Basis for Excluded Point Space
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Theorem
Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Consider the set $\BB$ defined as:
- $\BB = \set {\set x: x \in S \setminus \set p} \cup \set S$
Then $B$ is a basis for $S$.
Proof
Let $H \in \tau_{\bar p}$ be open in $T$.
If $H = S$ then trivially $H$ is the union of elements of $\BB$.
So suppose $H \ne S$.
Then by definition $p \notin H$ and so:
- $\forall y \in H: \exists \set y \in \BB$
Thus:
- $\ds H = \bigcup_{y \mathop \in H} \set y$
So $\BB$ is an analytic basis for $T$.
$\blacksquare$
It could equally well be shown that $\BB$ is also a synthetic basis for $T$.