Equivalence of Definitions of Closure of Topological Subspace

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Let $H \subseteq S$.

The following definitions of topological closure of $H$ in $T$ are equivalent:

Definition $1$ is equivalent to Definition $2$
This is proved in Set Closure as Intersection of Closed Sets.

Definition $2$ is equivalent to Definition $3$
This is proved in Set Closure is Smallest Closed Set.

Definition $1$ is equivalent to Definition $4$
By the definition of the interior, and Set is Subset of its Topological Closure, it easily follows that
 * $H^\circ \subseteq H \subseteq H^-$

Then:

Definition $1$ is equivalent to Definition $5$
Every isolated point of $H$ is a point of $H$.

So by Set Union Preserves Subsets:
 * $(1): \quad H^i \cup H' \subseteq H \cup H'$

If $S \setminus \left({H^i \cup H'}\right) = \varnothing$, then $(1)$ yields $H^i \cup H' = H \cup H'$ and so the proof is complete.

Otherwise, let $x \in S \setminus \left({H^i \cup H'}\right) \ne \varnothing$.

From De Morgan's Laws:
 * $S \setminus \left({H^i \cup H'}\right) = \left({S \setminus H^i}\right) \cap \left({S \setminus H'}\right)$

Thus $x$ is a point of $S$ that is neither an isolated point of $H$ nor a limit point of $H$.

Then there exist an open set $U$ that contains $x$ such that:
 * $U \cap H \ne \left\{{x}\right\}$

and:
 * $H \cap \left({U \setminus \left\{{x}\right\}}\right) = \varnothing$

This implies that
 * $U \cap H = \varnothing$

By Condition for Point being in Closure:
 * $x \notin H \cup H'$

It has been shown that:
 * $x \notin H^i \cup H' \implies x \notin H \cup H'$

From Rule of Transposition:
 * $x \in H \cup H' \implies x \in H^i \cup H'$

And so
 * $(2): \quad H \cup H' \subseteq H^i \cup H'$

Combining $(1)$ and $(2)$:
 * $H \cup H' = H^i \cup H'$

Hence the result.