Set Closure as Intersection of Closed Sets

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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:

$\mathbb K := \leftset {K \supseteq H: K}$ is closed in $\rightset T$

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:

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

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


What needs to be proved here is:

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


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

Let $K \in \mathbb K$.

From Topological Closure of Subset is Subset of Topological Closure, we have:

$H^- \subseteq K^-$

Thus from Closed Set Equals its Closure, we have:

$K^- = K$


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

since the choice of $K$ is arbitrary.

So from Intersection is Largest Subset:

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


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 $\ds \bigcap \mathbb K$ is the intersection of all closed sets in $T$ that contain $H$.

So from Intersection is Subset it follows that:

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


Finally, we have that:

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

So by definition of set equality:

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

which is what we needed to prove.