Excluded Point Space is First-Countable

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Theorem

Let $T = \left({S, \tau_{\bar p}}\right)$ be an excluded point space.


Then $T$ is first-countable.


Proof

Let $x \in S: x \ne p$.

By definition, the set $\left\{{x}\right\}$ is open in $T$.

Let $U \in \tau_{\bar p}: x \in U$.

Then $\left\{{x}\right\} \subseteq U$ and so $\left\{{\left\{{x}\right\}}\right\}$ is a local basis at $x$ which is trivially countable.


Now if $x = p$ there is only one $U \in \tau_{\bar p}: p \in U$, and that is $S$.

So $\left\{{S}\right\}$ is a local basis at $x$ which is trivially countable.


Hence the result, by definition of first-countable.

$\blacksquare$


Sources