The Gold Standard of Topology | Sequential Compactness and Compactness in Metric Spaces

A metric space X is compact if every family of open subsets covering X has a finite subfamily still covering X. The family can be infinite or uncountable. The selected subfamily must consist of members of the original cover.

X is sequentially compact if every sequence of points in X has a subsequence converging to a point of X. A subsequence keeps strictly increasing original indices. Finding one well-behaved sequence does not establish the property for every sequence.

The limit condition matters. The sequence 1/n in the space (0,1), with its usual metric, approaches a real number but not a point of that space. Convergence in an ambient space is not enough.

We work in ordinary metric spaces with the usual countable selection arguments of classical analysis. The empty space satisfies both properties immediately. In the substantive proofs below, assume X is nonempty.

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Take an arbitrary sequence in compact X. If a value occurs infinitely often, its corresponding constant subsequence converges to that value, so the conclusion follows.

Otherwise no value occurs infinitely often. Suppose there is no point x such that every neighbourhood of x contains terms with arbitrarily large indices. Then for each x choose an open neighbourhood Ux containing only finitely many sequence terms, counted by their indices.

The Ux form an open cover. Compactness gives finitely many of them covering X. A finite union of sets each containing only finitely many sequence indices can contain only finitely many indices.

But every term of the sequence lies in X and hence in that finite union. This contradicts the infinitely many indices. Therefore some x has arbitrarily late sequence terms in every neighbourhood. We have not yet extracted the subsequence; that is the metric step in the next chapter.

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For the point x obtained above, choose n₁ with x at distance less than one from xₙ₁. Then choose n₂ greater than n₁ with distance less than one half, and continue: d(xₙₖ,x) less than 1/k with increasing indices.

Every ball about x contains arbitrarily late terms, so each choice is possible. Given any ε greater than zero, choose K with 1/K less than ε. For every k at least K, d(xₙₖ,x) is less than ε.

Thus the extracted subsequence converges to x, which belongs to X. The original sequence was arbitrary, proving sequential compactness.

The balls of radii 1, 1/2, 1/3 and so on give a countable way to enter every metric neighbourhood. In an arbitrary topological space, a compactness argument can still yield accumulation information while this sequence extraction need not work.

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A metric space is totally bounded if for every ε greater than zero, finitely many open ε-balls cover it. The number of balls may depend on ε; it need not stay bounded as the radius shrinks.

Suppose X is sequentially compact but not totally bounded. For some ε, no finite collection of ε-balls covers X. Choose x₁, then choose x₂ outside the ε-ball about x₁, then x₃ outside both balls, continuing indefinitely.

All distinct selected points have distance at least ε. Any convergent subsequence would be Cauchy: sufficiently late pairs would have distance less than ε, by the triangle inequality around its limit. That is impossible.

This contradicts sequential compactness, so X is totally bounded. Ordinary boundedness only says the space fits inside one sufficiently large ball. Total boundedness demands finite coverage at every positive scale, a much stronger requirement.

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Let U be an arbitrary open cover of sequentially compact metric X. We claim there is a δ greater than zero such that for every x in X, some member of U contains the ball B(x,δ).

If not, for each positive integer n choose xn whose ball B(xn,1/n) is contained in no member of U. Sequential compactness supplies a subsequence xₙₖ converging to x.

Choose a cover member U₀ containing x. Since it is open, some ball B(x,r) lies inside U₀. For sufficiently large k, both d(xₙₖ,x) and 1/nk are less than r/2.

The triangle inequality then places B(xₙₖ,1/nk) inside B(x,r), hence inside U₀. This contradicts how xn was chosen. The desired uniform δ exists. It is a ball-containment form of the Lebesgue-number argument; we derived the form needed rather than assuming compactness.

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Use total boundedness with radius δ/2. Choose finitely many centres c₁ through cm whose δ/2-balls cover X.

For each centre cj, the preceding chapter provides an original cover member Uj containing B(cj,δ). It therefore also contains the smaller ball B(cj,δ/2).

Since those smaller balls cover X, the finitely many selected Uj cover X. They are members of the original arbitrary open cover, exactly as the compactness definition requires. Thus X is compact.

The logic has two distinct tasks: total boundedness supplies finitely many centres, while the uniform ball size turns each local ball into a member of the given cover. Finite coverage by some convenient balls alone would not yet prove finite coverage by the original family.

