Questions: Sequential Compactness in Metric Spaces

Question 1

You want to prove that a continuous function f: X → ℝ on a compact metric space X attains its maximum. Which characterization of compactness is most directly useful?

• A. Open-cover compactness, because it directly gives a finite subcover of the preimages of an open cover of ℝ
• B. Sequential compactness: take a sequence (xₙ) with f(xₙ) → sup f; extract a convergent subsequence xₙₖ → x*; continuity gives f(x*) = sup f
• C. Complete plus totally bounded, because totally bounded spaces have finite ε-nets
• D. The Lebesgue number lemma, which directly bounds how large f can be

Answer: (B)

Sequential compactness is the natural tool here. Let M = sup{f(x) : x ∈ X}. Choose xₙ with f(xₙ) → M. By sequential compactness, (xₙ) has a convergent subsequence xₙₖ → x* ∈ X. By continuity of f, f(x*) = lim f(xₙₖ) = M, so the supremum is attained. Open-cover compactness can also prove this result but requires constructing a cover of ℝ and working backward — more indirect. Sequential compactness aligns naturally with the sequence-based argument.

Question 2

What fails in general topological spaces that makes the equivalence between sequential and open-cover compactness specific to metric spaces?

• A. General topological spaces do not have a well-defined notion of distance, so sequences cannot be defined
• B. In uncountable product spaces with the product topology, Tychonoff's theorem gives compactness but the space can fail to be sequentially compact — no metric is available to extract subsequences via a diagonal argument
• C. Open covers do not exist in non-metrizable spaces
• D. Subsequences always converge in compact spaces regardless of topology

Answer: (B)

Sequences can be defined in any topological space (using nets or filters to generalize, though sequences themselves are well-defined). The issue is that in spaces without a countable base (like uncountable products), sequences are too 'thin' to detect all the topology. Tychonoff's theorem says arbitrary products of compact spaces are compact (open-cover sense), yet {0,1}^ℝ is compact but not sequentially compact — some sequences have no convergent subsequences. In metric spaces, second-countability and total boundedness are available to extract subsequences via diagonal arguments, making the equivalence possible.

Question 3

A topological space is sequentially compact if and only if it is compact in the open-cover sense.

• T. True
• F. False

Answer: False

This equivalence holds for metric spaces but not for general topological spaces. Counterexamples exist in both directions: (1) uncountable products of compact spaces are compact by Tychonoff's theorem but may fail to be sequentially compact (no convergent subsequences for certain sequences); (2) the ordinal space [0, ω₁) is sequentially compact (every countable sequence has a convergent subsequence) but is not compact in the open-cover sense. The equivalence in metric spaces relies on special properties: separability, total boundedness, and the ability to use diagonal arguments on sequences.

Question 4

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

• T. True
• F. False

Answer: True

This is exactly the metric-space equivalence theorem. The proof uses two key metric-space tools: total boundedness (sequential compactness implies the space can be covered by finitely many ε-balls for any ε) and the Lebesgue number lemma (for a sequentially compact metric space, every open cover has a Lebesgue number δ > 0 such that every δ-ball fits inside some cover element). Combining these gives a finite subcover: cover the space with finitely many δ/2-balls, each contained in a cover element.

Question 5

Why does the Heine-Borel theorem characterize compact subsets of ℝⁿ as exactly the closed and bounded sets, and how does sequential compactness underlie this?

Model answer: Closed and bounded implies sequentially compact in ℝⁿ: any sequence in a bounded set has a convergent subsequence by the Bolzano-Weierstrass theorem (applied coordinate by coordinate), and closedness ensures the limit remains in the set. Sequential compactness implies open-cover compactness in metric spaces (including ℝⁿ). Conversely, compact implies closed (compact subsets of Hausdorff spaces are closed) and bounded (the cover by unit balls has a finite subcover, bounding the diameter). The argument runs through sequential compactness as the natural intermediate: boundedness gives subsequences, closedness captures limits, and the metric-space equivalence converts this to open-cover compactness.

The Heine-Borel theorem is the prototype of compactness criteria, but its proof is cleaner when routed through sequential compactness. The Bolzano-Weierstrass theorem (every bounded real sequence has a convergent subsequence) is the foundational fact, applied dimension by dimension in ℝⁿ. This is why Heine-Borel fails in infinite-dimensional spaces: in ℓ² or C([0,1]), bounded sequences need not have convergent subsequences (the unit ball is closed and bounded but not compact). Compactness in infinite dimensions requires additional conditions — like the Arzelà-Ascoli theorem's equicontinuity — beyond mere boundedness and closedness.