Basis for Either-Or Topology

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

Let $\BB$ be the set:


 * $\BB := \set {\set x: x \in S, x \ne 0} \cup \set {\openint {-1} 1}$

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

Then $\BB$ is a basis for $T$.

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

Then:


 * $\ds U = \bigcup_{x \mathop \in U} \set x$

where $x \ne 0$.

Hence for all $x \in U$, we have $\set x \in \BB$.

Thus $U$ is the union of elements of $\BB$.

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

Then $\openint {-1} 1 \subseteq U$ by definition.

So one of four cases holds:
 * $U = \openint {-1} 1$


 * $U = \hointr {-1} 1 = \openint {-1} 1 \cup \set {-1}$


 * $U = \hointl {-1} 1 = \openint {-1} 1 \cup \set 1$


 * $U = \closedint {-1} 1 = \openint {-1} 1 \cup \set {-1} \cup \set 1$

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

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