Closed Sets of Either-Or Topology

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

Then the closed sets of $T$ are:
 * $\varnothing$
 * $S$
 * $\left\{{-1}\right\}$
 * $\left\{{1}\right\}$
 * $\left\{{-1, 1}\right\}$
 * Any subset of $\left[{-1 . . 1}\right]$ containing $\left\{{0}\right\}$ as a subset.

Proof
From Open and Closed Sets in a Topological Space we have that $\varnothing$ and $S$ are closed sets trivially.

From the definition of a closed set, we have:
 * $U$ is open in $T$ iff $S \setminus U$ is closed in $T$
 * $U$ is closed in $T$ iff $S \setminus U$ is open in $T$

where $S \setminus U$ denotes the relative complement of $U$ in $S$.

Now we have that:
 * $\left({-1 . . 1}\right)$ is open in $T$ so $S \setminus \left({-1 . . 1}\right) = \left\{{-1, 1}\right\}$ is closed in $T$.


 * Both $\left[{-1 . . 1}\right) \supseteq \left({-1 . . 1}\right)$ and $\left({-1 . . 1}\right] \supseteq \left({-1 . . 1}\right)$ therefore are open in $T$, so $\left\{{-1}\right\}$ and $\left\{{1}\right\}$ are closed in $T$.

Apart from $S$ itself, there are no other subsets of $S = \left[{-1. . 1}\right]$ which have $\left({-1 . . 1}\right)$ as a subset.

Finally, let $U \in \tau: \left\{{0}\right\} \nsubseteq U$.

Then by the definition of relative complement $\left\{{0}\right\} \subseteq S \setminus U$.

Now suppose $\left\{{0}\right\} \subseteq V$.

Then again by the definition of relative complement $\left\{{0}\right\} \nsubseteq S \setminus V$.

So $S \setminus V$ is open in $T$ and so $V$ is closed in $T$.