Sequentially Compact Metric Space is Lindelöf

This article contains the proof that a sequentially compact metric space is Lindelöf (i.e., from every cover one can extract a countable subcover).

This result is part of a possible proof of a more general result, namely, that a sequentially compact metric space is compact (i.e., from every cover one can extract a finite subcover). This is why it is stated as a lemma and not a theorem.

Lemma
A sequentially compact metric space is Lindelöf.

Proof
Let $$(X,d)$$ be a metric space, and take any open cover $$C$$.

We need to find a countable subset of $$C$$ which still covers $$X$$.

We will use the result that the topology of a sequentially compact metric space has a countable base.

So, take a countable base $$\mathcal{B}$$ for the topology $$(X,d)$$.

For each $$x \in X$$:
 * Take $$U_x \in C$$ such that $$x \in U_x$$ (which can be done, as $$C$$ covers all of $$X$$).
 * Take $$B_x \in \mathcal{B}$$ such that $$x \in B_x \subseteq U_x$$ (which can be done because $$\mathcal{B}$$ is a base).

Then, consider the family $$\Sigma := \{B_x \mid x \in X \}$$.

This family is a subset of $$\mathcal{B}$$, and hence is countable (notice there must be many repetitions here if the space $$X$$ is large; $$B_x$$ may be the same for many points $$x$$).

Also, $$\Sigma$$ covers $$X$$ (it contains every point $$x \in X$$, as $$x \in B_x$$).

Now, for each (open) set $$B \in \Sigma$$, choose one $$U_B \in C$$ such that $$B \subseteq U_B$$ (this can be done because every open set in $$\Sigma$$ is contained in some $$U \in C$$, by construction).

This family:
 * $$\{ U_B \mid B \in \Sigma \}$$

is what we want:


 * It is countable (as it does not have more sets than $$\Sigma$$).
 * It is an open cover of $$X$$, as it covers more than $$\Sigma$$ (for each $$B \in \Sigma$$, $$B \subseteq U_B$$), and $$\Sigma$$ already covers $$X$$.
 * It is a subcover of $$C$$, as we chose each $$U_B$$ among the sets in $$C$$.

This proves that $$X$$ is Lindelöf.