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

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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 $P$ if and only if $A$ is everywhere dense.


Proof

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:

$U \cap A \ne \O$

$\Box$


By Characterization of Closure by Open Sets:

$A^- = S$

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

Hence $A$ is everywhere dense.

$\Box$


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:

$B \wedge A \ne \bot_P$

$\blacksquare$


Sources

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