Topological Closure of Subset is Subset of Topological Closure

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Let $T = \left({S, \tau}\right)$ be a topological space.

Let $H \subseteq K$ and $K \subseteq S$.


$\operatorname{cl}\left({H}\right) \subseteq \operatorname{cl}\left({K}\right)$

where $\operatorname{cl}\left({H}\right)$ denotes the closure of $H$.

Proof 1

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

From Set is Subset of its Topological Closure:

$K \subseteq \operatorname{cl} \left({K}\right)$

By Subset Relation is Transitive, it follows that:

$H \subseteq \operatorname{cl} \left({K}\right)$

Hence, by definition of closure as the smallest closed set that contains $H$:

$\operatorname{cl} \left({H}\right) \subseteq \operatorname{cl} \left({K}\right)$


Proof 2

From the definition of closure:

$\operatorname{cl}\left({H}\right)$ is the union of $H$ and its limit points.

Let $x \in \operatorname{cl} \left({H}\right)$.

If $x \in H$ then $x \in K \implies x \in \operatorname{cl} \left({K}\right)$.

Otherwise $x$ is a limit point of $H$.

That is, every open set $U$ of $T$ such that $x \in U$ contains $y \in H$ such that $y \ne x$.

But as $y \in H$ it follows that $y \in K$.

So every open set $U$ of $T$ such that $x \in U$ contains $y \in K$ such that $y \ne x$.

This is the definition for a limit point of $K$.

Thus $x \in \operatorname{cl} \left({K}\right)$.

A part of this page has to be extracted as a theorem:
Extract the above paragraph into something like "Limit Point of Subset is Limit Point" (which proves that $H \subseteq K \implies H' \subseteq K'$). Then use Set Union Preserves Subsets to prove this result.