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 $(C\_k)\_\{k\; \backslash geq\; 0\}$ is a sequence of non-empty compact, closed subsets of S satisfying

:$C\_0\; \backslash supset\; C\_1\; \backslash supset\; \backslash cdots\; \backslash supset\; C\_n\; \backslash supset\; C\_\{n+1\}\; \backslash supset\; \backslash cdots,$

it follows that

:$\backslash bigcap\_\{k\; =\; 0\}^\backslash infty\; C\_k\; \backslash neq\; \backslash emptyset.$

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 $\{\backslash textstyle\; \backslash bigcap\_\{k\; =\; 0\}^\backslash infty\; C\_k\}=\backslash emptyset$. For each $k$, let $U\_k=C\_0\backslash setminus\; C\_k$. Since $\{\backslash textstyle\; \backslash bigcup\_\{k\; =\; 0\}^\backslash infty\; U\_k\}=C\_0\backslash setminus\; \{\backslash textstyle\; \backslash bigcap\_\{k\; =\; 0\}^\backslash infty\; C\_k\}$ and $\{\backslash textstyle\; \backslash bigcap\_\{k\; =\; 0\}^\backslash infty\; C\_k\}=\backslash emptyset$, we have $\{\backslash textstyle\; \backslash bigcup\_\{k\; =\; 0\}^\backslash infty\; U\_k\}=C\_0$. 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 $C\_0\backslash subset\; S$ is compact and $\backslash \{U\_k\; \backslash vert\; k\; \backslash geq\; 0\backslash \}$ is an open cover (on $C\_0$) of $C\_0$, a finite cover $\backslash \{U\_\{k\_1\},\; U\_\{k\_2\},\; \backslash ldots,\; U\_\{k\_m\}\backslash \}$ can be extracted. Let $M=\backslash max\_\{1\backslash leq\; i\backslash leq\; m\}\; \{k\_i\}$. Then $\{\backslash textstyle\; \backslash bigcup\_\{i\; =\; 1\}^m\; U\_\{k\_i\}\}=U\_M$ because $U\_1\backslash subset\; U\_2\backslash subset\backslash cdots\backslash subset\; U\_n\backslash subset\; U\_\{n+1\}\backslash cdots$, by the nesting hypothesis for the collection $(C\_k)\_\{k\; \backslash geq\; 0\}$. Consequently, $C\_0=\{\backslash textstyle\; \backslash bigcup\_\{i\; =\; 1\}^m\; U\_\{k\_i\}\}\; =\; U\_M$. But then $C\_M=C\_0\backslash setminus\; U\_M=\backslash emptyset$, a contradiction. ∎

The theorem in real analysis draws the same conclusion for closed and bounded subsets of the set of real numbers $\backslash mathbb\{R\}$. It states that a decreasing nested sequence $(C\_k)\_\{k\; \backslash geq\; 0\}$ of non-empty, closed and bounded subsets of $\backslash mathbb\{R\}$ 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 $(C\_k)\_\{k\; \backslash geq\; 0\}$ is $\backslash \{0\backslash \}$. On the other hand, both the sequence of open bounded sets $C\_k=(0,1/k)$ and the sequence of unbounded closed sets $C\_k=[k,\backslash infty)$ have empty intersection. All these sequences are properly nested.

This version of the theorem generalizes to $\backslash mathbf\{R\}^n$, 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

: $C\_k\; =\; [\backslash sqrt\{2\},\; \backslash sqrt\{2\}+1/k]\; =\; (\backslash sqrt\{2\},\; \backslash sqrt\{2\}+1/k)$

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 $(C\_k)\_\{k\; \backslash geq\; 0\}$ be a sequence of non-empty, closed, and bounded subsets of $\backslash mathbb\{R\}$ satisfying

:$C\_0\; \backslash supset\; C\_1\; \backslash supset\; \backslash cdots\; C\_n\; \backslash supset\; C\_\{n+1\}\; \backslash cdots.$

Then,

:$\backslash bigcap\_\{k\; =\; 0\}^\backslash infty\; C\_k\; \backslash neq\; \backslash emptyset.$

:

Proof. Each nonempty, closed, and bounded subset $C\_k\backslash subset\backslash mathbb\{R\}$ admits a minimal element $x\_k$. Since for each $k$, we have

:$x\_\{k+1\}\; \backslash in\; C\_\{k+1\}\; \backslash subset\; C\_k$,

it follows that

:$x\_k\; \backslash le\; x\_\{k+1\}$,

so $(x\_k)\_\{k\; \backslash geq\; 0\}$ 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

:$x=\backslash lim\_\{k\backslash to\; \backslash infty\}\; x\_k.$

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

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

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

:$\backslash lim\_\{k\backslash to\backslash infty\}\; \backslash operatorname\{diam\}(C\_k)\; =\; 0,$

where $\backslash operatorname\{diam\}(C\_k)$ is defined by

:$\backslash operatorname\{diam\}(C\_k)\; =\; \backslash sup\backslash \{d(x,y)\; \backslash mid\; x,y\backslash in\; C\_k\backslash \}.$

Then the intersection of the $C\_k$ contains exactly one point:

:$\backslash bigcap\_\{k=1\}^\backslash infty\; C\_k\; =\; \backslash \{x\backslash \}$

for some $x\; \backslash in\; X$.

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 $x\_k\backslash in\; C\_k$ 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 $(x\_k)\_\{k\; \backslash geq\; 1\}$ be a Cauchy sequence in $X$, and let $C\_k$ be the closure of the tail $(x\_j)\_\{j\; \backslash geq\; k\}$ of this sequence.)

• Kuratowski's intersection theorem
• Helly's theorem - another theorem on intersection of sets.

• {{MathWorld | urlname=CantorsIntersectionTheorem | title=Cantor's Intersection Theorem}}
• Jonathan Lewin. An interactive introduction to mathematical analysis. Cambridge University Press. {{ISBN|0-521-01718-1}}. Section 7.8.

Category:Articles containing proofs

Category:Real analysis

Category:Compactness theorems

Category:Theorems in calculus