Either-Or Topology is not T1

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Theorem

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


Then $T$ is not a $T_1$ (Fréchet) space.


Proof

Let $x = \dfrac 1 2$.

We have that $V = \left\{{x}\right\}$ such that $x \in V, 0 \notin V$.

However, by definition of the either-or topology, the only open sets of $T$ containing $0$ also contain $\left({-1 \,.\,.\, 1}\right)$, and so must also contain $x$.

So we have that $\not \exists U, V \in \tau: 0 \in U, x \notin U, x \in V, 0 \notin V$.

Hence the result by definition of $T_1$ (Fréchet) space.

$\blacksquare$


Sources