Sequentially Compact Implies Compact

The Converse Direction

Compactness and sequential compactness begin with different kinds of data. Compactness concerns open covers and finite subcovers; sequential compactness concerns sequences and convergent subsequences. The previous tutorial proved that compactness implies sequential compactness in metric spaces. Here we prove the converse, completing the equivalence for metric spaces.

The proof turns sequence control into cover control in two stages. First, sequential compactness gives total boundedness, so the space can be covered by finitely many small balls. Second, sequential compactness ensures that an arbitrary open cover has a ball radius that works uniformly at every point. A finite collection of sufficiently small balls can then be matched with finitely many members of the given cover.

Definition: A metric space \(X\) is sequentially compact if every sequence in \(X\) has a subsequence that converges to a point of \(X\). It is compact if every open cover of \(X\) has a finite subcover.

We will use a consequence established earlier in the course. By the Sequential Characterization of Total Boundedness, every sequentially compact metric space is totally bounded. In particular, for every \(\varepsilon>0\), there are finitely many points \(a_1,\ldots,a_m\) such that the balls \(B_\varepsilon(a_i)\) cover the space. This supplies the finite collection of centers needed at the last stage of the argument.

A Uniform Ball Radius for an Open Cover

An open cover may consist of sets with very different shapes and sizes. The key fact is that in a sequentially compact metric space, there is nevertheless a single positive radius such that every point-centered ball of that radius is contained in some member of the cover. The member may depend on the point; the radius does not.

Theorem (Uniform Ball Radius for an Open Cover): Let \((X,d)\) be a nonempty sequentially compact metric space, and let \(\mathcal{U}\) be an open cover of \(X\). There is a number \(\delta>0\) such that for every \(x\in X\), the ball \(B_\delta(x)\) is contained in some member of \(\mathcal{U}\).

Proof. Suppose no such \(\delta\) exists. Then for every positive integer \(n\), there is a point \(x_n\in X\) such that \(B_{1/n}(x_n)\) is not contained in any member of \(\mathcal{U}\). Since \(X\) is sequentially compact, there are increasing indices \(n_k\) and a point \(x\in X\) such that \(x_{n_k}\to x\).

Because \(\mathcal{U}\) covers \(X\), some \(U\in\mathcal{U}\) contains \(x\). The set \(U\) is open, so there is an \(r>0\) such that \(B_r(x)\subseteq U\). Convergence gives an index \(K_1\) such that

d(x_{n_k},x)<r/2 \qquad\text{for every }k\geq K_1.

Also, \(n_k\to\infty\), so there is an index \(K_2\) such that \(1/n_k<r/2\) for every \(k\geq K_2\). Choose \(k\) at least as large as both \(K_1\) and \(K_2\). If \(y\in B_{1/n_k}(x_{n_k})\), the triangle inequality gives

d(y,x)\leq d(y,x_{n_k})+d(x_{n_k},x) <1/n_k+r/2 <r.

Thus \(y\in B_r(x)\subseteq U\), and so \(B_{1/n_k}(x_{n_k})\subseteq U\). This contradicts the choice of \(x_{n_k}\), whose ball was not contained in any member of \(\mathcal{U}\). The contradiction proves that a uniform \(\delta>0\) exists. \(\square\)

The metric is essential to this argument: convergence puts the centers \(x_{n_k}\) near \(x\), and the triangle inequality then places an entire small ball around \(x_{n_k}\) inside the open set containing \(x\). The contradiction would not follow merely from knowing that the centers converge if there were no way to control nearby points.

Sequential Compactness Implies Compactness

Theorem (Sequentially Compact Metric Spaces Are Compact): Every sequentially compact metric space is compact.

Proof. The empty space is compact, since the empty family is a finite subcover of its empty open cover. Now suppose \(X\) is nonempty and sequentially compact, and let \(\mathcal{U}\) be an open cover of \(X\). By the uniform-ball-radius theorem, there is a \(\delta>0\) such that for every \(x\in X\), the ball \(B_\delta(x)\) is contained in some member of \(\mathcal{U}\).

By the Sequential Characterization of Total Boundedness, \(X\) is totally bounded. Apply total boundedness with radius \(\delta/2\). There are finitely many points \(a_1,\ldots,a_m\in X\) such that

X\subseteq B_{\delta/2}(a_1)\cup\cdots\cup B_{\delta/2}(a_m).

For each \(i\), the uniform-radius property gives a member \(U_i\in\mathcal{U}\) such that \(B_\delta(a_i)\subseteq U_i\). Since \(\delta/2<\delta\), we have \(B_{\delta/2}(a_i)\subseteq B_\delta(a_i)\subseteq U_i\). Consequently,

X\subseteq \bigcup_{i=1}^m B_{\delta/2}(a_i) \subseteq \bigcup_{i=1}^m U_i.

The finitely many members \(U_1,\ldots,U_m\) therefore cover \(X\). Since \(\mathcal{U}\) was arbitrary, every open cover has a finite subcover, so \(X\) is compact. \(\square\)

Together with the result from the previous tutorial, this proves that for metric spaces compactness and sequential compactness are equivalent. The two implications use different arguments: compactness gives a cluster point by ruling out a particular open cover, while sequential compactness gives a finite subcover by combining a uniform ball radius with a finite net.

Worked Applications

Worked Example: A Sequence Together with Its Limit

Consider the subset of the real line

S=\{0\}\cup\{1/n:n\in\mathbb{N},\ n\geq1\},

with the usual metric. We verify sequential compactness directly. Take any sequence \((s_j)\) in \(S\). If some value \(q\in S\) occurs for infinitely many indices, those indices give a constant subsequence, which converges to \(q\in S\).

