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 $T$.