Cantor's intersection theorem

Cantor's intersection theorem refers to two closely related theorems in general topology and real analysis, named after Georg Cantor, about intersections of decreasing nested sequences of non-empty compact sets.

Topological Statement

Let $S$ be a topological space. A decreasing nested sequence of non-empty compact, closed subsets of $S$ has a non-empty intersection. In other words, supposing (Ck) is a sequence of non-empty compact, closed subsets of $S$ satisfying

$C_{0}\supseteq C_{1}\supseteq \cdots C_{k}\supseteq C_{k+1}\cdots ,$ it follows that

$\left(\bigcap _{k}C_{k}\right)\neq \emptyset .$ Note: We may rule out the closed condition provided that every compact subset of $S$ is closed, for example when $S$ is Hausdorff.

Proof

Assume, by way of contradiction, that $\bigcap C_{n}=\emptyset$ . For each n, let $U_{n}=C_{0}\setminus C_{n}$ . Since $\bigcup U_{n}=C_{0}\setminus \bigcap C_{n}$ and $\bigcap C_{n}=\emptyset$ , we have $\bigcup U_{n}=C_{0}$ .

Since $C_{0}$ $S$ is compact and $(U_{n})$ is an open cover (on $C_{0}$ ) of $C_{0}$ , we can extract a finite cover $\{U_{n_{1}},U_{n_{2}},\ldots ,U_{n_{m}}\}$ . Let $U_{k}$ be the largest set of this cover, which exists by the ordering hypothesis on the collection $(C_{n}).$ Then $C_{0}$ $U_{k}$ . But then $C_{k}=C_{0}\setminus U_{k}=\emptyset$ , a contradiction.

Statement for Real Numbers

The theorem in real analysis draws the same conclusion for closed and bounded subsets of the set of real numbers R. It states that a decreasing nested sequence (Ck) of non-empty, closed and bounded subsets of 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 Ck = [0, 1/k], the intersection over {Ck} is {0}. On the other hand, both the sequence of open bounded sets Ck = (0, 1/k) and the sequence of unbounded closed sets Ck = [k, ∞) have empty intersection. All these sequences are properly nested.

This version of the theorem generalizes to Rn, the set of n-element vectors of real numbers, but does not generalize to arbitrary metric spaces. For example, in the space of rational numbers, the sets

$C_{k}=[{\sqrt {2}},{\sqrt {2}}+1/k]=({\sqrt {2}},{\sqrt {2}}+1/k)$ are closed and bounded, but their intersection is empty.

Note that this contradicts neither the topological statement, as the sets Ck 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.

Proof

Each closed bounded non-empty subset Ck of R admits a minimal element xk. Since for each k, we have

$x_{k+1}\in C_{k+1}\subseteq C_{k}$ ,

it follows that

$x_{k}\leq x_{k+1}$ ,

so (xk) is an increasing sequence contained in the bounded set C0. The Monotone convergence theorem now guarantees the existence of a limit point

$x=\lim _{k\to \infty }x_{k}.$ For fixed k, we have that xjCk for all jk and since Ck was closed, it follows that xCk. Our choice of k was arbitrary, hence x belongs to the intersection of all Ck 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 Cn is a sequence of non-empty closed nested subsets of X whose diameters tend to zero:

$\lim _{n\to \infty }\operatorname {diam} (C_{n})=0$ where diam(Cn) is defined by

$\operatorname {diam} (C_{n})=\sup\{d(x,y)|x,y\in C_{n}\}.$ Then the intersection of the Cn contains exactly one point:

$\bigcap _{n=1}^{\infty }C_{n}=\{x\}$ for some x in X.

Proof

A proof goes as follows. Since the diameters tend to zero, the diameter of the intersection of the Cn 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 xn of Cn for each n. Since the diameter of Cn tends to zero and the Cn are nested, the xn form a Cauchy sequence. Since the metric space is complete this Cauchy sequence converges to some point x. Since each Cn is closed, and x is a limit of a sequence in Cn, x must lie in Cn. This is true for every n, and therefore the intersection of the Cn 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 xn be a Cauchy sequence in X, and let Cn be the closure of the tail of this sequence.)