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.
 A: 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.
A: 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 $S$.
