Either-Or Topology is First-Countable

Theorem

Let $T = \struct {S, \tau}$ be the either-or space.

Then $T$ is a first-countable space.

Proof

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

Then $\set x$ 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 $\openint {-1} 1$.

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

Hence the result by definition of first-countable space.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $17$. Either-Or Topology: $3$