Characterization of Derivative by Local Basis

Theorem

Let $T = \struct {S, \tau}$ be a topological space.

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

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

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

Then

$x \in A'$

if and only if:

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

where:

$A'$ denotes the derivative of $A$.

Proof

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 $\BB$ are all open sets, it follows that:

For every open set $U \in \BB$:

if $x \in U$

then there exists a point $y$ of $T$ such that $y \in A \cap U$ and $x \ne y$

$\Box$

Necessary Condition

Let $x$ be such that: $(1): \quad$ for every subset $U \in \BB$, 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 $V \in \BB$ 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.

$\blacksquare$

Sources

• 1977: Ryszard Engelking: General Topology

• Mizar article TOPGEN_1:18