Derived set

Remarks

The closure of $A$ is $\overline{A} = A \cup A'$. In a T1 space, the derived set $A'$ is always closed.

Example: in $\R$, let $A = \set{1/n : n \ge 1}$. Each $1/n$ has a neighbourhood that meets $A$ only in $1/n$, so $1/n \notin A'$. Every interval $(-\varepsilon, \varepsilon)$ contains $1/n$ for large $n$, so $0 \in A'$. Thus $A' = \set{0}$.

Example: the derived set of $\Q$ in $\R$ is $\R$, and the derived set of $\Z$ in $\R$ is $\emptyset$.

Non-example: in $\R$, the point $2$ is not in the derived set of $A = \set{2} \cup [3, 4]$. The neighbourhood $(1, 5/2)$ of $2$ contains no point of $A$ other than $2$, so the condition fails.