Either-Or Topology is Topology

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

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

Proof
From the definition:


 * $H \in \tau \iff \paren {\set 0 \nsubseteq H \lor \openint {-1} 1 \subseteq H}$

for any $H \subseteq S$.

First note that:
 * $\set 0 \nsubseteq \O$ and so $\O \in \tau$


 * $\openint {-1} 1 \subseteq S$ and so $S \in \tau$

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

If either $\set 0 \nsubseteq H_1$ or $\set 0 \nsubseteq H_2$ then from De Morgan's Laws: Difference with Intersection (indirectly):
 * $\set 0 \nsubseteq H_1 \cap H_2$

Otherwise:
 * $\openint {-1} 1 \subseteq H_1$ and $\openint {-1} 1 \subseteq H_2$

and so from Intersection is Largest Subset:
 * $\openint {-1} 1 \subseteq H_1 \cap H_2$

So:
 * $H_1, H_2 \in \tau \implies H_1 \cap H_2 \in \tau$

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

Then either:
 * $\forall H \in \HH: \set 0 \nsubseteq H$ in which case $\set 0 \nsubseteq \bigcup \HH$

or:
 * $\exists H \in \HH: \openint {-1} 1 \subseteq H$ in which case $\openint {-1} 1 \subseteq \bigcup \HH$

In both cases $\displaystyle \bigcup \HH \in \tau$.

Note that $\set 0 \nsubseteq H \implies \openint {-1} 1 \nsubseteq H$ so there does not exist the confusion of what happens if the conditions are contradictory.

Hence the result, from the definition of topology.