Compactness in Metric Spaces

Compactness and Sequences in Metric Spaces

Total boundedness controls a space at every small scale, but its sequence consequence is only the existence of a Cauchy subsequence. To ensure that the subsequence converges to a point of the space, we also need completeness. This pairing leads to a powerful criterion for compactness in metric spaces.

There are two useful formulations to keep distinct. The open-cover definition is the standard definition of compactness. Sequential compactness describes a property of sequences. In metric spaces these ideas are closely connected, and the sequential formulation can be tested using the two properties developed in earlier tutorials: completeness and total boundedness.

Definition: A metric space $X$ is sequentially compact if every sequence in $X$ has a subsequence that converges to a point of $X$. A metric space $X$ is compact in the open-cover sense if every collection of open subsets of $X$ whose union is $X$ has a finite subcollection whose union is still $X$.

The open-cover definition asks for a finite selection from an arbitrary cover. Sequential compactness instead asks for a convergent subsequence from each sequence. Neither definition should be replaced by the statement that a space is merely bounded or complete. The central metric-space criterion is that sequential compactness is exactly completeness plus total boundedness.

The Sequential Compactness Criterion

Theorem (Sequential Compactness and Completeness with Total Boundedness): A metric space $X$ is sequentially compact if and only if it is complete and totally bounded.

Proof. Suppose first that $X$ is sequentially compact. We show that it is complete. Let $(x_n)$ be a Cauchy sequence in $X$. By sequential compactness, it has a subsequence $(x_{n_k})$ converging to some $x\in X$. The earlier theorem A Cauchy Sequence with a Convergent Subsequence Converges now implies that the whole sequence $(x_n)$ converges to $x$. Thus every Cauchy sequence in $X$ converges in $X$, so $X$ is complete.

We next show that $X$ is totally bounded. Suppose otherwise. By the Sequential Characterization of Total Boundedness, there is a sequence in $X$ with no Cauchy subsequence. Sequential compactness would give it a convergent subsequence. Every convergent sequence in a metric space is Cauchy, so this is a contradiction. Therefore $X$ is totally bounded.

Conversely, suppose $X$ is complete and totally bounded, and let $(x_n)$ be any sequence in $X$. The Sequential Characterization of Total Boundedness gives a Cauchy subsequence $(x_{n_k})$. Since $X$ is complete, this subsequence converges to a point of $X$. Thus every sequence has a convergent subsequence in $X$, and $X$ is sequentially compact. $\square$

The proof separates the two jobs. Total boundedness extracts a Cauchy subsequence; completeness turns that subsequence into a convergent one. In the other direction, sequential compactness supplies enough convergent subsequences to force both properties.

How Open-Cover Compactness Enters

The open-cover definition also has direct consequences for sequences. We first show that a compact metric space is totally bounded. Fix $\varepsilon>0$. The collection of all open balls $B_\varepsilon(x)$, with $x\in X$, covers $X$, since each point belongs to its own ball. Compactness gives a finite subcover $$ X\subseteq B_\varepsilon(x_1)\cup\cdots\cup B_\varepsilon(x_m). $$ The finite set $\{x_1,\ldots,x_m\}$ is an $\varepsilon$-net for $X$. Since this works for every $\varepsilon>0$, $X$ is totally bounded.

Compactness also forces every sequence to have a convergent subsequence. Here is a direct argument. Let $(x_n)$ be a sequence in a compact metric space $X$. If its range is finite, at least one value occurs infinitely many times, and those occurrences give a constant, hence convergent, subsequence.

Now suppose its range $E=\{x_n:n\in\mathbb{N}\}$ is infinite. The set $E$ must have a limit point in $X$. Otherwise, for each $x\in X$, there would be a ball $B_{r_x}(x)$ meeting $E\setminus\{x\}$ in no points. These balls cover $X$. A finite subcover would imply that $E$ is contained in a finite union of sets each containing at most one point of $E$, making $E$ finite. This contradicts the assumption that $E$ is infinite.

Let $p\in X$ be a limit point of $E$. Every ball centered at $p$ contains infinitely many points of $E\setminus\{p\}$: if one such ball contained only finitely many, a smaller positive-radius ball could be chosen to avoid all of those finitely many points, contradicting that $p$ is a limit point. We can therefore choose increasing indices $n_k$ such that $$ d(x_{n_k},p)<\frac{1}{k}. $$ At each step, the ball $B_{1/k}(p)$ contains infinitely many values of the sequence, while only finitely many terms occur among the first $n_{k-1}$ indices. A value in that ball consequently occurs at some index larger than $n_{k-1}$. The displayed bound shows that $x_{n_k}\to p$. This proves sequential compactness.

