A finite set has no limit points I am looking for an easy to remember proof to the statement "A finite set has no limit points". The best proof I found say that

Any finite set is composed of isolated points only. Since for any
  isolated point there exists a neighborhood that does not contain any
  other element of the set, a finite set cannot have any limit points.

Is this proof correct and is there a way to represent this mathematically?
 A: I'm not convinced.
Definition: s is a limit point of S if every open set that contains s contains a second distinct element of S.
Consider a finite  non-empty set X of two or more elements with the indiscrete topology. Every point is a limit point. (there are only two open sets in the indiscrete topology, {}, and X itself).   
A: The point $a$ is a limit point of the set $A$ if every open neighborhood of $a$ (i.e. every open set of which $a$ is a member) contains some member of $A$ other than $A$.
No matter how small you make an open neighborhood of $a$ it still contains some member of $A$ other than $a$.  That means $a$ can be approached from within $A$.
In order to prove the desired result, you must be working in a space in which, for every point $b$ other than $a$, there is some open neighborhood of $A$ that excludes $b$.  That happens in metric spaces and in many other topological spaces.
Let's say your finite set is $\{a=a_1,a_2,a_3,\ldots,a_n\}$.
Then some open neighborhood of $a$ excludes $a_2$;
and some open neighborhood of $a$ excludes $a_3$;
and some open neighborhood of $a$ excludes $a_4$;
and so on $\cdots$
and some open neighborhood of $a$ excludes $a_n$.
The intersection of those open neighborhoods of $a$ is an open neighborhood of $a$ because the intersection of finitely many open sets is open.  There you have an open neighborhood of $a$ that excludes all other members of $A$, so $a$ is not a limit point of $A$.
