Equivalence of Definitions of Everywhere Dense

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

Let $H \subseteq S$ be a subset.

$(1)$ implies $(2)$
Let $H$ be a subset of $S$ which is everywhere dense in $T$ by definition $1$.

Then by definition:
 * $H^- = S$

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

there exists $U \in \tau$ such that $U \cap H = \O$.

Let $x \in S$ such that $x \in U$.

Thus $U$ is an open set of $T$ which does not contain an element of $H$ which is distinct from $x$.

Indeed, $U$ is an open set of $T$ which does not contain any elements of $H$ at all.

Hence, by definition, $x$ is not a limit point of $H$.

By definition, the closure of $H$ is the union of $H$ with its limit points.

But $x$ is not in $H$, nor is it a limit point of $H$.

That is:
 * $x \notin H^-$

So $H^- \ne S$.

But this contradicts our definition of everywhere dense in $T$ by definition $1$.

Hence our assertion that there exists $U \in \tau$ such that $U \cap H = \O$ must be false.

So, by Proof by Contradiction:
 * $\forall U \in \tau: H \cap U \ne \O$

Thus $H$ is a subset of $S$ which is everywhere dense in $T$ by definition $2$.

$(2)$ implies $(1)$
Let $H$ be a subset of $S$ which is everywhere dense in $T$ by definition $2$.

Then by definition:
 * $\forall U \in \tau \setminus \set \O: H \cap U \ne \O$

It is taken for granted that $H^- \subseteq S$.

$H^- \ne S$.

Then:
 * $\exists x \in S: x \notin H^-$

Thus:
 * $x \notin H$

and also, $x$ is not a limit point of $H$.

Hence, by definition of limit point:
 * $\exists U \in \tau: \not \exists y \in U: y \in H$

Hence by definition of set intersection:


 * $U \cap H = \O$

But this contradicts our definition of everywhere dense in $T$ by definition $2$.

So, by Proof by Contradiction:
 * $H^- = S$

Thus $H$ is a subset of $S$ which is everywhere dense in $T$ by definition $1$.