# Basis for Either-Or Topology

## Theorem

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

Let $\mathcal B$ be the set:

$\mathcal B := \left\{{\left\{{x}\right\}: x \in S, x \ne 0}\right\} \cup \left\{{ \left({-1 \,.\,.\, 1}\right) }\right\}$

... that is, the set of all singleton subsets of $S$ less $\left\{{0}\right\}$ and including the open real interval $\left({-1 \,.\,.\, 1}\right)$.

Then $\mathcal B$ is a basis for $T$.

## Proof

Let $U \in \tau$ such that $0 \notin U$.

Then:

$\displaystyle U = \bigcup_{x \mathop \in U} \left\{{x}\right\}$

where $x \ne 0$.

Hence for all $x \in U$, we have $\left\{{x}\right\} \in \mathcal B$.

Thus $U$ is the union of elements of $\mathcal B$.

Now suppose $U \in \tau$ such that $0 \in U$.

Then $\left({-1 \,.\,.\, 1}\right) \subseteq U$ by definition.

So one of four cases holds:

$U = \left({-1 \,.\,.\, 1}\right)$
$U = \left[{-1 \,.\,.\, 1}\right) = \left({-1 \,.\,.\, 1}\right) \cup \left\{{-1}\right\}$
$U = \left({-1 \,.\,.\, 1}\right] = \left({-1 \,.\,.\, 1}\right) \cup \left\{{1}\right\}$
$U = \left[{-1 \,.\,.\, 1}\right] = \left({-1 \,.\,.\, 1}\right) \cup \left\{{-1}\right\} \cup \left\{{1}\right\}$

All of these sets are in $\mathcal B$, so that $U$ is the union of elements of $\mathcal B$.

Hence, by definition, $\mathcal B$ is a basis for $T$.

$\blacksquare$