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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?2012-08-09
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    @ThomasAndrews yes2012-08-09
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    It is true in any metric space.2012-08-09

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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.2012-08-09
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    @ThomasAndrews, Thank you, but I was just waiting for Siddhant, If he can manage from my argument.2012-08-09
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    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.2012-08-09
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    @NieldeBeaudrap Thank you.2012-08-09
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    @Patience what is $z \in S\{x}$? does it mean that deleted nbhd of x?2012-08-09
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    @SiddhantTrivedi yes, exactly2012-08-09
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    ok so, now tell me the $d$ we have chosen, if you take any deleted nbd of $x$ with radius our chosen $d$, will this nbd contains any point from $S$?2012-08-09
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    @Patience i understand whatever u ask by example is it right or not check plzzz....if i take S={i/n}={i,i/5,i/10,...} here i take d as u defined |x-z| in this case zero is limit point and |0-i/5| or any point of S={i/n} must containing is S.2012-08-09
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    Can u explain more?2012-08-09
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    @SiddhantTrivedi: in the example you give above, you seem to describe an infinite set. Can you achieve $d=0$ when $S$ is finite?2012-08-09
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    @NieldeBeaudrap how can i show d=0 whenever S is finite?2012-08-10
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    @SiddhantTrivedi: Hint -- you can't; it's impossible. Can you prove that, though? And what does this imply?2012-08-10
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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.