# Discrete Space is First-Countable

Jump to navigation
Jump to search

## Theorem

Let $T = \left({S, \tau}\right)$ be a discrete topological space.

Then $T$ is first-countable.

## Proof

From Point in Discrete Space is Neighborhood, every point $x \in S$ is contained in an open set $\left\{{x}\right\}$.

From the definition of local basis, it is clear that $\left\{{\left\{{x}\right\}}\right\}$ is (trivially) a local basis at $x$.

That is, that every open set of $S$ containing $x$ also contains at least one of the sets of $\left\{{\left\{{x}\right\}}\right\}$.

Equally trivially, we have that $\left\{{\left\{{x}\right\}}\right\}$ is countable.

Hence the result by definition of first-countable.

$\blacksquare$

## Sources

- 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*... (previous) ... (next): $\text{II}: \ 1 - 3: \ 7$