V.8 · Compactness & Connectedness — Problem 8 #

← Back to V. Compactness & Connectedness, Problem 8

Problem. Let \(K_1 \supseteq K_2 \supseteq \ldots\) be non-empty compact subsets of a metric space. (a) Show that \(\bigcap_{n=1}^\infty K_n \neq \varnothing\). (b) Show by example that this can fail if the \(K_n\) are only closed. (c) Show by example that it can fail if the \(K_n\) are only bounded.

Hint #

Suppose the intersection is empty and let \(U_n = S \setminus K_n\). Each \(U_n\) is open because compact sets are closed. Every point of \(K_1\) is missing from some \(K_n\), so the \(U_n\) cover \(K_1\). The \(U_n\) increase with \(n\), so a finite subcover is really just its largest member. What does \(K_1 \subseteq U_N\) say about \(K_N\)? For (b) and (c), look at half-lines and at shrinking open intervals.

Solution #

(a) Suppose, for contradiction, that \(\bigcap_{n} K_n = \varnothing\). Let \(U_n = S \setminus K_n\). Since compact sets are closed, each \(U_n\) is open, and since the \(K_n\) decrease, the \(U_n\) increase: \(U_1 \subseteq U_2 \subseteq \ldots\). If \(x \in K_1\), then \(x\) is not in every \(K_n\), so \(x \in U_n\) for some \(n\); hence \(\{U_n\}\) is an open cover of \(K_1\). By compactness, finitely many \(U_{n_1}, \ldots, U_{n_k}\) cover \(K_1\), and because the sets are increasing their union is \(U_N\) with \(N = \max\{n_1, \ldots, n_k\}\). Thus \(K_1 \subseteq U_N = S \setminus K_N\). But \(K_N \subseteq K_1\), so \(K_N \subseteq S \setminus K_N\), which forces \(K_N = \varnothing\), contradicting the hypothesis that every \(K_n\) is non-empty.

(b) In \((\mathbb{R}, |\cdot|)\) let \(K_n = [n, \infty)\). Each is closed and non-empty, and \(K_{n+1} \subseteq K_n\). But no real number \(x\) lies in every \(K_n\), since \(x \notin K_n\) once \(n > x\). The sets are not bounded, so not compact, and the conclusion fails.

(c) In \((\mathbb{R}, |\cdot|)\) let \(K_n = (0, \tfrac{1}{n})\). Each is bounded and non-empty, and \(K_{n+1} \subseteq K_n\). But if \(x > 0\) then \(x \notin K_n\) once \(n > 1/x\), and \(x \leq 0\) is in none of them; so the intersection is empty. The sets are not closed, so not compact.

Long solution #

Strategy. The natural first attempt, “pick \(x_n \in K_n\) and take a limit,” needs tools about sequences that come later in the course. The open-cover definition gives a cleaner argument: assuming the intersection is empty, the complements of the \(K_n\) form an open cover of \(K_1\), and a finite subcover of an increasing chain is just its biggest member. This forces one of the \(K_n\) to be empty. The counterexamples in (b) and (c) show that each half of “closed and bounded” is needed: dropping boundedness lets the sets run off to infinity, and dropping closedness lets them shrink to a point that none of them contains.

Facts used. (1) Every compact set in a metric space is closed (proved in class and in Problem V.3). (2) De Morgan: \(x \notin \bigcap_n K_n\) iff \(x \notin K_n\) for some \(n\), i.e. \(\big(\bigcap_n K_n\big)^c = \bigcup_n K_n^c\). (3) A union of finitely many members of an increasing chain of sets equals the largest of them.

Part (a). Assume for contradiction that \(\bigcap_{n=1}^\infty K_n = \varnothing\). Define \(U_n = S \setminus K_n = K_n^c\).

Step 1: the \(U_n\) are open. \(K_n\) is compact, hence closed by fact (1), so its complement \(U_n\) is open.

