Either-Or Topology is First-Countable
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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: $3$