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Since the set, S, turns out to be a finite set consisting of just two elements, i.e. $\{-1,1\}$, therefore there should be no limit points to the set.

But the solution given in the book is, the set S has two limit points $-1$ and $1$.

Which solution is correct?

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    $\begingroup$ What is your definition of a limit point? $\endgroup$
    – Dave
    Dec 21, 2017 at 19:33
  • $\begingroup$ A limit point of a set, S is a point, x such that every neighbourhood of x (open sets containing x) contains a member of S other than x. $\endgroup$
    – Ajay Singh
    Dec 21, 2017 at 19:36
  • $\begingroup$ It depends on what topological space you are working in (and in particular what topology you are using), and how you define a limit point. Under the usual topology of $\Bbb R$ and your definition of limit points that you gave in your comment just now, any finite set has no limit points. $\endgroup$
    – JMoravitz
    Dec 21, 2017 at 19:40
  • $\begingroup$ In your definition of a limit point, the set $S$ has no limit points, because the ball of radius $\frac{1}{2}$ around $1$ or $-1$ does not contain the other point. One can define (as is done at my school) a limit point as one in which any neighbourhood of the point contains a point in $S$, and under this definition both $1$ and $-1$ are limit points of $S$. $\endgroup$
    – Dave
    Dec 21, 2017 at 19:40
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    $\begingroup$ Your book is probably talking about the sequence $\{(-1)^n\}$, not the set $S$. $\endgroup$
    – user9464
    Dec 21, 2017 at 19:41

1 Answer 1

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The notation $\{(-1)^n\mid {n\in{\bf N}}\}$, understood as a set, is the same as $\{-1,1\}$, which has no limit point.

However, in some context, $\{(-1)^n\mid {n\in{\bf N}}\}$ is used as a (bad) notation for the real sequence $(a_n)_{n=1}^\infty$ with $a_n:=(-1)^n$. In this case,

$x$ is a "limit point" of the sequence $(a_n)_{n=1}^\infty$

means

$x$ is the limit of some convergent subsequence of $(a_n)_{n=1}^\infty$.

Note that both $1$ and $-1$ are limits of some subsequences of the sequence $((-1)^n)_{n=1}^\infty$.

For more general discussions, see the Wikipedia article on Limit point.

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  • $\begingroup$ But still the neighbourhood of say 1 does not contain any of the point of the set except for 1. $\endgroup$
    – Ajay Singh
    Dec 21, 2017 at 20:10
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    $\begingroup$ @AjayChoudhary The idea here is that the definition of limit point for a set and limit point for a sequence are different - in particular, $\{-1,1\}$ has no limit points, but the sequence $((-1)^n)_{n=1}^{\infty}$ does have a limit point, because "limit point" means two different things in either situation. $\endgroup$ Dec 21, 2017 at 20:14

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