Cantor's intersection theorem

Theorem. 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 (C_k)_{k \geq 0} is a sequence of non-empty compact, closed subsets of S satisfying :C_0 \supset C_1 \supset \cdots \supset C_n \supset C_{n+1} \supset \cdots, it follows that :\bigcap_{k = 0}^\infty C_k \neq \emptyset. The closedness condition may be omitted in situations where every compact subset of S is closed, for example when S is Hausdorff. Proof. Assume, by way of contradiction, that {\textstyle \bigcap_{k = 0}^\infty C_k}=\emptyset. For each k, let U_k=C_0\setminus C_k. Since {\textstyle \bigcup_{k = 0}^\infty U_k}=C_0\setminus {\textstyle \bigcap_{k = 0}^\infty C_k} and {\textstyle \bigcap_{k = 0}^\infty C_k}=\emptyset, we have {\textstyle \bigcup_{k = 0}^\infty U_k}=C_0. Since the C_k are closed relative to S and therefore, also closed relative to C_0, the U_k, their set complements in C_0, are open relative to C_0. Since C_0\subset S is compact and \{U_k \vert k \geq 0\} is an open cover (on C_0) of C_0, a finite cover \{U_{k_1}, U_{k_2}, \ldots, U_{k_m}\} can be extracted. Let M=\max_{1\leq i\leq m} {k_i}. Then {\textstyle \bigcup_{i = 1}^m U_{k_i}}=U_M because U_1\subset U_2\subset\cdots\subset U_n\subset U_{n+1}\cdots, by the nesting hypothesis for the collection (C_k)_{k \geq 0}. Consequently, C_0={\textstyle \bigcup_{i = 1}^m U_{k_i}} = U_M. But then C_M=C_0\setminus U_M=\emptyset, a contradiction. ∎ ==Alternate version of the topological statement==

Theorem. Let S be a compact space. A sequence of non-empty nested closed subsets of S has non-empty intersection. This follows immediately from the characterization of compactness as sets for which any family of closed subsets satisfying the finite intersection property has non-trivial intersection. We can also recover the version above by noting that for subsets of C_1 , being closed via the Subspace topology with respect to C_1 and S are equivalent (because C_1 is closed). Then set S = C_1 . ==Statement for real numbers==

The theorem in real analysis draws the same conclusion for closed and bounded subsets of the set of real numbers \mathbb{R}. It states that a decreasing nested sequence (C_k)_{k \geq 0} of non-empty, closed and bounded subsets of \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 C_k=[0,1/k], the intersection over (C_k)_{k \geq 0} 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,\infty) have empty intersection. All these sequences are properly nested. This version of the theorem generalizes to \mathbf{R}^n, 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 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 (C_k)_{k \geq 0} be a sequence of non-empty, closed, and bounded subsets of \mathbb{R} satisfying :C_0 \supset C_1 \supset \cdots C_n \supset C_{n+1} \cdots. Then, :\bigcap_{k = 0}^\infty C_k \neq \emptyset. : Proof. Each nonempty, closed, and bounded subset C_k\subset\mathbb{R} admits a minimal element x_k. Since for each k, we have :x_{k+1} \in C_{k+1} \subset C_k, it follows that :x_k \le x_{k+1}, so (x_k)_{k \geq 0} is an increasing sequence contained in the bounded set C_0. The monotone convergence theorem for bounded sequences of real numbers now guarantees the existence of a limit point :x=\lim_{k\to \infty} x_k. For fixed k, x_j\in C_k for all j\geq k, and since C_k is closed and x is a limit point, it follows that x\in C_k. Our choice of k is arbitrary, hence x belongs to {\textstyle \bigcap_{k = 0}^\infty 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. Theorem. Suppose that X is a complete metric space, and (C_k)_{k \geq 1} is a sequence of non-empty closed nested subsets of X whose diameters tend to zero: :\lim_{k\to\infty} \operatorname{diam}(C_k) = 0, where \operatorname{diam}(C_k) is defined by :\operatorname{diam}(C_k) = \sup\{d(x,y) \mid x,y\in C_k\}. Then the intersection of the C_k contains exactly one point: :\bigcap_{k=1}^\infty C_k = \{x\} for some x \in X. Proof (sketch). Since the diameters tend to zero, the diameter of the intersection of the 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 x_k\in C_k for each k. Since the diameter of C_k tends to zero and the C_k are nested, the x_k form a Cauchy sequence. Since the metric space is complete this Cauchy sequence converges to some point x. Since each C_k is closed, and x is a limit of a sequence in C_k, x must lie in C_k. This is true for every k, and therefore the intersection of the C_k 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 (x_k)_{k \geq 1} be a Cauchy sequence in X, and let C_k be the closure of the tail (x_j)_{j \geq k} of this sequence.) == See also ==