Condition for Open Extension Space to be First-Countable

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $T^*_{\bar p} = \left({S^*_p, \tau^*_{\bar p}}\right)$ be the open extension space of $T$.


Then $T^*_{\bar p}$ is a first-countable space if and only if $T$ is.


Proof

Let $T = \left({S, \tau}\right)$ be a first-countable space.

Then every point in $S$ has a countable local basis.

Every open set of $T$ is an open set of $T^*_{\bar p}$ by definition.

So if $x$ has a countable local basis in $T$, then it has one in $T^*_{\bar p}$ as well.

Finally we note that $p$ is by definition in exactly one open set of $T^*_{\bar p}$, that is, $S^*_p$, and thus (trivially) has a countable local basis in $T^*_{\bar p}$.

Then every point in $S^*_p$ has a countable local basis in $T^*_{\bar p}$.

So if $T$ is a first-countable space, then $T^*_{\bar p}$ is also a first-countable space.


Now suppose $T = \left({S, \tau}\right)$ is not a first-countable space.

Then $\exists x \in S$ such that $x$ has no countable local basis in $T$.


As every open set of $T$ is an open set of $T^*_{\bar p}$ by definition, it follows that $x$ has no countable local basis in $T^*_{\bar p}$ either.


So if $T $ is not a first-countable space, then neither is $T^*_{\bar p}$ a first-countable space.

$\blacksquare$


Sources