Metric Space is First-Countable

Theorem

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

Then $M$ is a first-countable space.

Proof

Let $x \in A$.

Consider the set:

$\mathcal B = \left\{{N_{1/n} \left({x}\right): n \in \N^*}\right\}$

where $N_{1/n} \left({x}\right)$ is the open $\epsilon$-ball neighborhood of $x$.

That is:

$\mathcal B = \left\{{N_1 \left({x}\right), N_{1/2} \left({x}\right), N_{1/3} \left({x}\right), \ldots}\right\}$

Then $\mathcal B$ is a countable local basis at $x$.

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

$\blacksquare$

Sources

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