# Set Closure as Intersection of Closed Sets

## Theorem

Let $T$ be a topological space.

Let $H \subseteq T$.

Let the closure of $H$ (in $T$) be defined as:

$H^- := H \cup H'$

where $H'$ is the derived set of $H$.

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 $H$ (in $T$) can be defined as:

$\displaystyle H^- := \bigcap \mathbb K$

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

## Proof

What needs to be proved here is:

$\displaystyle H^- = \bigcap \mathbb K$

where:

$H^- = H \cup H'$
$H'$ denotes the set of all limit points of $H$
$\displaystyle \bigcap \mathbb K$ is the intersection of all closed sets of $T$ which contain $H$.

Let $K \in \mathbb K$.

$H^- \subseteq K^-$

Thus from Closed Set Equals its Closure, we have:

$K^- = K$

Thus:

$\displaystyle \forall K \in \bigcap \mathbb K: H^- \subseteq K$

since the choice of $K$ is arbitrary.

So from Intersection is Largest Subset:

$\displaystyle H^- \subseteq \bigcap \mathbb K$

$\Box$

Conversely, from Topological Closure is Closed, $H^-$ is closed.

From Set is Subset of its Topological Closure, $H \subseteq H^-$.

So $H^-$ is, by definition, a closed set of $T$ which contains $H$.

But we have by its definition that $\displaystyle \bigcap \mathbb K$ is the intersection of all closed sets in $T$ that contain $H$.

So from Intersection is Subset it follows that:

$\displaystyle \bigcap \mathbb K \subseteq H^-$

$\Box$

Finally, we have that:

$\displaystyle H^- \subseteq \bigcap \mathbb K$
$\displaystyle \bigcap \mathbb K \subseteq H^-$

So by definition of set equality:

$\displaystyle H^- = \bigcap \mathbb K$

which is what we needed to prove.

$\blacksquare$