Lindelöf Metric Space is Second-Countable

Theorem

Let $M = \left({A, d}\right)$ be a metric space.

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

Proof

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.

So all we need to do is demonstrate that if $M$ is Lindelöf then it is second-countable.

Suppose $M$ is Lindelöf.

Let us define the open covers on $X$:

$\mathcal C_k = \left\{{N_{1/k} \left({x}\right): x \in X}\right\}$

for all $k \in \N^*$.

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 $X$.

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

$\blacksquare$

Sources

• Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (1970)... (previous)... (next): $\text{I}: \ \S 5$