Derivative is Included in Closure

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

Let $A$ be a subset of $T$.

Then
 * $\operatorname{Der} A \subseteq \operatorname{Cl} A$

where
 * $\operatorname{Der} A$ denotes the derivative of $A$,
 * $\operatorname{Cl} A$ denotes the closure of $A$.

Proof
Let $x \in \operatorname{Der} A$.

According to Condition for Point being in Closure it is enough to prove that for every open subset $G$ of $T$ if $x \in G$, then $A \cap G \neq \varnothing$.

Let $G$ be an open subset of $T$.

Assume $x \in G$.

Then there exists a point $y$ of $T$ such that $y \in A \cap G$ and $x \ne y$ by Characterization of Derivative by Open Sets. Hence $A \cap G \neq \varnothing$.