Questions: Compact Sets and the Heine-Borel Theorem

Question 1

The half-open interval [0, 1) is bounded. Is it compact?

• A. Yes — it is bounded, and boundedness is sufficient for compactness in ℝ
• B. No — it is not closed, so it fails the Heine-Borel condition; an open cover can slide off the missing boundary at 1
• C. Yes — it is connected and contains its left endpoint, which is sufficient
• D. No — it contains infinitely many points, so no finite subcover can work

Answer: (B)

Heine-Borel requires BOTH closed and bounded. [0, 1) is bounded but not closed — the limit point 1 is missing. Concretely, the open cover {[0, 1 − 1/n) : n = 1, 2, 3, …} covers [0, 1) but no finite subcollection does, since any finite sublist leaves a neighborhood of 1 uncovered. Option D is a common misunderstanding: containing infinitely many points has nothing to do with compactness — [0, 1] is also infinite and is compact.

Question 2

A student claims that every closed set in ℝ is compact. Which example best refutes this?

• A. The empty set ∅ — it is closed but trivially has no points
• B. The entire real line ℝ — it is closed but unbounded, and the cover {(−n, n) : n = 1, 2, …} has no finite subcover
• C. The rationals ℚ — they are not closed in ℝ, so this is not a counterexample
• D. The open interval (0, 1) — it is not closed, so this is not a counterexample

Answer: (B)

ℝ itself is closed (it contains all its limit points) but is not bounded — you can cover it with {(−n, n) : n = 1, 2, 3, …}, and no finite subcollection covers all of ℝ. This shows that closedness alone is insufficient: you also need boundedness. Options C and D are not counterexamples to the student's claim because they fail to be closed; you need a set that IS closed but fails compactness.

Question 3

A set in ℝ is compact if and only if it is bounded.

• T. True
• F. False

Answer: False

Boundedness alone is not sufficient. The Heine-Borel Theorem requires the set to be both closed AND bounded. The open interval (0, 1) is bounded but not compact (it is not closed). Conversely, [0, ∞) is closed but not compact (it is not bounded). Both conditions are independently necessary.

Question 4

The Extreme Value Theorem — that a continuous function on a closed interval [a, b] attains its maximum and minimum — holds precisely because [a, b] is compact.

• T. True
• F. False

Answer: True

This is the key application of Heine-Borel. The proof runs: [a, b] is compact (closed and bounded, by Heine-Borel). Continuous images of compact sets are compact. Therefore f([a, b]) is compact in ℝ — closed and bounded — so it contains its supremum and infimum. The EVT is not a standalone result; it is a corollary of compactness and the continuity of f.

Question 5

Why does the open interval (0, 1) fail to be compact, even though it is bounded? Use the open-cover definition.

Model answer: The open cover {(1/n, 1) : n = 1, 2, 3, …} covers every point in (0, 1) — any x > 0 eventually satisfies x > 1/n. But no finite subcollection covers (0, 1): any finite sub-list has a largest n, say N, and the set (1/N, 1) leaves points in (0, 1/N] uncovered. The failure occurs because 0 is a limit point of (0, 1) that is not in the set — the cover can 'slide' toward the missing boundary point without ever being trapped by finitely many sets.

Closedness in the Heine-Borel theorem plays exactly this role: it prevents the escape toward missing boundary points. Adding 0 to get [0, 1] traps the cover — any open set covering 0 covers a neighborhood of 0, and from there finitely many of the (1/n, 1) suffice. The open interval's incompactness is a direct consequence of its missing boundary.