Characterization of Derivative by Local Basis

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

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

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

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

Then
 * $x \in A'$


 * for every $U \in \mathcal B$, there exists a point $y$ of $T$ such that $y \in A \cap U$ and $x \ne y$
 * for every $U \in \mathcal B$, there exists a point $y$ of $T$ such that $y \in A \cap U$ and $x \ne y$

where:
 * $A'$ denotes the derivative of $A$.

Sufficient Condition
Let $x \in A'$.

By Characterization of Derivative by Open Sets:

For every open set $U$ of $T$:
 * if $x \in U$
 * then there exists a point $y$ of $T$ such that $y \in A \cap U$ and $x \ne y$

As the elements of $\mathcal B$ are all open sets, it follows that:

For every open set $U \in \mathcal B$:
 * if $x \in U$
 * then there exists a point $y$ of $T$ such that $y \in A \cap U$ and $x \ne y$

Necessary Condition
Let $x$ be such that: $(1): \quad$ for every subset $U \in \mathcal B$, there exists a point $y$ of $T$ such that $y \in A \cap U$ and $x \ne y$.

By Characterization of Derivative by Open Sets, to prove that $x \in A'$ it is enough to prove:
 * for every open set $U$ of $T$:
 * if $x \in U$
 * then there exists a point $y$ of $T$ such that $y \in A \cap U$ and $x \ne y$.

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

Let that $x \in U$.

Then by definition of local basis:
 * there exists $V \in \mathcal B$ such that:
 * $x \in V \subseteq U$

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

By the corollary to Set Intersection Preserves Subsets:


 * $A \cap V \subseteq A \cap U$

and so:
 * $y \in A \cap V \implies y \in A \cap U$

and so $y$ fulfils the conditions of the hypothesis.