Topological Closure of Subset is Subset of Topological Closure

Theorem
Let $$T$$ be a topological space.

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

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

Proof
Follows directly from the definition of closure.

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


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


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