Metric Space is First-Countable

From ProofWiki
Jump to: navigation, search


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

Then $M$ is first-countable.


Let $x \in A$.


$\mathcal B = \left\{{B_{1/n} \left({x}\right): n \in \N_{>0}}\right\}$

where $B_\epsilon \left({x}\right)$ denotes the open $\epsilon$-ball of $x$ in $M$.

By Surjection from Natural Numbers iff Countable, we have that $\mathcal B$ is countable.

By the definition of a first-countable space, it suffices to show that $\mathcal B$ is a local basis at $x$.

By Open Ball is Open Set, every element of $\mathcal B$ is an open neighborhood of $x$.

Let $U$ be an open neighborhood of $x$.

By the definition of an open set, there exists a strictly positive real number $\epsilon$ such that $B_\epsilon \left({x}\right) \subseteq U$.

By the Archimedean Principle, there exists a natural number $n > \dfrac 1 \epsilon$.

That is, $\dfrac 1 n < \epsilon$, and so $B_{1/n} \left({x}\right) \subseteq B_\epsilon \left({x}\right) \subseteq U$.

The result follows from Subset Relation is Transitive.