# The closure of a set A is the union of A and the derived set of A?

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 $\bar S = \{ x \in \mathbb R^n \space | \space \forall r>0, B(x;r)\cap S\ne \emptyset \}$

• What is your definition of closure? – Math1000 Nov 7 '17 at 7:56
• The closure of a subset of a topological space is the subset itself plus all of its limit points. That's exactly $A\cup A^\prime$, so this is a mere definition. – B. Pasternak Nov 7 '17 at 7:56
• It's not a theorem, it's the definition of a closure. – Oria Gruber Nov 7 '17 at 7:57
• @B.Pasternak Another common definition is that it is the intersection of all closed sets containing $A$ – eepperly16 Nov 7 '17 at 7:58
• @eepperly16 True, but let's use the Axiom of Choice to choose this as our definition.. – B. Pasternak Nov 7 '17 at 8:01

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 $U \cap A = \{x\}$.