Accumulation points: an example of a sequence that has exactly two accumulation points The definition in my text defines A.P. as: A point p is an A.P. of a set S if it is the limit of a sequence of points of S-{p}.
I am confused by my classnotes and information about A.P. in my textbook is not enough for me to understand the topic. 
 A: Exactly two accumulation points can be done using the sequence
$$\frac{1}{3}, \frac{2}{3},\frac{1}{4},\frac{3}{4}, \frac{1}{5},\frac{4}{5},\frac{1}{6}, \frac{5}{6},\frac{1}{7}, \frac{6}{7},\dots,$$
and in many other ways. In this case the accumulation points are $0$ and $1$.
Here I intend the set $S$ of the definition to be $\{\frac{1}{3}, \frac{2}{3},\frac{1}{4},\frac{3}{4}, \frac{1}{5},\frac{4}{5},\frac{1}{6}, \frac{5}{6},\frac{1}{7}, \frac{6}{7},\dots\}$. We could work with a somewhat larger set for $S$, by, in addition to the points mentioned, throwing in all the integers, or something else that would not produce additional accumulation points.
A: The sequence $(-1)^n\left(1+\frac{1}{n}\right)$ has exactly two A.Ps $1$ and $-1$.
More generally if $(a_n)_{n\in \mathbb{N}}$ is a sequence s.t. $a_n \to a \neq0 $ and the set $\{a_n:n \in \mathbb{N}\}$ is infinite then the sequence $((-1)^na_n)_{n\in \mathbb{N}}$ has exactly two accumulation points $a$ and $-a$.
A: This seems a little vague.
What is your question? Do you want an example of the sequence or do you want more info. about accumulation points?
Intuitively, accumulations points are the points of the set S which are not isolated.
e.g. 1 is an A.P. of (0,1) but 2 is not an A.P of $(0,1) \cup$ {2} (because it's isolated.) To prove this, the set $X = $ ($(0,1) \cup$ {2} ) - {2} = $(0,1)$ but 2 is not the limit of a sequence in $(0,1)$ so 2 is not an A.P. of $(0,1) \cup$ {2}
from there its pretty easy to construct such a sequence.
A: Point $p$ is accumulation point of a set $S$, if every punctured neighborhood  of the point $p$ contains points of $S.$ (In other words, if there exists sequence $\{x_n\} \subset S$ that converges to $p.$)
Set $$S={\left \lbrace \sin{\dfrac{1}{n}}, \cos{\dfrac{1}{n}}  \right \rbrace}_{n \in \mathbb{N}}=\left \lbrace \sin{1}, \cos{1},\sin{\dfrac{1}{2}}, \cos{\dfrac{1}{2}}, \ldots, \sin{\dfrac{1}{n}}, \cos{\dfrac{1}{n}} , \ldots \right \rbrace$$ has exactly two accumulation points.
