Open Set Characterization of Denseness/Analytic Basis

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Theorem

Let $\struct {X, \tau}$ be a topological space.

Let $S \subseteq X$.

Let $\BB \subseteq \tau$ be an analytic basis for $\tau$.


Then $S$ is (everywhere) dense in $X$ if and only if every non-empty open set of $\BB$ contains an element of $S$.

Proof

Necessary Condition

Let $S$ be everywhere dense in $X$.

By Open Set Characterization of Denseness then every non-empty open set contains an element of $S$.

Every non-empty set of an analytic basis is an open set by definition.

Hence every non-empty set of non-empty open set of $\BB$ contains an element of $S$.

$\Box$


Sufficient Condition

Suppose that every non-empty open set of $\BB$ contains an element of $S$.

Let $U$ be any non-empty open set.

Let $x \in U$.

By definition of an analytic basis then:

$\exists B \in \BB: x \in B \subseteq U$

By assumption $B$ contains an element $s$ of $S$.

Hence $s \in U$.

Since $U$ was an arbitrary non-empty open set then every non-empty open set contains an element of $S$

By Open Set Characterization of Denseness, $S$ is everywhere dense in $X$.

$\blacksquare$