Either-Or Topology is Topology

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

Then $\tau$ is a topology on $T$.

Proof
From the definition:


 * $H \in \tau \iff \left({\left\{{0}\right\} \nsubseteq H \lor \left({-1 \,.\,.\, 1}\right) \subseteq H}\right)$

for any $H \subseteq S$.

First note that:
 * $\left\{{0}\right\} \nsubseteq \varnothing$ and so $\varnothing \in \tau$


 * $\left({-1 \,.\,.\, 1}\right) \subseteq S$ and so $S \in \tau$

Now suppose $H_1, H_2 \in \tau$.

If either $\left\{{0}\right\} \nsubseteq H_1$ or $\left\{{0}\right\} \nsubseteq H_2$ then $\left\{{0}\right\} \nsubseteq H_1 \cap H_2$ from De Morgan's Laws (indirectly).

Otherwise $\left({-1 \,.\,.\, 1}\right) \subseteq H_1$ and $\left({-1 \,.\,.\, 1}\right) \subseteq H_2$, and so $\left({-1 \,.\,.\,1}\right) \subseteq H_1 \cap H_2$ from Intersection is Largest Subset.

So $H_1, H_2 \in \tau \implies H_1 \cap H_2 \in \tau$.

Now let $\mathcal H \subseteq \tau$ be a set of elements of $\tau$.

Then either:
 * $\forall H \in \mathcal H: \left\{{0}\right\} \nsubseteq H$ in which case $\left\{{0}\right\} \nsubseteq \bigcup \mathcal H$

or:
 * $\exists H \in \mathcal H: \left({-1 \,.\,.\, 1}\right) \subseteq H$ in which case $\left({-1 \,.\,.\, 1}\right) \subseteq \bigcup \mathcal H$

In both cases $\bigcup \mathcal H \in \tau$.

Hence the result, from the definition of topology.

Note that $\left\{{0}\right\} \nsubseteq H \implies \left({-1 \,.\,.\, 1}\right) \nsubseteq H$ so there does not exist the confusion of: "What happens if the conditions are contradictory?"