MATH 4630

Chapter 3: Compact Sets

Compact Sets

Throughout our study of analysis, we have worked with sequences in infinite sets such as (0,1), ℝ, etc. A difficulty is that all the infinite processes—limits, series, integrals—can behave badly. The concept of a compact set is designed precisely to impose enough structure so that infinite processes can be controlled as if they were finite.

Question: Suppose (an) is a sequence that does not converge. Can we at least guarantee some "convergent part" of it?

Recall the Bolzano–Weierstrass Theorem: every bounded sequence has a convergent subsequence.

The key feature of [0,1] is: it is bounded (so B-W applies) and closed (so limits stay inside). This is exactly what compactness captures!

Definition of Compactness:

Definition

A set K ⊆ ℝ is called compact if every sequence (an) contained in K has a subsequence (ank) that converges to a limit that is also in K: $$(a_n) \subseteq K \implies \exists\; (a_{n_k}) \text{ with } a_{n_k} \to L \in K.$$

Compactness is a "self-contained" property: not only do convergent subsequences exist, but their limits remain trapped inside K.

Your turn: For each set K, determine whether it is compact. If K is not compact, produce a sequence (xn) ⊆ K such that every convergent subsequence of (xn) converges to a limit not contained in K. If K is compact, explain why every sequence in K has a subsequence converging to a limit in K.
  1. K = [0, 1] — Compact?
  1. K = (0,1) — Compact?
  2. K = [1, ∞) — Compact?

    (Hint: Can you find an unbounded sequence in K?)

  3. K = {0} ∪ {1/n : n ∈ ℕ} — Compact?

The Characterization Theorem:

The examples above reveal a pattern: compact sets in ℝ appear to be exactly the sets that are both closed and bounded. Let us record this precisely.

Definition

A set A ⊆ ℝ is bounded if there exists M > 0 such that

Theorem

A set K ⊆ ℝ is compact if and only if it is closed and bounded.

The proof has two parts. We show compactness implies closed and bounded, and then we show that closed and bounded implies compactness.

Proof sketch (⟹): We will prove this implication in two steps.

Step 1: Suppose, for contradiction, that K is compact but not bounded.

Show that for each n ∈ ℕ there exists xn ∈ K with |xn| > n.
Explain why the sequence (xn) cannot have any convergent subsequence:
Why does this contradict compactness of K?

Step 2: Now suppose K is compact and (xn) ⊆ K with xn → L.

Use compactness to prove that (xn) has a convergent subsequence (xnk) with (xnk) → L.
Conclude that L ∈ K. Why does this show K is closed?

Proof sketch (⟸): Let K be closed and bounded, and let (an) ⊆ K.

Prove that (an) has a convergent subsequence (ank) converging to L.
Argue why L ∈ K. Does it imply K is compact?
Your turn: Decide whether each set is compact by checking if it is closed and bounded.
SetBounded?Closed?Compact?
[2, 7]
[2, 7)
ℤ
{1/n : n ∈ ℕ}
[0,1] ∪ [3,5]

Nested Compact Sets:

Recall the Nested Interval Property from Chapter 1: a nested sequence of closed intervals [a1,b1] ⊇ [a2,b2] ⊇ ⋯ always has nonempty intersection. Compactness allows us to generalize this property.

Theorem

If K1 ⊇ K2 ⊇ K3 ⊇ ⋯ is a nested sequence of nonempty compact sets, then $$\bigcap_{n=1}^{\infty} K_n \; \underline{\hspace{2cm}}.$$

Proof: The proof is left as an Exercise. Write down a formal argument on your own.

Note: Closedness is not enough in Theorem 2, we need the compactness.

Consider Fn = [n, ∞). Each Fn is closed, and F1 ⊇ F2 ⊇ ⋯

What is ⋂n=1∞ Fn?

Is the conclusion of the Theorem satisfied?

What hypothesis of the Theorem does the family {Fn} fail?

