Let $A$ be a subset of a metric space. $A'$ be the set of limit points of $A$ and $A^i$ be the set of isolated points of $A$.
Show $A \subset A' \cup A^i$.
The picture on my mind is that I draw a disk plus a point outside of the disk, call it a set $A$, if the point falls on the disk, it is automatically a limit point of $A$. If the point is the point that lies outside of the disk, it is of course an isolated point. But I got trouble formalizing it in a rigorous way. Any hint?
EDIT: Definition added:
Limit Point: $A \subset X$, $X$ a metric space. $b \in X$ is a limit point if every neighborhood (topology sense) of $b$ contains a point of $A$ different from $b$.
Isolated point: $b$ is an isolated point of $A$ if there exists a neighborhood around $b$ that contains no point from $A$.