Cantor Intersection Theorem

The Cantor Intersection Theorem states that a nested sequence of nonempty compact sets has nonempty intersection. This elegant result follows from compactness and is fundamental in analysis, topology, and fractal geometry. It guarantees that "shrinking" compact sets always leave at least one point behind.

Statement

Theorem3.1Cantor Intersection Theorem

Let $(K_n)$ be a sequence of nonempty compact sets in $\mathbb{R}$ with $K_1 \supseteq K_2 \supseteq K_3 \supseteq \cdots$ (nested). Then

$\bigcap_{n=1}^\infty K_n \neq \varnothing.$

RemarkCompactness is essential

The theorem requires compactness. If the $K_n$ are merely closed (not compact), the intersection can be empty. For example, $K_n = [n, \infty)$ are closed, nested, and nonempty, but $\bigcap_{n=1}^\infty [n, \infty) = \varnothing$ .

Proof

Suppose, for contradiction, that $\bigcap_{n=1}^\infty K_n = \varnothing$ . Then

$K_1 = K_1 \setminus \bigcap_{n=1}^\infty K_n = \bigcup_{n=1}^\infty (K_1 \setminus K_n).$

Each $U_n = K_1 \setminus K_n$ is open in $K_1$ (since $K_n$ is closed and $K_1$ is a subspace). Thus $\{U_n\}_{n=1}^\infty$ is an open cover of $K_1$ .

Since $K_1$ is compact, there exist finitely many indices $n_1, \ldots, n_k$ such that

$K_1 = U_{n_1} \cup \cdots \cup U_{n_k} = (K_1 \setminus K_{n_1}) \cup \cdots \cup (K_1 \setminus K_{n_k}).$

Let $N = \max(n_1, \ldots, n_k)$ . Then $K_N \subseteq K_{n_i}$ for all $i$ (by nesting), so

$K_1 \setminus K_N \supseteq K_1 \setminus K_{n_i} \quad \text{for all } i.$

Thus $K_1 \setminus K_N \supseteq U_{n_1} \cup \cdots \cup U_{n_k} = K_1$ , which implies $K_N = \varnothing$ , contradicting the assumption that $K_N$ is nonempty.

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RemarkAlternative proof via sequences

By sequential compactness, choose $x_n \in K_n$ for each $n$ . Then $(x_n)$ is a sequence in $K_1$ , which is compact, so $(x_n)$ has a convergent subsequence $x_{n_k} \to L$ with $L \in K_1$ (by closedness of $K_1$ ). Since $K_m$ is closed and $x_{n_k} \in K_m$ for all $k$ with $n_k \geq m$ , we have $L \in K_m$ for all $m$ . Thus $L \in \bigcap K_n$ .

Applications

ExampleNested interval theorem

The Nested Interval Theorem is a special case: if $I_n = [a_n, b_n]$ with $I_{n+1} \subseteq I_n$ and each $I_n$ is nonempty, then $\bigcap_{n=1}^\infty I_n \neq \varnothing$ . (Each $[a_n, b_n]$ is compact.)

Moreover, if $|I_n| = b_n - a_n \to 0$ , then $\bigcap_{n=1}^\infty I_n$ contains exactly one point.

ExampleCantor set construction

The Cantor set $C$ is constructed by repeatedly removing middle thirds from $[0, 1]$ :

$C_0 = [0, 1], \quad C_1 = [0, 1/3] \cup [2/3, 1], \quad C_2 = [0, 1/9] \cup [2/9, 1/3] \cup [2/3, 7/9] \cup [8/9, 1], \quad \ldots$

Each $C_n$ is a finite union of closed intervals, hence compact. By Cantor's Intersection Theorem, $C = \bigcap_{n=0}^\infty C_n \neq \varnothing$ . In fact, $C$ is uncountable and has Lebesgue measure zero — a remarkable set!

ExampleLimit of shrinking sets

Let $K_n = [0, 1/n]$ . Then $(K_n)$ is a nested sequence of compact sets with $\bigcap_{n=1}^\infty K_n = \{0\}$ (a singleton). This shows the intersection need not have the same "size" as the individual sets.

ExampleFailure for open sets

Let $U_n = (0, 1/n)$ (open intervals). Then $U_1 \supseteq U_2 \supseteq \cdots$ , but $\bigcap_{n=1}^\infty (0, 1/n) = \varnothing$ . The sets must be closed (or compact) for the theorem to hold.

Summary

The Cantor Intersection Theorem ensures nested compact sets have nonempty intersection:

Essential hypothesis: compactness (or at least one compact set in the nesting).
Proof uses compactness to extract a finite subcover.
Applications: Nested Interval Theorem, Cantor set, existence proofs.

See Compactness and Heine-Borel for related results.