Sequentially Compact Metric Space is Lindelöf

Theorem
Let $M = \struct {X, d}$ be a sequentially compact metric space.

Then $M$ is also a Lindelöf space.

That is, from every open cover of $M$, it is possible to extract a countable subcover.

Proof
Take any open cover $C$ of $M$.

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

We have that a Sequentially Compact Metric Space is Second-Countable.

Thus, by definition, the topology of $M$ has a countable basis.

Let $\BB$ be a countable basis for the topology induced by $d$ of $M$.

Let $x \in X$.

As $C$ covers $X$:


 * $\exists U_x \in C: x \in U_x$

As $\BB$ is a basis:


 * $\exists B_x \in \BB: x \in B_x \subseteq U_x$

Thus:
 * $(1): \quad \forall x \in X: \exists B_x \in \BB: x \in B_x$

Consider the set $\Sigma := \set {B_x: x \in X}$.

$\Sigma$ is a subset of $\BB$.

Hence $\Sigma$ is countable.

As $\Sigma$ contains every $x \in X$ from $(1)$, $\Sigma$ covers $X$.

By construction of $\Sigma$, every open set in $\Sigma$ is contained in some $U \in C$.

For each open set $B \in \Sigma$, choose one $U_B \in C$ such that $B \subseteq U_B$.

Let $\UU$ be the set defined as:
 * $\UU = \set {U_B: B \in \Sigma}$

As $\UU$ does not have more sets than $\Sigma$, $\UU$ is countable.

We have that:
 * $\forall B \in \Sigma: B \subseteq U_B$

Thus it follows that $\UU$ covers $X$.

As:
 * $\forall U_B \in \UU: U_B \in C$

$U$ is a countable subcover of $C$.

Thus a countable subcover has been obtained from $C$.

As $C$ is arbitrary, it follows that $M$ is a Lindelöf space.

Also see

 * Sequentially Compact Metric Space is Compact