Prove that a finite set of points $z_1,z_2,…,z_n$ cannot have any accumulation points.

Question

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.

Answer 1

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?

Answer 2

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.