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

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

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