1
$\begingroup$

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?

$\endgroup$
8
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

5
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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
  • 2
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.