I'm having trouble understanding why
$\bar A = A \cup A'$, where $A\subseteq\mathbb R^n$, $\space\bar A$ is the closure of $A$ , and $A'$ is the derived set of $A$.
EDIT: Here are some definitions I am working with:
(i) Definition of adherent point– Let $S \subseteq \mathbb R^n $, and $x$ a point in $\mathbb R^n$, not necessarily in $S$. Then $x$ is a adherent to $S$ if every ball n-ball $B(x;r)$ contains at least one point of $S$.
(ii) A set $S \subseteq \mathbb R^n $ is closed, if and only if, it contains all its adherent points.
(iii) The closure of a set $S$ is the set $\bar S = \{ x \in \mathbb R^n \space | \space \forall r>0, B(x;r)\cap S\ne \emptyset \}$
Best Answer
First show $A\cup A' \subseteq \overline A$.
Now assume $x \in \overline A$. If x not in A' then
there is some open U nhood x with $U \cap A = \{x\}$.
Thus x in A. Equality follows.
This is true of all topological spaces including R.
Looking for a solution to a homework problem is not research.