It remains to consider the case in which every value occurs only finitely many times. Then \(0\) occurs only finitely often, and each value \(1/n\) occurs only finitely often. We can choose, recursively, indices \(j_1<j_2<\cdots\) for which \(s_{j_k}=1/m_k\) and \(m_k>k\). Indeed, after any finite number of indices have been chosen, only finitely many terms have denominator at most \(k\): there are finitely many such values, each occurring finitely often. There are infinitely many remaining indices, so a later term with denominator greater than \(k\) can be selected.

The selected subsequence satisfies

0\leq s_{j_k}=\frac{1}{m_k}<\frac{1}{k}.

As \(1/k\to0\), it follows that \(s_{j_k}\to0\), and \(0\in S\). Thus every sequence in \(S\) has a subsequence converging in \(S\). The theorem shows that \(S\) is compact. In particular, this conclusion applies to every open cover of \(S\), even if its cover sets have no simple interval form.

Worked Example: A Compact Space of Binary Sequences

Let \(X\) consist of all sequences \(a=(a_1,a_2,\ldots)\) with \(a_j\in\{0,1\}\), and define

d(a,b)=\sum_{j=1}^{\infty}2^{-j}|a_j-b_j|.

This is a metric: the series converges because \(|a_j-b_j|\leq1\) and \(\sum_{j=1}^{\infty}2^{-j}=1\); nonnegativity and symmetry follow term by term; if \(d(a,b)=0\), every nonnegative summand is zero, so \(a_j=b_j\) for each \(j\); and the triangle inequality follows by summing \(|a_j-c_j|\leq|a_j-b_j|+|b_j-c_j|\).

Take any sequence of points \(a^{(1)},a^{(2)},\ldots\) in \(X\). Since the first coordinate takes only two values, infinitely many terms agree in that coordinate. Keep an infinite set of indices on which the first coordinate is constant. From those indices, keep an infinite subset on which the second coordinate is constant, and continue. Choose increasing indices \(n_k\) diagonally, with \(n_k\) from the infinite set fixed through coordinate \(k\). For each fixed \(j\), the \(j\)-th coordinate of \(a^{(n_k)}\) is then eventually constant. Let \(a_j\) be that eventual value, and let \(a=(a_1,a_2,\ldots)\in X\).

To verify convergence in the metric, fix \(\varepsilon>0\). Choose \(m\) such that \(2^{-m}<\varepsilon\). For all sufficiently large \(k\), the first \(m\) coordinates of \(a^{(n_k)}\) agree with those of \(a\). Therefore

d(a^{(n_k)},a) =\sum_{j=m+1}^{\infty}2^{-j}|a^{(n_k)}_j-a_j| \leq\sum_{j=m+1}^{\infty}2^{-j} =2^{-m} <\varepsilon.

So \(a^{(n_k)}\to a\), proving that \(X\) is sequentially compact. The theorem now guarantees that this binary sequence space is compact in the open-cover sense.

Worked Example: A Compact Image of Binary Sequences

Use the binary sequence space \(X\) from the preceding example and define

f(a)=\sum_{j=1}^{\infty}\frac{a_j}{3^j}.

The series converges for every \(a\in X\), since \(0\leq a_j/3^j\leq1/3^j\) and \(\sum_{j=1}^{\infty}1/3^j=1/2\). For \(a,b\in X\), termwise comparison gives

|f(a)-f(b)| \leq\sum_{j=1}^{\infty}\frac{|a_j-b_j|}{3^j} \leq\sum_{j=1}^{\infty}\frac{|a_j-b_j|}{2^j} =d(a,b),

because \(3^{-j}\leq2^{-j}\) for every positive integer \(j\). Thus \(f\) is Lipschitz, and in particular continuous. By the earlier theorem Continuous Images Preserve Sequential Compactness, the image \(f(X)\), with its usual metric as a subset of \(\mathbb{R}\), is sequentially compact. Applying the theorem of this tutorial shows that \(f(X)\) is compact. The argument does not require identifying the image more explicitly: sequential compactness of the domain, continuity, and the metric-space theorem suffice.

What the Proof Does—and Does Not—Say

A common pitfall is to try to get a finite subcover directly from convergence of subsequences. Convergence controls one sequence of points, while an open cover must be handled at every point of the space. The uniform-ball-radius theorem bridges that gap: it converts sequential compactness into a scale that works across the entire cover. Total boundedness then reduces the space to finitely many centers at that scale.

Another important point is that the theorem is stated for metric spaces. The proof uses metric balls, distances, and the ability to place a small ball around a convergent sequence term inside an open set containing its limit. The equivalence between compactness and sequential compactness should not be transferred to arbitrary topological spaces without additional hypotheses.

The reverse implication is also useful in practice. To prove that a metric space is compact, it can be easier to start with an arbitrary sequence and extract a convergent subsequence than to analyze every possible open cover. Once sequential compactness is established, the theorem supplies the open-cover conclusion automatically.

Check Your Understanding

Use the uniform-ball-radius argument and the finite-net step to answer the following questions.

Why does failure of a uniform ball radius allow the construction of points \(x_n\) whose balls \(B_{1/n}(x_n)\) lie in no single cover member?
In the uniform-radius proof, why does convergence of a subsequence of centers eventually place its small balls inside the open set containing the limit?
Where is total boundedness used in the proof that sequential compactness implies compactness?
Why does a finite \(\delta/2\)-net, together with a uniform radius \(\delta\), produce members of the original open cover that cover the whole space?
Why must the limit obtained in the definition of sequential compactness belong to the space for this proof to apply?