Can $A\subset \mathbb{R}$ with non-empty set of limit points, be a finite set? I was wondering if it is possible to say that given $A\subset \mathbb{R}$ such that the set of limit points is not empty, $A$ can be a finite set?
I was thinking about it as consequence of Weierstrass theorem (if A is infinite and bounded,then it has at least one limit point), but is it also true the inverse? Am I on the right track?
Thanks
 A: No, you're on the wrong track.
Consider the definition of limit point: $x$ is a limit point of the set $A$ if (and only if) every neighborhood $U$ of $x$ contains a point of $A$ different from $x$.
Instead of generic neighborhoods you can use intervals of the form $(x-\delta,x+\delta)$. Now what can we do? Start from $\delta_0=1$ and select $a_0\in (x-\delta_0,x+\delta_0)\cap A$, $a_0\ne x$.
Next we choose $\delta_1=|x-a_0|/2$ and we select
$$
a_1\in(x-\delta_1,x+\delta_1)\cap A,\quad a_1\ne x
$$
For obvious reasons, $a_1\ne a_0$. This starts a recursion! Suppose you have found distinct points $a_0,\dots,a_{n-1}$ in $A$ all different from $x$; then fix
$$
\delta_n=\frac{1}{2}\min\{|x-a_0|,|x-a_1|,\dots,|x-a_{n-1}|\}
$$
and you can choose
$$
a_n\in (x-\delta_n,x+\delta_n)\cap A,\quad a_n\ne x
$$
so that $a_n$ is different from the previous ones. Since you can go on forever, you conclude that $A$ is infinite.
Note that the same proof can be adapted to any metric space, where Weierstrass theorem doesn't necessarily hold.
