Questions: Compact Metric Spaces and Characterizations

Question 1

Is the open interval (0, 1) with the usual metric compact? Why or why not?

• A. Yes — it is bounded, and bounded metric spaces are compact
• B. No — it is not complete, because the sequence 1/n is Cauchy but its limit 0 is not in (0, 1)
• C. Yes — every sequence in (0, 1) has a convergent subsequence, by the Bolzano-Weierstrass theorem
• D. No — it is not totally bounded because it contains infinitely many points

Answer: (B)

(0, 1) is totally bounded but not complete, so it fails the complete-and-totally-bounded characterization and is not compact. Option A is the classic misconception: boundedness does not imply compactness — the open interval is bounded but not compact. Option C sounds appealing but is wrong: 1/n → 0, which is outside (0, 1), so this sequence has no convergent subsequence *in (0, 1)*, confirming non-compactness.

Question 2

A metric space is complete. Does completeness guarantee that it is compact?

• A. Yes — completeness means no Cauchy sequence escapes to a missing point, which is exactly what compactness requires
• B. No — the real line ℝ is complete but not compact, because it is not totally bounded
• C. Yes — in metric spaces, completeness and compactness coincide by the Heine-Borel theorem
• D. No — completeness and compactness are entirely unrelated properties in metric spaces

Answer: (B)

Completeness is necessary but not sufficient for compactness. ℝ is complete (every Cauchy sequence converges) but not compact — the sequence 1, 2, 3, ... has no convergent subsequence. Compactness requires both completeness AND total boundedness. Total boundedness (every ε > 0 admits a finite ε-cover) is what prevents 'escape to infinity'; completeness prevents 'converging to a missing point.' Both conditions must hold.

Question 3

In a metric space, sequential compactness (every sequence has a convergent subsequence) and open-cover compactness (every open cover has a finite subcover) are equivalent.

• T. True
• F. False

Answer: True

True — and this is one of the central theorems about metric spaces. In general topological spaces, these notions can diverge. In metric spaces, all four notions (open-cover compactness, sequential compactness, countable compactness, and the Bolzano-Weierstrass property) coincide. This equivalence is what makes metric spaces so tractable and compact metric spaces so structurally rich.

Question 4

A totally bounded metric space is compact.

• T. True
• F. False

Answer: False

False. Total boundedness says that for every ε > 0, the space can be covered by finitely many ε-balls — it cannot 'escape to infinity.' But it does not prevent sequences from converging to missing points. The open interval (0, 1) is totally bounded (it fits in a ball of radius 1) but not compact, because it is not complete: the sequence 1/n is Cauchy but its limit 0 is not in the space. Compactness requires total boundedness AND completeness.

Question 5

Explain why compactness in a metric space requires both completeness and total boundedness. Use a counterexample to show what goes wrong when either condition is dropped alone.

Model answer: Completeness prevents sequences from 'converging to a missing point': a complete space has no gaps. Total boundedness prevents sequences from 'running to infinity': any sequence in a totally bounded space can be refined to stay in smaller and smaller regions. Drop completeness: (0, 1) is totally bounded but not complete — the sequence 1/n is Cauchy but its limit is missing, so no convergent subsequence exists in the space. Drop total boundedness: ℝ is complete but not totally bounded — the sequence 1, 2, 3, ... has no convergent subsequence. Both conditions together seal both escape routes, guaranteeing compactness.

The completeness+total boundedness characterization is structurally illuminating because it separates two distinct ways compactness can fail. Each condition closes exactly one failure mode. Together they ensure every sequence is 'trapped': total boundedness forces subsequences into smaller and smaller cells, and completeness guarantees the limit of those subsequences is actually in the space.