Condition for Open Extension Space to be Second-Countable

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Theorem

Let $T = \struct {S, \tau}$ be a topological space.

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


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


Proof

Let $T = \struct {S, \tau}$ 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 $\tau^*_{\bar p} = \tau \cup \set {S^*_p}$ has one as well.

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


Now suppose $T = \struct {S, \tau}$ 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 $\tau^*_{\bar p} = \tau \cup \set {S^*_p}$ 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

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