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$p$ is a cluster point of $S\subset M$ if each neighborhood of $p$ contains infinitely many points. Here is my confusion, a cluster point is also a limit point of $S$, right?

If so, then how does the sequence $((-1)^n)$, ${n\in \mathbb N}$ has two cluster points namely $1, -1$ especially since the sequence does not have a limit as n approaches infinity.

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    @Dejan’s comment isn’t really$a$one-liner: he’s directing you to the definitions of the two comments and suggesting that you see for yourself that they really aren’t the same thing.2012-11-01

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Limit point of a set is not the same thing as limit point of a sequence. Look at the actual definitions:

  • $x$ is a limit point of the set $S$ if every open nbhd of $x$ contains a point of $S\setminus\{x\}$;

  • $x$ is a limit point of the sequence $\langle x_k:k\in\Bbb N\rangle$ if for every open set nbhd $U$ of $x$, $\{k\in\Bbb N:x_k\in U\}$ is infinite. (I prefer to use the term cluster point: it’s less likely to result in confusion with the very different notion of the limit of the sequence.)

Now consider the sequence $\langle (-1)^k:k\in\Bbb N\rangle$. Let $U$ be an open nbhd of $1$;

$\begin{align*} \{k\in\Bbb N:(-1)^k\in U\}&=\{k\in\Bbb N:(-1)^k=1\}\\ &=\{k\in\Bbb N:k\text{ is even}\}\;. \end{align*}$

The set of even natural numbers is certainly an infinite set, so $1$ is a limit point of the sequence.

If instead we let $U$ be an open nbhd of $-1$, we have

$\begin{align*} \{k\in\Bbb N:(-1)^k\in U\}&=\{k\in\Bbb N:(-1)^k=-1\}\\ &=\{k\in\Bbb N:k\text{ is odd}\}\;. \end{align*}$

The set of odd natural numbers is also infinite, so $-1$ is also a limit point of the sequence.

However, neither $1$ nor $-1$ is a limit point of the set $S=\{-1,1\}$: $(0,2)$ is an open nbhd of $1$ that contains no point of $S\setminus\{1\}$, and $(-2,0)$ is an open nbhd of $-1$ that contains no point of $S\setminus\{-1\}$.

Nor is either $-1$ or $1$ the limit of the sequence: that has yet a different definition.

  • $x$ is the limit of the sequence $\langle x_k:k\in\Bbb N\rangle$ if for every open nbhd $U$ of $x$ there is an $n_U\in\Bbb N$ such that $x_k\in U$ whenever $k\ge n_U$.

Neither $-1$ nor $1$ satisfies this definition for the sequence $\langle(-1)^k:k\in\Bbb N\rangle$. Take $U=(0,2)$, for instance; this is an open nbhd of $1$, and no matter how big you set the cutoff $n$, there will be a $k\ge n$ such that $k$ is odd and therefore $(-1)^k=-1\notin U$. A very similar argument shows that $-1$ is not the limit of the sequence.

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    @user43901: There is, but in all honesty I’m not very keen on chat as a medium for discussing mathematics; my experiences with it (when used for that purpose) have been only so-so. It also doesn’t mesh well with my work habits.2012-11-01
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A point $p\in M$ is a cluster point of the sequence $(a_n)_n$ if every neighborhood of $p$ contains $a_n$ for infinitely many $n$. In your case, $1$ is a cluster point, because for each neighborhood $V$ of $1$ (for example $V = (1-\epsilon,1+\epsilon)$), we have $(-1)^n\in V$ for all even $n$ (i.e. infinitely many $n$). The point $-1$ is a cluster point of this sequence for the same reason (this time because there are infinitely many odd numbers).

A point $p\in M$ is a limit of the sequence $(a_n)_n$ if every neighborhood of $p$ contains $a_n$ for all big enough $n$, more precisely: for every neighborhood $V$ there exists a $N$ such that $a_n\in V$ for all $n\geq N$. In your case, a limit doesn't exist. The only candidates for a limit would be $-1$ and $1$, since every other number has a neighborhood that doesn't contain $a_n$ for any $n$. But $(-2,0)$ is a neighborhood of $-1$ that contains no even numbers and thus cannot contain every positive integer from some place on. For similar reasons $1$ is not a limit.

On the other hand, limit points are defined for sets (in contrast with sequences). A limit point of a set $S$ is a point $p\in M$ such that every neighborhood of $p$ a point $s\in S$ such that $s\neq p$. Since $-1$ is the only element of the open set $(-2,0)$ that lies in $S =\{-1,1\}$, it is not a limit point of $S$.

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    Thank you so much. This explanation helps a lot. I was confusing the limit of a sequence to a limit point of a set. Hence, all the crap. Here is something that really confused me: The set $S$ clusters at $p$ iff each nbd of $p$ contains at least one point of $S$ other than $p$.2012-11-01