Set Closure is Smallest Closed Set

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Theorem

Let $T$ be a topological space.

Let $H \subseteq T$.

Let $H^-$ denote the closure of $H$ in $T$.


Then $H^-$ is the smallest superset of $H$ that is closed in $T$.


Proof 1

Define:

$\mathbb K := \left\{{K \supseteq H: K}\right.$ is closed in $\left.{T}\right\}$

That is, let $\mathbb K$ be the set of all supersets of $H$ that are closed in $T$.

The claim is that $H^-$ is the smallest set of $\mathbb K$.


From Set is Subset of its Topological Closure:

$H \subseteq H^-$

From Topological Closure is Closed, $H^-$ is closed in $T$.

Thus $H^- \in \mathbb K$.


Let $K \in \mathbb K$.

From Set Closure as Intersection of Closed Sets:

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

Therefore, from Intersection is Subset: General Result:

$H^- \subseteq K$

Thus by definition $H^-$ is the smallest set of $\mathbb K$.

$\blacksquare$


Proof 2

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 from Set Closure as Intersection of Closed Sets, we have:

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

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


Let $K \subseteq T$ such that $K$ is closed in $T$ and $H \subseteq K$.

That is, let $K \in \mathbb K$.

Then from Intersection is Subset it follows directly that $H^- \subseteq K$.

So $H^-$ is a subset of any closed set in $T$ which contains $H$, and so is the smallest closed set that contains $H$.

$\Box$


Let $V$ be the smallest closed set that contains $H$.


Let $K$ be closed in $T$ such that $H \subseteq K$.

Then by definition $V \subseteq K$.

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

That is:

$\displaystyle V = \bigcap \mathbb K$

where $\mathbb K := \left\{{K \supseteq H: K}\right.$ is closed in $\left.{T}\right\}$.

From Set Closure as Intersection of Closed Sets it follows that $V = H^-$.

$\Box$


The implication has been demonstrated to hold in both directions, so the closure of $H$ in $T$ can be defined as the smallest closed set that contains $H$.

$\blacksquare$


Sources