Everywhere Dense iff Interior of Complement is Empty

Theorem

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

Let $A \subset S$.

Then $A$ is everywhere dense if and only if:

$\paren {\relcomp S A}^\circ = \O$

where $A^\circ$ is the interior of $A$.

Proof

By definition of everywhere dense, $A$ is everywhere dense if and only if:

$A^- = S$

where $A^-$ is the closure of $A$.

That happens if and only if:

\(\ds \paren {\relcomp S A}^\circ\)

\(=\)

\(\ds \relcomp S {A^-}\)

Complement of Interior equals Closure of Complement

\(\ds \)

\(=\)

\(\ds \relcomp S S\)

\(\ds \)

\(=\)

\(\ds \O\)

Relative Complement with Self is Empty Set

Hence the result.

$\blacksquare$