Equivalence of Definitions of Closure of Topological Subspace

Theorem
Let $T$ be a topological space.

Let $H \subseteq T$.

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


 * $(1): \quad \operatorname{cl} \left({H}\right)$ is the union of $H$ and its limit points
 * $(2): \quad \operatorname{cl} \left({H}\right) = \displaystyle \bigcap_{H \subseteq K \subseteq T: K \text{ closed}} K$
 * $(3): \quad \operatorname{cl} \left({H}\right)$ is the smallest closed set that contains $H$
 * $(4): \quad \operatorname{cl} \left({H}\right)$ is the union of $H$ and its boundary
 * $(5): \quad \operatorname{cl} \left({H}\right)$ is the union of all isolated points of $A$ and all limit points of $H$

$(1) \implies (2)$
Let $\displaystyle V = \bigcap_{H \subseteq K \subseteq T: K \text{ closed}} K$.

That is, $V$ is the intersection of all closed sets in $T$ that contain $H$.

Let $K$ be closed, and let $H \subseteq K$.

From statement $(1)$:
 * $\operatorname{cl} \left({H}\right)$ is the union of $H$ and its limit points

these proofs have been demonstrated:
 * Closure of Subset is Subset of Closure
 * Closed Set Equals its Closure
 * Closure is Closed.

From Closure of Subset is Subset of Closure, we have:
 * $\operatorname{cl}\left({H}\right) \subseteq \operatorname{cl}\left({K}\right)$

From Closed Set Equals its Closure, we have $\operatorname{cl}\left({K}\right) = K$.

So $\operatorname{cl}\left({H}\right) \subseteq V$.

Conversely, from Closure is Closed, $\operatorname{cl}\left({H}\right)$ is closed.

Hence $V \subseteq \operatorname{cl}\left({H}\right)$.

So $V = \operatorname{cl}\left({H}\right)$ and hence statement $(1)$ implies statement $(2)$.

$(2) \implies (3)$
Let $V = \operatorname{cl}\left({H}\right)$ be defined as in statement $(2)$.

If $K$ is closed in $T$ and $H \subseteq K$, then $V \subseteq K$ from Intersection Subset.

So $V = \operatorname{cl}\left({H}\right)$ is a subset of any closed set in $T$ which contains $H$, and so is the smallest closed set that contains $H$.

$(3) \implies (2)$
Let $K$ be closed set in $T$ such that $H \subseteq K$.

If $V$ is the smallest closed set that contains $H$, then $V \subseteq K$.

It follows from Intersection Largest: General Result that $V$ is the intersection of all closed sets in $T$ that contain $H$.

$(3) \implies (1)$
Let $V = \operatorname{cl}\left({H}\right)$ be defined as in statement $(3)$:
 * $V$ is the smallest closed set that contains $H$.

Let $U$ be the union of $H$ and its limit points.

Let $x \in V$.

If $x \in H$ then $x \in U$.

However, suppose $x \notin H$.

Then