# Characterization of Derivative by Open Sets

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## Theorem

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

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

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

Then

$x \in A'$
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 $x \ne y$

where

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

## Proof

### Sufficient Condition

Let $x \in A'$.

Then $x$ is an accumulation point of $A$ by Definition:Set Derivative.

Then by definition of accumulation point:

$(1): \quad x \in \left({A \setminus \left\{ {x}\right\} }\right)^-$

where $A^-$ denotes the closure of $A$.

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

Let $x \in U$.

Then by $(1)$ and Condition for Point being in Closure:

$\left({A \setminus \left\{ {x}\right\}}\right) \cap U \ne \varnothing$

Then there exists $y$ being a point such that:

$(2): \quad y \in A \setminus \left\{ {x}\right\}$ and $y \in U$.

Then $y \in A$ and $y \notin \left\{ {x}\right\}$ by definition of set difference.

Hence by $(2)$ and the definitions of set intersection and singleton:

$y \in A \cap U$ and $x \ne y$

$\Box$

### Necessary Condition

Let $x$ be such that:

$(3): \quad$ 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 $x \ne y$.

It will be proved that:

for every open set $G$ of $T$:
if $x \in G$
then $\left({A \setminus \left\{ {x}\right\} }\right) \cap U \ne \varnothing$.

Let $G$ be an open set of $T$ such that $x \in G$.

Then by $(3)$ there exists a point $y$ of $T$ such that:

$(4): \quad y \in A \cap G$ and $x \ne y$

Then by the definitions of set intersection and singleton:

$y \in A$ and $y \notin \left\{ {x}\right\}$

By definition of set difference:

$y \in A \setminus \left\{ {x}\right\}$

By $(4)$ and the definition of set intersection:

$y \in G$

Hence:

$\left({A \setminus \left\{ {x}\right\} }\right) \cap U \ne \varnothing$

$\Box$

$x \in \left({A \setminus \left\{ {x}\right\} }\right)^-$

Then, by definition, $x$ is an accumulation point of $A$.

Hence by definition of set derivative:

$x \in A'$

$\blacksquare$