Compactness Implies Sequential Compactness in Metric Spaces

February 17, 2026 · Luciano Muratore

Compactness Implies Sequential Compactness in Metric Spaces

In metric spaces, compactness is equivalent to sequential compactness.

Here we prove one direction:

If $(X,d)$ is compact, then every sequence in $X$ admits a convergent subsequence whose limit lies in $X$ .

Goal

Let $(X,d)$ be a compact metric space and let $(x_n) \subset X$ be a sequence.

We want to construct a convergent subsequence of $(x_n)$ .

Step 1 — The Trivial Case

If some value of the sequence appears infinitely many times, then we can extract a constant subsequence.
Such a subsequence is convergent, and we are done.

So assume that no value appears infinitely many times.

Define the set

A = \{ x_n : n \in \mathbb{N} \}.

Under our assumption, $A$ is infinite.

Step 2 — Compactness of the Closure

Since $X$ is compact, every closed subset of $X$ is compact.

Consider the closure $\overline{A} \subset X$ .
Because it is closed in a compact space, it is compact.

This allows us to use the finite subcover property.

Step 3 — First Open Cover

Consider the open cover of $A$ given by

\{ B(x_n, 1) \}_{n \in \mathbb{N}}.

By compactness of $\overline{A}$ , there exists a finite subcover:

A \subset B(x_{n_1},1) \cup \cdots \cup B(x_{n_k},1).

Since $A$ is infinite, at least one of these balls contains infinitely many points of $A$ .

Call this ball

B_1 = B(x_{n_1}, 1).

Choose $x_{m_1} \in A \cap B_1$ as the first element of our subsequence.

Step 4 — Iterative Construction

Now consider the compact set

A \cap B_1.

Cover it with open balls of radius $\frac{1}{2}$ centered at points of $A$ .

By compactness, extract a finite subcover.
At least one of these balls contains infinitely many points of $A$ .

Call this ball

B_2 \subset B_1,

with radius $\frac{1}{2}$ .

Choose

x_{m_2} \in A \cap B_2, \qquad m_2 > m_1.

Proceed inductively.

We construct a nested sequence of open balls:

B_1 \supset B_2 \supset B_3 \supset \cdots

such that:

The radius of $B_k$ is $\frac{1}{k}$ .
Each $B_k$ contains infinitely many points of $A$ .
We choose $x_{m_k} \in A \cap B_k$ with $m_k > m_{k-1}$ .

This defines a subsequence $(x_{m_k})$ .

Step 5 — Cauchy Property

Because the balls are nested and their radii tend to zero:

\text{radius}(B_k) = \frac{1}{k} \to 0,

the subsequence $(x_{m_k})$ is Cauchy.

Indeed, for $j,k$ sufficiently large, both $x_{m_j}$ and $x_{m_k}$ lie in a ball of arbitrarily small radius.

Step 6 — Convergence

Since $\overline{A}$ is compact, it is complete.

Therefore, every Cauchy sequence in $\overline{A}$ converges.

Hence,

x_{m_k} \to x

for some $x \in \overline{A} \subset X$ .

Conclusion

Every sequence in $X$ admits a convergent subsequence with limit in $X$ .

Therefore,

X \text{ is sequentially compact.}

\blacksquare

Systematic Structure of the Proof

Define $A = \{x_n : n \in \mathbb{N}\}$ .
If a value appears infinitely many times, extract a constant subsequence.
Otherwise, $A$ is infinite.
The closure $\overline{A}$ is compact.
Cover $A$ with balls of radius $1$ and extract a finite subcover.
Choose a ball containing infinitely many points.
Repeat with radii $\frac{1}{2}, \frac{1}{3}, \dots$
Obtain nested balls with radii tending to zero.
Select one point from each ball to build a subsequence.
The subsequence is Cauchy and hence convergent in the compact space.

Why This Construction Is Important

This proof is constructive:

It does not assume completeness.
It does not rely on Heine–Borel in $\mathbb{R}^n$ .
It works in any compact metric space.

It reveals the deep mechanism behind compactness:

Compactness allows infinite data to be forced into arbitrarily small regions.

This mechanism is foundational in analysis, topology, and functional analysis.