Confusion on limit points and accumulation points of a sequence I'm reading through the text book "Theory of Statistics" by James E. Gentle. I'm starting with chapter 0 which includes a bunch of prerequisite math. I've gotten to the topology section of the chapter and have some confusion about the definitions the author gives for limits and accumulation points. I'm going to copy the definitions word for word:
Limit point: A sequence $\{x_n\}$ is said to converge to the point $x$, or to have a limit $x$, if given any open set T containing $x$, there is an integer N such that $x_n\in T\forall n\geq N$
accumulation point or cluster point: a point $x$ is said to be an accumulation point or cluster point of the sequence $\{x_n\}$ if given any open set T containing $x$ and any integer N, there is an integer $n\geq N \ni x_n\in T$
I was wondering if anyone could give some examples of sequences and their respective limit and accumulation points. Thank you.
A pdf of the book can be obtained from https://mason.gmu.edu/~jgentle/books/MathStat.pdf as of Tuesday October 15, 2019 this link works. The definitions in question are on page 617.
Edit: I would have posted this as a comment under @Henno Brandsma's answer but it was too long to be a comment. I want to make sure I'm understanding this correctly. let $(\mathbb{R} , \mathcal{T})$ be the the topological space where $\mathcal{T}$ is the standard euclidean topology. If I have the sequence $x_n = (-1)^n$ then I can say that -1 and 1 are accumulation points of $x_n$ 
First I'll start with 1: any open set T of the form $(1 - \epsilon, 1 +\epsilon)$ $\forall \epsilon > 0$ contains 1. Now, for any arbitrary integer $N$, I can choose an integer $n \geq N$ such that $x_n\in T$ Either $x_N = 1$ and the condition for accumulation point holds for $n=N$ or $x_N=-1$ and I can let $n=N+1$ and the condition holds.
Showing that -1 is an accumulation point can be done in the same way.
Even though 1 and -1 are accumulation points of $x_n=(-1)^n$ in $(\mathbb{R},\mathcal{T})$, I can't say they're limits since there's no integer N such that $x_n\in T\forall n\geq N$ Is this correct?
 A: Taking examples in the (hopefully) familiar $\Bbb R$ and its usual topology: 
The sequence $1,-1,1,-1,1,-1,1,-1,\ldots$, so $x_n = (-1)^n$ for $n=1,2,3$ etc. has no limit but two accumulation/cluster points $1$ and $-1$: the sequence gets close to them infinitely many often, but not eventually, which is what you need for a limit: like $x_n = \frac{1}{n}$ converging to $0$ (or having $0$ as its limit point); in a metric space (or more generally a Hausdorff space) a sequence can only have at most one limit point, but it can have lots of cluster points: if $f: \Bbb N \to \Bbb Q$ is a bijection, $x_n = f(n)$ is a sequence in $\Bbb R$ that has every $x \in \Bbb R$ as a cluster point, because in every open neighbourhood $O$ of $x$ will be infinitely many rationals, so infinitely many times we will have $x_n \in O$, and that is what a cluster point means. A sequence like $x_n = n$ has no cluster point at all, so also no limit point ( a limit point is certainly always a cluster point, but not vice versa as we saw). But you will learn (hopefully) that any bounded sequence in $\Bbb R$ will have at least one cluster point. Normally we find convergent sequences (so the ones with a limit point) more important and they play in role in approximations and continuity etc. But cluster points can be important too.