Open Covers and the Heine–Borel Theorem:

There is a third, and in many ways most powerful, way to think about compactness. It uses the concept of open covers.

Definition

Let A ⊆ ℝ. An open cover for A is a (possibly infinite) collection of open sets {Oλ : λ ∈ Λ} such that $$A \subseteq \bigcup_{\lambda \in \Lambda} O_\lambda.$$ A finite subcover is a finite subcollection {Oλ1, …, Oλn} from the original cover whose union still contains A.

Think of an open cover as a family of "patches" that together cover A. The question is: can we always get away with only finitely many patches?

Example: Let A = (0,1] and for each n ∈ ℕ, define $$O_n = \left(\frac{1}{n}, 2\right).$$
  1. Show that {On : n ∈ ℕ} is an open cover of (0,1].
  2. Let {On1, …, Onk} be any finite subcollection of {On : n ∈ ℕ}. Can this be a subcover for A?
  3. From your previous knowledge, is the interval (0,1] compact? Why or why not?
Heine–Borel Theorem

Let K ⊆ ℝ. The following are equivalent:

  1. K is compact: every sequence in K has a subsequence converging to a limit in K.
  2. K is closed and bounded.
  3. K has the finite cover property: every open cover of K has a finite subcover.

The equivalence of (i) and (ii) was established in the previous theorem. Let us examine the implications (iii) ⟹ (ii) and (ii) ⟹ (iii).

Proof that (iii) ⟹ (ii): Suppose K has the finite cover property, that is, every open cover of K has a finite subcover.

Show that K is bounded. We will show the boundedness in steps.

Step 1: For each x ∈ K, let Ox = V1(x) = (x−1, x+1). Explain why {Ox : x ∈ K} is an open cover of K.

Step 2: By (iii), there is a finite subcover {Ox1, …, Oxn}. Prove that this implies K is bounded.

Show that K is closed: We will show the closedness in steps.

Step 1: Suppose (yn) ⊆ K with yn → y. Prove that y ∈ K. Assume for contradiction y ∉ K. For each x ∈ K, let Ox = V|x−y|/2(x). Prove that {Ox : x ∈ K} is an open cover for K.

By (iii), there is a finite subcover {Ox1, …, Oxn}. Let $$\varepsilon_0 = \min\left\{\frac{|x_i - y|}{2} : 1 \leq i \leq n\right\}.$$

Step 2: Write the ε-N definition of yn → y.
For N ∈ ℕ, show that yN ∉ ⋃ni=1 Oxi.
Does it produce any contradiction?
Have we proved all the implications required for the proof?
Your turn: For each of the following sets K, determine whether K is compact and cite appropriate Theorem or other resources. If K is not compact, then produce an open cover with no finite subcover.
  1. K = [−5, 5]:
  2. K = ℕ:
  3. K = {x ∈ ℝ : |x2 − 3| ≤ 1} = [−2, −√2] ∪ [√2, 2]: Compact?

Activity:

The following Exercise shows the power of compactness and connects it to the concepts we studied earlier this semester.

Exercise: Prove that if K is compact and nonempty, then sup K exists and sup K ∈ K.

Proof: The following hints will help you prove the result:

1 Why does sup K exist?

2 Use the definition of supremum to justify: for each n ∈ ℕ there exists an ∈ K with $$\sup K - \frac{1}{n} < a_n \leq \sup K.$$

What does the Squeeze Theorem tell you about (an)?

3 Use compactness of K to conclude sup K ∈ K.

Follow-up question: Does sup(0,1) belong to (0,1)? Why does the argument break down for open sets?

Summary: Three equivalent definitions of compactness in ℝ

Sequential: Every sequence has a convergent subsequence with limit in K

⟷ (Thm 1)

Closed & Bounded: K is closed and bounded

⟷ (Heine-Borel)

Finite Cover: Every open cover of K has a finite subcover

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