Condition for Open Extension Space to be Second-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 second-countable space if and only if $T$ is.


Proof

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

Then $\tau$ has a countable basis.

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

So if $\tau$ has a countable basis in $T$, then $\tau^*_{\bar p} = \tau \cup \left\{{S^*_p}\right\}$ has one as well.

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


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

Then $\tau$ has no countable basis in $T$.

As every open set of $T$ is an open set of $T^*_{\bar p}$ by definition, it follows that $\tau^*_{\bar p} = \tau \cup \left\{{S^*_p}\right\}$ has no countable basis either.

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

$\blacksquare$


Sources