# Cantor's intersection theorem

In real analysis, a branch of mathematics, Cantor's intersection theorem, named after Georg Cantor, is a theorem related to compact sets of a compact space $S$. It states that a decreasing nested sequence of non-empty compact subsets of $S$ has nonempty intersection. In other words, supposing {Ck} is a sequence of non-empty, closed and bounded sets satisfying

$C_0 \supseteq C_1 \supseteq \cdots C_k \supseteq C_{k+1} \cdots, \,$

it follows that

$\left(\bigcap_{k} C_k\right) \neq \emptyset. \,$

The result is typically used as a lemma in proving the Heine–Borel theorem, which states that sets of real numbers are compact if and only if they are closed and bounded. Conversely, if the Heine–Borel theorem is known, then it can be restated as: a decreasing nested sequence of non-empty, compact subsets of a compact space has nonempty intersection.

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.

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.

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

Assume, by way of contradiction, that $\bigcap C_n=\emptyset$. For each n, let $U_n=X\setminus C_n$. Since $\bigcup U_n=X\setminus\bigcap C_n$ and $\bigcap C_n=\emptyset$, thus $\bigcup U_n=X$.

Since $X$ is compact and $(U_n)$ is an open cover of it, we can extract a finite cover. Let $U_k$ be the largest set of this cover; then $X=U_k$. But then $C_k=X\setminus U_k=\emptyset$, a contradiction.

## Variant in complete metric spaces

In a complete metric space, the following variant of Cantor's intersection theorem holds. Suppose that X is a non-empty complete metric space, and Cn is a sequence of 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:

$\cap_{n=1}^\infty C_n = \{x\}$

for some x in X.

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 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.)

## References

• Jonathan Lewin. An interactive introduction to mathematical analysis. Cambridge University Press. ISBN 0-521-01718-1. Section 7.8.