Derivative of Subset is Subset of Derivative

Theorem

Let $T = \struct {S, \tau}$ 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:

$A' \subseteq B'$

$\blacksquare$

Sources

• 1977: Ryszard Engelking: General Topology

• Mizar article TOPGEN_1:30