TITLE: A point in subset $A $of metric space is either limit point or isolated point.
QUESTION [5 upvotes]: 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 $A$.

REPLY [4 votes]: This is just a matter of using the definitions of isolated point and limit point, so it’s hard to give a useful hint without doing the problem, but I’ll try.
Let $x\in A$; if $x\in A^i$, we’re done with $x$, so suppose that $x\notin A^i$. This means that there is no open set $U$ in $X$ such that $U\cap A=\{x\}$. In other words, if $U$ is open, and $x\in U$, then $U$ also contains at least one other point of $A$. What does that tell you about $x$?