displaystyle x\in X}$ .

Proof (sketch). Since the diameters tend to zero, the diameter of the intersection of the ${\displaystyle C_{k}}$  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 ${\displaystyle x_{k}\in C_{k}}$  for each ${\displaystyle k}$ . Since the diameter of ${\displaystyle C_{k}}$  tends to zero and the ${\displaystyle C_{k}}$  are nested, the ${\displaystyle x_{k}}$  form a Cauchy sequence. Since the metric space is complete this Cauchy sequence converges to some point ${\displaystyle x}$ . Since each ${\displaystyle C_{k}}$  is closed, and ${\displaystyle x}$  is a limit of a sequence in ${\displaystyle C_{k}}$ , ${\displaystyle x}$  must lie in ${\displaystyle C_{k}}$ . This is true for every ${\displaystyle k}$ , and therefore the intersection of the ${\displaystyle C_{k}}$  must contain ${\displaystyle x}$ . ∎

A converse to this theorem is also true: if ${\displaystyle X}$  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 ${\displaystyle X}$  is a complete metric space. (To prove this, let ${\displaystyle (x_{k})_{k\geq 1}}$  be a Cauchy sequence in ${\displaystyle X}$ , and let ${\displaystyle C_{k}}$  be the closure of the tail ${\displaystyle (x_{j})_{j\geq k}}$  of this sequence.)

• Kuratowski's intersection theorem

• Weisstein, Eric W. "Cantor's Intersection Theorem". MathWorld.
• Jonathan Lewin. An interactive introduction to mathematical analysis. Cambridge University Press. ISBN 0-521-01718-1. Section 7.8.