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A metric space is complete if every Cauchy sequence converges to a point of the space. Sequential compactness implies completeness: take any Cauchy sequence, extract a convergent subsequence, and use the Cauchy condition to show the full sequence has the same limit.

For the final step, fix ε. Choose a tail whose terms are mutually within ε/2, and choose a subsequence term in that tail within ε/2 of the subsequence limit. Every sufficiently late original term is within ε of the limit by the triangle inequality.

Completeness and total boundedness together also yield sequential compactness. Repeatedly cover by finitely many balls of radii tending to zero, retaining an infinite set of sequence indices in one ball at each stage. Nested retained index sets allow increasing-index choices.

The selected subsequence is Cauchy because its sufficiently late terms lie together in one arbitrarily small ball. Completeness supplies its limit in X. This gives the familiar three-way equivalence, but neither completeness nor total boundedness alone is enough.

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Consider X = (0,1) with the usual real metric. It is totally bounded: at each ε, finitely many sufficiently closely spaced centres cover the interval by ε-balls.

It is not complete. The sequence xn = 1/(n+1) is Cauchy in X, while its only possible real limit is zero, outside X. No subsequence can converge to a point of X because every subsequence still approaches zero in the ambient metric.

It is not compact either. For integers n at least two, let Un = (1/n,1), interpreted as an open subset of X. These sets cover X: any positive x exceeds 1/n for sufficiently large n.

A finite subfamily has a largest index N and union (1/N,1). It misses points of X near zero. Hence no finite subcover exists. The sequence failure and cover failure expose the same missing boundary through different definitions.

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Give the positive integers the discrete metric: d(m,n) is zero when m = n and one otherwise. The space is bounded because every distance is at most one.

It is complete. A Cauchy sequence must eventually have all pairwise distances less than one half, which forces all sufficiently late terms to be equal. Such a sequence converges to its eventual value in the space.

It is not totally bounded: every open ball of radius one half is a singleton, so finitely many such balls cover only finitely many integers. The sequence xn = n has no convergent subsequence because any convergent sequence in this metric is eventually constant.

The singleton cover also has no finite subcover, proving noncompactness directly. This example shows why the Euclidean closed-and-bounded test cannot be transported to every metric space without its finite-dimensional setting.

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Let K = {0} together with {1/n : n is a positive integer}, with the usual real metric. Prove compactness directly from an arbitrary open cover.

Choose a member U containing zero. Some relative ε-neighbourhood of zero lies in U, so all sufficiently late points 1/n belong to U. Only finitely many points of K remain uncovered by that one set.

Choose one cover member for each remaining point. Together with U, these finitely many members cover K. This completes the open-cover proof without first invoking a general theorem about closed bounded subsets of the real line.

For the sequential version, a repeated value occurring infinitely often gives a constant subsequence. If no value repeats infinitely often, successively choose sequence terms from reciprocals with denominators tending to infinity; they converge to zero. Removing zero destroys the limit condition and compactness, which is the changed-condition check.

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Must the entire sequence converge in a compact metric space? No. The sequence alternating between zero and one in [0,1] fails to converge, but its even and odd subsequences each converge.

Does a finite cover by small balls prove compactness? It proves total boundedness when available at every scale. The interval example shows why another condition is needed.

Are compactness and sequential compactness interchangeable in every topology? No. The proof above uses metric balls and countable extraction. Outside the metric setting, consult the actual hypotheses before replacing open-cover arguments by sequences.

Does compact mean finite? No. The interval [0,1] and the reciprocal set K are infinite compact metric spaces. Compactness concerns finite subcovers for every open cover, not a finite number of points.

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To disprove sequential compactness in a metric space, find one sequence with no convergent subsequence whose limit stays in the space. To disprove compactness directly, find one open cover with no finite subcover.

To prove the properties, retain the universal quantifier. Start with an arbitrary sequence or cover. A demonstration involving a favourite sequence or a specially convenient cover cannot establish the property for all of them.

The university references give background in the standard compactness framework. The proofs and contrasting examples here were laid out to make each metric step and each missing hypothesis explicit. The sequence-limit guide explains why general topological spaces require more care.

For transfer, ask which structural feature turns a manageable test into a complete test. Here shrinking metric balls connect local neighbourhoods to sequences, and total boundedness connects every small scale to finitely many centres. Those mechanisms carry the equivalence.

References checked 22 October 2026

John Hunter, UC Davis: metric spaces and compactness

Michael Fochler, Binghamton University: analysis notes with compactness proofs

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