Set Closure as Intersection of Closed Sets

Theorem
Let $T$ be a topological space, and let $H \subseteq T$.

Let $\mathbb K$ be defined as:
 * $\mathbb K := \left\{{K \supseteq H: K}\right.$ is closed in $\left.{T}\right\}$

That is, let $\mathbb K$ be the set of all closed sets of $T$ which contain $H$.

Then the closure of $H$ can be defined as:
 * $\displaystyle \operatorname{cl} \left({H}\right) := \bigcap \mathbb K$

That is, as the intersection of all the closed sets of $T$ which contain $H$.

Proof
What needs to be proved here is:


 * $\displaystyle \operatorname{cl} \left({H}\right) = \bigcap \mathbb K$

where:
 * $\mathbb K$ is as defined above: $\mathbb K$ is the set of all closed sets of $T$ which contain $H$.
 * $\operatorname{cl} \left({H}\right)$ is the closure of $H$ defined as the union of $H$ and its limit points.

Let us define:
 * $\displaystyle V := \bigcap \mathbb K$

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

Let $K \in \mathbb K$.

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

From Set is Subset of Closure, $H \subseteq \operatorname{cl}\left({H}\right)$.

So $\operatorname{cl}\left({H}\right)$ is, by definition, a closed set of $T$ which contains $H$.

By we have by its definition that $V$ is the intersection of all closed sets in $T$ that contain $H$.

So from Intersection Subset it follows that $V \subseteq \operatorname{cl}\left({H}\right)$.

Finally, we have that:


 * $\displaystyle \operatorname{cl}\left({H}\right) \subseteq \bigcap \mathbb K$
 * $\displaystyle \bigcap \mathbb K \subseteq \operatorname{cl}\left({H}\right)$

So by definition of set equality:
 * $\displaystyle \operatorname{cl}\left({H}\right) = \bigcap \mathbb K$

which is what we needed to prove.