Either-Or Topology is First-Countable

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Theorem

Let $T = \left({S, \tau}\right)$ be the either-or space.


Then $T$ is a first-countable space.


Proof

Let $x \in S$ such that $x \ne 0$.

Then $\left\{{x}\right\}$ is open in $T$ and so on its own forms a local basis of $x$ which is (trivially) countable.


Let $x = 0$.

Let $U \in \tau$ be open in $T$ such that $x \in U$.

Then by definition of the either-or space, $U$ contains the open set $\left({-1 \,.\,.\, 1}\right)$.

So $\left({-1 \,.\,.\, 1}\right)$ forms a local basis of $0$ which is (trivially) countable.


Hence the result by definition of first-countable space.

$\blacksquare$


Sources