Cantor's intersection theorem

Topological Statement

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

it follows that

Proof

Assume, by way of contradiction, that ⋂ C n = ∅ . For each n, let U n = C 0 ∖ C n . Since ⋃ U n = C 0 ∖ ⋂ C n and ⋂ C n = ∅ , thus ⋃ U n = C 0 .

Since C 0 ⊂ S 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 C 0 ⊂ U k . But then C k = C 0 ∖ U k = ∅ , 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 (C_k) 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 C_k = [0, 1/k], the intersection over {C_k} is {0}. On the other hand, both the sequence of open bounded sets C_k = (0, 1/k) and the sequence of unbounded closed sets C_k = [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

are closed and bounded, but their intersection is empty.

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

Proof

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

it follows that

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

For fixed k, we have that x_j∈C_k for all j≥k and since C_k was closed, it follows that x∈C_k. Our choice of k was arbitrary, hence x belongs to the intersection of all C_k 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 non-empty complete metric space, and C_n is a sequence of closed nested subsets of X whose diameters tend to zero:

where diam(C_n) is defined by

Then the intersection of the C_n contains exactly one point:

for some x in X.

Proof

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