Derivative of Subset is Subset of Derivative

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

Let $A$, $B$ be subsets of $S$.

Then
 * $A \subseteq B \implies A' \subseteq B'$

where $A'$ denotes the derivative of $A$ in $T$.

Proof
Let $A \subseteq B$.

Let $x \in A'$.

By Characterization of Derivative by Open Sets it is enough to prove that:
 * for every open set $G$ of $T$:
 * if $x \in G$
 * then there exists $y$ such that $y \in B \cap G$ and $x \ne y$.

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

Let $x \in G$.

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

By the corollary to Set Intersection Preserves Subsets:
 * $A \cap G \subseteq B \cap G$

Hence:
 * $y \in B \cap G$ and $x \ne y$.

The conditions of the hypothesis are thus fulfilled, and:
 * $x \in B'$

Thus by definition of subset:


 * $A' \subseteq B'$