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.

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Presumably, you mean points in the complex plane? –  Thomas Andrews Aug 9 '12 at 16:15
@ThomasAndrews yes –  Siddhant Trivedi Aug 9 '12 at 16:16
It is true in any metric space. –  lhf Aug 9 '12 at 17:01

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?

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You also need to show that no $z_i$ is an accumulation point, but that is essentially the same argument. –  Thomas Andrews Aug 9 '12 at 16:19
@ThomasAndrews, Thank you, but I was just waiting for Siddhant, If he can manage from my argument. –  La Belle Noiseuse Aug 9 '12 at 16:21
I have suggested a revision which emphasizes features which are important for the case where $S$ is not necessarily finite, and which removes the proof by contradiction. –  Niel de Beaudrap Aug 9 '12 at 16:25
@NieldeBeaudrap Thank you. –  La Belle Noiseuse Aug 9 '12 at 16:26
@SiddhantTrivedi: in the example you give above, you seem to describe an infinite set. Can you achieve $d=0$ when $S$ is finite? –  Niel de Beaudrap Aug 9 '12 at 22:43

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.

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