Prove that a finite set of points $z_1,z_2,......,z_n$ cannot have any accumulation points. How can I prove a finite set of points $z_1,z_2,......,z_n$ on the complex plane cannot have any accumulation points.please give me some hints.
 A: Hint: Consider $S = \bigl\{z_1,\ldots,z_n\bigr\}$, and take any element $x \in S$. Consider $$d=\inf\;\Bigl\{|x-z| \;:\; z \in S \smallsetminus \{x\} \Bigr\} \;,$$ the infimum of the distances of all other points in $S$ from $x$. What conditions have to apply to $d$ for $x$ to be a limit point? Can they hold for $S$ finite?
A: For the sake of completion I will include the proof using @CityOfGod's assertion.
Let $ S = \{z_1, z_2, ... , z_n\} $, 
Let $z$ be some arbitrary point (may or may not be contained in $S$), and let $d = \inf \{ |z - z_i|  : z_i \in S\}$
This establishes that there is a minimum distance between any arbitrary point $z$ and a nearby point $z_i :  z_i \in S$
Now we can let $B_\epsilon(z) $ be a neighbourhood around the arbitrary point $z$. If the set $S$ were infinite then there would always be some radius $\epsilon$ in which a point in $S \setminus \{z\}$ is included, therefore proving there is an accumulation point. However because $S$ is finite, an $\epsilon$ can be provided such that $\epsilon < d$. 
In other words, if $\epsilon$ is less than the minimum distance between some arbitrary point $z$ and a point $z_i$ then $\forall x \in B_\epsilon(z), x \notin S$. Therefore there is no accumulation point. 
