Either-Or Topology is T0

Theorem

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

Then $T$ is a $T_0$ (Kolmogorov) space.

Proof

Let $x, y \in S$ such that $x \ne y$.

Without loss of generality, let $x \ne 0$.

Then $U = \set x$ is open in $T$ from the definition of the either-or topology.

We have that $x$ and $y$ are arbitrary.

So:

$\forall x, y \in S: \exists U \in \tau: x \in U, y \notin U$

and the result follows by definition of $T_0$ (Kolmogorov) 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: $1$