Topological Closure of Subset is Subset of Topological Closure/Proof 1

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

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

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

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

Proof
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 $H$:


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