Metric Space is Lindelöf iff Second-Countable

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

Then $M$ is Lindelöf $M$ is second-countable.

Sufficient Condition
We have from Second-Countable Space is Lindelöf that second-countability implies Lindelöf in all topological spaces, regardless of whether they are metric spaces or not.

Necessary Condition
Suppose $M = \struct {A, d}$ is Lindelöf.

Let us define the open covers on $A$:
 * $\CC_k = \set {\map {N_{1/k} } x: x \in S}$

for all $k \in \N_{>0}$.

As $M$ is Lindelöf, each one of these has a countable subcover.

The union of all these subcovers is a countable basis for the topology on $A$.

Hence the result, by definition of second-countable space.