Either-Or Topology is Topology

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

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

Let $\UU \subseteq \tau$.

Then either:
 * $\forall U \in \UU: \set 0 \nsubseteq U$ in which case $\set 0 \nsubseteq \bigcup \UU$

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

In both cases:
 * $\ds \bigcup \UU \in \tau$

Note that:
 * $\set 0 \nsubseteq U \implies \openint {-1} 1 \nsubseteq U$

so there does not exist the confusion of what happens if the conditions are contradictory.

Let $U_1, U_2 \in \tau$.

By definition of either-or topology, either:
 * $\openint {-1} 1 \subseteq U_1$ and $\openint {-1} 1 \subseteq U_2$

or:
 * $\set 0 \nsubseteq U_1$ or $\set 0 \nsubseteq U_2$

Suppose:
 * $\openint {-1} 1 \subseteq U_1$ and $\openint {-1} 1 \subseteq U_2$

From Intersection is Largest Subset:
 * $\openint {-1} 1 \subseteq U_1 \cap U_2$

By definition of either-or topology:
 * $U_1 \cap U_2 \in \tau$

Suppose:
 * $\set 0 \nsubseteq U_1$ or $\set 0 \nsubseteq U_2$

Then:

By definition of either-or topology:
 * $U_1 \cap U_2 \in \tau$

Thus by Proof by Cases:
 * $\forall U_1, U_2 \in \tau: U_1 \cap U_2 \in \tau$

From Open Real Interval is Subset of Closed Real Interval:
 * $\openint {-1} 1 \subseteq \closedint {-1} 1 = S$

By definition of either-or topology:
 * $S \in \tau$

Hence the result, from the definition of topology.