Thus open-cover compactness implies sequential compactness in a metric space. The criterion above then implies that every compact metric space is complete and totally bounded. The reverse connection between the open-cover and sequential formulations is another reason the completeness–total-boundedness criterion is useful; the open-cover formulation will be examined further in its own right.

Worked Examples: Applying the Criterion

Worked Example: The Closed Interval from Completeness and Nets

Consider $X=[-2,3]$ with the usual metric. This set is complete: it is a closed subset of the complete metric space $\mathbb{R}$, so the Closed Subset of a Complete Space theorem applies.

To verify total boundedness directly, fix $\varepsilon>0$, and choose a positive integer $N$ so large that $5/N<\varepsilon$. Use the finite set $$ F=\left\{-2+\frac{5j}{N}:j=0,1,\ldots,N\right\}. $$ The mesh points divide $[-2,3]$ into intervals of length $5/N$. For any $x\in[-2,3]$, there is a mesh point $z\in F$ in the same interval, so $$ |x-z|\leq\frac{5}{N}<\varepsilon. $$ Thus $F$ is an $\varepsilon$-net. The interval is totally bounded, and the criterion shows that it is sequentially compact.

Worked Example: The Open Interval Is Not Sequentially Compact

Let $X=(0,1)$ with the usual metric, and consider $x_n=1/n$ for $n\geq2$. The sequence converges to $0$ in $\mathbb{R}$, and every subsequence also converges to $0$ in $\mathbb{R}$. None can converge to a point of $(0,1)$: if a subsequence converged to $x\in(0,1)$ in the subspace, it would also converge to $x$ in $\mathbb{R}$, contradicting uniqueness of metric limits. Therefore $(0,1)$ is not sequentially compact.

The criterion identifies the failed condition. The sequence $(1/n)$ is Cauchy in $(0,1)$ but has no limit there, so this metric space is not complete. This example shows why total boundedness alone cannot guarantee sequential compactness.

Worked Example: An Infinite Discrete Space Is Not Sequentially Compact

Let $X$ be an infinite set with the discrete metric $\delta(x,y)=0$ if $x=y$ and $\delta(x,y)=1$ if $x\ne y$. This space is complete: by the earlier Cauchy Sequences in a Discrete Metric theorem, every Cauchy sequence is eventually constant and therefore converges in $X$.

It is not totally bounded. For $\varepsilon=1/2$, every ball $B_{1/2}(x)$ is the singleton $\{x\}$. No finite collection of these balls covers an infinite set. The criterion therefore shows that $X$ is not sequentially compact. Explicitly, choose a sequence of distinct points in $X$. Every subsequence still consists of distinct points, so the distance between any two of its different terms is $1$; no subsequence can converge.

Together, this and the open-interval example show the separate roles of the two conditions. The open interval fails completeness, while the infinite discrete space fails total boundedness.

What the Criterion Tells Us

In metric spaces, compactness is a strong finiteness principle, even when the space contains infinitely many points. Total boundedness says that finitely many centers suffice at every requested accuracy. Completeness ensures that Cauchy behavior does not lead outside the space. Together, these properties ensure that every sequence has a convergent subsequence.

A common pitfall is to stop after obtaining a Cauchy subsequence. A Cauchy subsequence need not converge in an incomplete space, as the sequence $1/n$ in $(0,1)$ illustrates. Another is to assume that completeness alone is enough: an infinite discrete space is complete but not totally bounded. Both hypotheses in the criterion are necessary.

Check total boundedness.
At an arbitrary positive accuracy, construct a finite net, or use the sequential characterization to test whether one must exist.

Check completeness.
Verify that every Cauchy sequence converges to a point in the space, not merely in a larger ambient space.

Combine the conclusions.
Total boundedness gives a Cauchy subsequence; completeness makes it converge in the space.

Check Your Understanding

Use the definitions and criterion to answer the following questions.

What is the difference between sequential compactness and compactness in the open-cover sense?
Where does total boundedness enter the proof that completeness plus total boundedness gives sequential compactness?
Why does sequential compactness imply completeness?
Which condition fails for the open interval $(0,1)$, and which condition fails for an infinite discrete space?
Why does a compact metric space have a finite $\varepsilon$-net for every $\varepsilon>0$?
Does finding a Cauchy subsequence alone prove sequential compactness? What additional property is needed?

What You'll Learn