# Can anyone clarify how a diverging sequence can have cluster points?

$p$ 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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The phrase "limit of a sequence" cares about the order of the terms in the sequence, but the phrase "cluster point" doesn't. Your sequence doesn't have a limit because it oscillates for arbitrarily large $n$. But it does have cluster points, as you say. It may help you to prove the following: if $a$ is a cluster point of $\{a_n\}$, then there is some subsequence $\{a_{n_k}\}$ with $\{a_{n_k}\} \to a$. – user29743 Nov 1 '12 at 0:50
I mean if $a$ is a cluster point of $\{a_n\}$, then $a$ is a limit point of the sequence; thus the subsequence must converge to a. @countinghaus: If you can, just expand on the topic. As yo u can see, I am pretty lost. – user43901 Nov 1 '12 at 0:57
@user43901: "limit point" and "limit" are two different things. – Dejan Govc Nov 1 '12 at 1:04
@DejanGovc: elaborate on that please. Isn't a limit point $p \text{ of } S$ the limit of a sequence say $(p_n)$ that converges in $S$? I just don't see the connection; hence, if anyone can take their valuable time and explain this scenario more fulling instead of one-liners, it will be much beneficial for the understanding. – user43901 Nov 1 '12 at 1:08
@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. – Brian M. Scott Nov 1 '12 at 1:28

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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@BrianMScott: Thanks for clarifying this, Brian. Much appreciated. I am still digesting this. I got myself into a bad state of confusion with the definitions. The thing 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 – user43901 Nov 1 '12 at 2:24
@user43901: Can you explain what it is about that definition that’s causing you trouble? – Brian M. Scott Nov 1 '12 at 2:35
@BrianMScott: I really don't know exactly what was about it, but it took me time to grasp the difference between a limit point of a set and a cluster point of a set. And I was wondering-- is the cluster point, a limit point? I know it's not the other way round. – user43901 Nov 1 '12 at 3:26
@BrianMScott: how do you start a chat with a user? Say, if I wanted to discuss a certain topic with you, is there a way we can just type instead of putting in posts? – user43901 Nov 1 '12 at 3:28
@user43901: Limit point and cluster point are generally used interchangeably, whether you’re talking about sets or about sequences; I prefer cluster point, because I think that it’s both more descriptive and less likely to be confused with the notion of the limit of a sequence. You may be thinking of the fact that there is a difference between ‘$x$ is in the closure of $A$’ and ‘$x$ is a limit point of $A$’: a point can be in the closure of $A$ either because it’s a limit (or cluster) point of $A$, or because it just happens to be in $A$. – Brian M. Scott Nov 1 '12 at 3:31

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 $p$. – user43901 Nov 1 '12 at 2:00