Step 2: the \(U_n\) increase. If \(x \in U_n\) then \(x \notin K_n \supseteq K_{n+1}\), so \(x \notin K_{n+1}\), i.e. \(x \in U_{n+1}\). Hence \(U_1 \subseteq U_2 \subseteq U_3 \subseteq \ldots\)

Step 3: the \(U_n\) cover \(K_1\). Let \(x \in K_1\). Since the intersection is empty, \(x \notin \bigcap_n K_n\), so by fact (2) there is some \(n\) with \(x \notin K_n\), i.e. \(x \in U_n\). Thus \(K_1 \subseteq \bigcup_n U_n\), and \(\{U_n\}_{n \in \mathbb{N}}\) is an open cover of \(K_1\).

Step 4: extract a finite subcover. \(K_1\) is compact, so there are indices \(n_1 < n_2 < \cdots < n_k\) with \(K_1 \subseteq U_{n_1} \cup \cdots \cup U_{n_k}\). By Step 2 and fact (3), this union equals \(U_{n_k}\). Writing \(N = n_k\),

Step 5: the contradiction. Since the \(K_n\) are nested, \(K_N \subseteq K_1\), and so \(K_N \subseteq S \setminus K_N\). A set contained in its own complement is empty: \(K_N = \varnothing\). This contradicts the assumption that every \(K_n\) is non-empty. Hence \(\bigcap_{n=1}^\infty K_n \neq \varnothing\). \(\blacksquare\)

What was really used. Only \(K_1\) needed to be compact; the remaining \(K_n\) were used only through their closedness (Step 1) and non-emptiness (Step 5). So the theorem holds for a decreasing sequence of non-empty closed subsets of a single compact set. Compare Problem V.3, where likewise one compact set carried the whole argument.

Part (b): closed but not bounded. Take \(K_n = [n, \infty) \subseteq \mathbb{R}\).

• Closed: the complement \((-\infty, n)\) is open.
• Non-empty and nested: \(n \in K_n\), and \([n+1, \infty) \subseteq [n, \infty)\).
• Empty intersection: given any real \(x\), the Archimedean property provides \(n \in \mathbb{N}\) with \(n > x\), and then \(x \notin K_n\).

The \(K_n\) are not bounded, hence not compact, and the conclusion of (a) fails. The sets have “escaped to infinity.”

Part (c): bounded but not closed. Take \(K_n = (0, \tfrac{1}{n}) \subseteq \mathbb{R}\).

• Bounded: \(K_n \subseteq (0,1)\).
• Non-empty and nested: \(\tfrac{1}{2n} \in K_n\), and \((0, \tfrac{1}{n+1}) \subseteq (0, \tfrac{1}{n})\).
• Empty intersection: if \(x \leq 0\) then \(x\) is in no \(K_n\); if \(x > 0\), choose \(n > 1/x\), so \(\tfrac{1}{n} < x\) and \(x \notin K_n\).

The \(K_n\) are not closed, since \(0\) is a limit point of each that none contains, hence not compact. The sets “shrink toward a missing point.” Closing them up, i.e. replacing \(K_n\) by \([0, \tfrac{1}{n}]\), makes them compact and the intersection becomes \(\{0\}\), exactly as (a) predicts.

Remarks. (i) Part (a) is the Cantor intersection theorem. In \(\mathbb{R}\) with \(K_n = [a_n, b_n]\) closed intervals, it is the nested interval property, which is equivalent to the least upper bound property. (ii) This is the result lurking behind the Cantor-set problem from past exams: the Cantor set is \(\bigcap_n S_n\) with each \(S_n\) a finite union of closed intervals, and (a) shows immediately that it is non-empty, while Problem V.3 shows it is compact. (iii) If in addition the diameters of the \(K_n\) tend to \(0\), the intersection is a single point: two points in it would be at some positive distance \(d\), but both lie in every \(K_n\), impossible once the diameter drops below \(d\). This one-point version is the engine of the contraction mapping theorem later in the course.