Characterization of Derivative by Local Basis

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

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

Let $x$ be a point of $T$.

Let $\mathcal B \subseteq \tau$ be a basis at $x$.

Then
 * $x \in \operatorname{Der} A$ iff for every subset $U \in \mathcal B$, there exists a point $y$ of $T$ such that $y \in A \cap U$ and $x \neq y$

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

Proof
First implication follows from Characterization of Derivative by Open Sets because an element of basis is open.

To prove the second implication let us assume

(1): for every subset $U \in \mathcal B$, there exists a point $y$ of $T$ such that $y \in A \cap U$ and $x \neq y$

To prove that $x \in \operatorname{Der} A$ according to Characterization of Derivative by Open Sets it is enough to prove that for every open subset $U$ of $T$ if $x \in U$, then there exists a point $y$ of $T$ such that $y \in A \cap U$ and $x \neq y$.

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

Assume that $x \in U$. Then there exists $V \in \mathcal B$ such that $x \in V \subseteq U$ by Definition:Basis (Topology)/Analytic Basis.

By assuption (1) there exists a point $y$ of $T$ such that $y \in A \cap V$ and $x \neq y$.

$y \in A \cap V \subseteq A \cap U$.