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 $(3)$ implies Definition $(4)$
Let $H^-$ be the closure of $H$ by definition 3.

Let $x$ be a point in the boundary of $H$.

Suppose $x \notin H^-$.

Since $H^-$ is closed, by definition $S \setminus H^-$ is open in $T$.

Thus by definition $S \setminus H^-$ is a neighbourhood of $x$ which contains no points in $H$.

This contradicts the fact that $x$ is a point in the boundary of $H$.

Definition $(5)$ implies Definition $(2)$
Let $H^-$ be the closure of $H$ by definition 5.

That is:
 * $H^- = H^i \cup H'$

First we show that $H^-$ is closed.

Let $y \in T \setminus H^-$.

Then:
 * $y$ is not an isolated point of $H$
 * $y$ is not a limit point of $H$.

So there exists a neighbourhood $V_y$ of $H$ such that:
 * $V_y \cap \left({H \setminus \left\{{x}\right\} }\right) = \varnothing$

So $H^-$ is a closed and we have one inclusion.

For the reverse inclusion, let $h$ be given.

If $h$ is an isolated point of $H$, the proof is finished.

If $h$ is a limit point of $H$, then every neighbourhood $V_h$ of $h$ has a non-empty intersection with $H \setminus \left\{{h}\right\}$.

Hence the same applies for every $K \setminus \left\{{h}\right\}$.

Since $K$ is closed, $h$ lies in every closed $K$ containing $H$.