Questions: Heine-Borel Theorem

Question 1

Which of the following subsets of ℝ is compact according to the Heine-Borel theorem?

• A. The open interval (0, 1)
• B. The integers ℤ
• C. The closed interval [0, 1]
• D. The real line ℝ

Answer: (C)

[0, 1] is both closed (contains all its limit points) and bounded (contained in [−2, 2]), so Heine-Borel guarantees it is compact. (0, 1) is bounded but not closed — it fails to contain the limit point 0. ℤ is closed but unbounded. ℝ is neither bounded. Both conditions are required.

Question 2

In an infinite-dimensional function space such as L²([0,1]), is every closed and bounded set compact?

• A. Yes — Heine-Borel applies to any complete metric space
• B. Yes — closed and bounded always implies compact regardless of the space
• C. No — Heine-Borel is specific to ℝⁿ and fails in infinite-dimensional spaces
• D. No — but only because function spaces are not metric spaces

Answer: (C)

The Heine-Borel theorem is a special feature of ℝⁿ, not a universal truth. In infinite-dimensional normed spaces such as L², the closed unit ball is closed and bounded but not compact — it contains sequences with no convergent subsequence. The theorem's power is precisely that it collapses an infinite verification into two geometric checks, but only because ℝⁿ has just the right structure.

Question 3

A closed subset of ℝ is typically compact.

• T. True
• F. False

Answer: False

Closedness alone is not sufficient — the set must also be bounded. The integers ℤ form a closed set (they contain all their limit points, since they have none outside the set), yet ℤ is unbounded. Any cover by open intervals of width 1 centered at each integer has no finite subcover. Both conditions — closed AND bounded — are required by Heine-Borel.

Question 4

The Heine-Borel theorem fails in general metric spaces: a closed and bounded set need not be compact outside of ℝⁿ.

• T. True
• F. False

Answer: True

This is a crucial limitation of the theorem. In infinite-dimensional Banach spaces, or even in unusual metric spaces, closed and bounded does not imply compact. For example, in the metric space of rationals ℚ with the standard metric, the set {q ∈ ℚ : 0 ≤ q ≤ 1} is closed and bounded in ℚ but not compact (the sequence of rational approximations to √2/2 has no rational limit in the set). Heine-Borel is a theorem about ℝⁿ specifically.

Question 5

Why are both closedness and boundedness necessary for a subset of ℝ to be compact? Give a counterexample for each condition failing.

Model answer: Boundedness keeps the set from stretching to infinity; without it, you can construct an open cover that cannot be reduced to finitely many sets (e.g., cover ℤ with intervals (n−0.5, n+0.5) — no finite subcollection covers all integers). Closedness keeps the set from leaking through its boundary; without it, a sequence of points can converge to a limit outside the set, and you can cover the set with intervals that each avoid that limit (e.g., cover (0,1) with (1/n, 1) — any finite subcollection misses points near 0).

Each condition blocks a different failure mode. Boundedness prevents the 'escape to infinity' failure (ℤ example). Closedness prevents the 'converging sequence escapes' failure ((0,1) example). The Heine-Borel theorem says these are the only two ways compactness can fail in ℝⁿ.