Either-Or Topology is T0

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Theorem

Let $T = \left({S, \tau}\right)$ 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 $U = \left\{{x}\right\}$ 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.

$\blacksquare$


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