Basis for Excluded Point Space

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Theorem

Let $T = \left({S, \tau_{\bar p}}\right)$ be an excluded point space.


Consider the set $\mathcal B$ defined as:

$\mathcal B = \left\{{\left\{{x}\right\}: x \in S \setminus \left\{{p}\right\}}\right\} \cup \left\{{S}\right\}$

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 $\mathcal B$.

So suppose $H \ne S$.

Then by definition $p \notin H$ and so:

$\forall y \in H: \exists \left\{{y}\right\} \in \mathcal B$

Thus:

$\displaystyle H = \bigcup_{y \in H} \left\{{y}\right\}$

So $\mathcal B$ is an analytic basis for $T$.

$\blacksquare$


It could equally well be shown that $\mathcal B$ is also a synthetic basis for $T$.