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I know the point $x\in X$ is the point of accumulation on the set $A$ ($A$ is a subset of the metrics space $(X,d)$), if $T(x,r)\cup(A\backslash\{x\})\neq \phi,$ $\forall r>0,$ but I didn`t now how to show that finite set no accumulation points.

Thank you for your solution. Thanks for your attention

Madrit Zhaku
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2 Answers2

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Hint:

Think about the result that $\forall r>0$ gives you. Indeed, if you make any ball $T(x,r)$ with radii $r$ while $r$ is an arbitrary positive real value, so that intersection would not be empty. But how many $r$ are there for you to take? Infinitely selection. So there are infinitely many points in shared $T(x,r)\cap A$. So it makes us to have infinite set $A$.

Mikasa
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(Assuming you're in a metric space). Let $S=\{x_1,\ldots,x_n\}\subset X$ be your finite set. Now the set $\{d(x_i,x_j):x_i,x_j\in S\}$ is of course finite, so you can choose its minimum, say $m$. What happens when you look at the ball $B(x_i,m/2)$ about each $x_i\in S$?

Nick D.
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  • @ Nick D. can you tell me please what happens when look at the ball $B(x_i, m/2)$ about each $x_i\in S$ – Madrit Zhaku Jan 03 '14 at 18:22
  • Figure it out. What is the definition of accumulation point? What is the intersection of $S$ with $B(x_i,m/2)$? – Nick D. Jan 03 '14 at 18:25
  • Please if you can help me about are because I do not understand good metric spaces, please – Madrit Zhaku Jan 03 '14 at 18:27
  • @MadritZhaku: You should imagine about $B(x_i,m/2)$. Is there any point in the intersection with $B(x_i,m/2)$? – Mikasa Jan 03 '14 at 18:28
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    So...how many points in $S$ are within a distance of $m/2$ from $x_i$? Remember that $m$ is the minimum distance between any two points in $S$. – Nick D. Jan 03 '14 at 18:29
  • @Nick D. Sir, isn't this 'only' shows that each $x_i\in S$ is not accumulation point of $S$. But, we know accumulation points of set may or may not belongs to the set. So what about the points which are not in $S$? are they are not accumulation points of $S$? How. Please elaborate. – Akash Patalwanshi Jul 31 '19 at 12:55
  • @AkashPatalwanshi Thanks. If $x\in X-S$, you can repeat the above with $S\cup{x}$. – Nick D. Jul 31 '19 at 21:58