Element of Ordered Set of Topology is Dense iff is Everywhere Dense

Theorem
Let $T = \struct {S, \tau}$ be a topological space.

Let $P = \struct {\tau, \preceq}$ be an ordered set where $\mathord \preceq = \mathord \subseteq \cap \paren {\tau \times \tau}$

Let $A \in \tau$.

Then $A$ is a dense element in $P$ $A$ is everywhere dense.

Sufficient Condition
Assume that
 * $A$ is a dense element in $P$.

By Bottom in Ordered Set of Topology:
 * $\bot_P = \O$

We will prove that
 * for every open subset $U$ of $S$: $U \ne \O \implies U \cap A \ne \O$

Let $U$ be an open subset of $S$ such that
 * $U \ne \O$

By definition of open set:
 * $U \in \tau$

By definition of topological space:
 * $U \cap A \in \tau$

By Meet in Inclusion Ordered Set:
 * $U \wedge A = U \cap A$

Thus by definition of dense element:
 * $U \cap A \ne \O$

By Characterization of Closure by Open Sets:
 * $A^- = S$

where $A^-$ denotes the topological closure of $A$.

Hence $A$ is everywhere dense.

Necessary Condition
Assume that
 * $A$ is everywhere dense.

Let $B \in \tau$ such that
 * $B \ne \bot_P$

By definition of non-empty set:
 * $\exists x: x \in B$

By definition:
 * $B$ is an open set.

By definition of everywhere dense:
 * $A^- = S$

By Characterization of Closure by Open Sets:
 * $B \cap A \ne \O$

By definition of topological space:
 * $B \cap A \in \tau$

Thus by Meet in Inclusion Ordered Set:
 * $B \wedge A \ne \bot_P$