Convergent sequences and accumulation points Definitions:
Let $a$ be an accumulation point of $A$. Then $\forall \ \epsilon >0$, $B_{\epsilon}(a) \setminus \{a\}$ contains an element of $A $.
Question:
I have two questions: if $(a_n)_{n\in N}$ is a convergent sequence in $\mathbb{R}$ then,

*

*Does the set $\{a_n\}$ have exactly one accumulation point? Or, could it have more than one?


*If so, does $(a_n)_{n\in N}$ necessarily converge to the said accumulation point?
I'm tempted to say no to (1), but I'm afraid that I'm missing something. My counter-example to (1) is $\{a_n\} = \{ 4, 3, 2, 1, 0,0,0,...\}$ (i.e. inserting $0$s after the 4th element). Then the set has no accumulation point and it converges to 0. Is that correct?
 A: If $\{a_{n}\}$ has an accumulation point, say, $a$, and $(a_{n})$ is convergent. Then choose some $n_{1}$ such that $a_{n_{1}}\in B_{1}(a)-\{a\}$. Then choose some $n_{2}$ such that $B_{1/2}(a)-\{a,a_{1},...,a_{n_{1}}\}$, proceed in this way we have $a_{n_{k}}\rightarrow a$. Since $(a_{n})$ is convergent, one has $a_{n}\rightarrow a$. 
Here I use the following definition:
$a$ is an accumulation point for $A$ if for every $\delta>0$, $(B_{\delta}(a)-\{a\})\cap A\ne\emptyset$.
And note that in the topology of ${\bf{R}}$, being such an accumulation point also implies that $(B_{\delta}(a)-\{a\})\cap A$ contains infinitely many points. 
A: I am calling $x$ an accumulation point of the set $A$ iff $(B(x,\epsilon) \cap A)\setminus \{x\} \neq \emptyset$ for all $\epsilon>0$.
Suppose $a_n \to a$.
The set $\{a_n\}_n$ can have at most one accumulation point which would have to be $a$. If $b \neq a$, then there is some $\epsilon>0$
such that $B(b,\epsilon)$ contains a finite number of points hence
$b$ cannot be an accumulation point.
Note that the sequence $a_n = 1$ has $\{a_n\}_n = \{1\}$ which has no accumulation points.
In general, the set $\{a_n\}_n$ will have $a$ as an accumulation point iff
for all $N$ there is some $n \ge N$ such that $a_n \neq a$.
A: The usual definition of an accumulation point (for a subset of $\mathbb{R}$) is as follows:

Let $A \subseteq \mathbb{R}$.  We say that $a$ is an accumulation point of $A$ if for all $r > 0$ the set $B(a,r) \cap A \setminus \{a\}$ is nonempty.  That is, every ball centered at $a$ contains a point of $A$ other than $a$ itself.

Even if the sequence $(a_n)$ converges, the set $\{ a_n \}$ needn't have any accumulation points.  For example, any constant set $\{a, a, a, \dotsc, \} = \{a\}$ does not have any accumulation points (as no ball centered at $a$ contains any point of the set other than $a$, but the sequence $(a,a,a,\dotsc)$ converges to $a$.  On the other hand, if $\{a_n\}$ has an accumulation point, and $(a_n)$ is convergent, then the accumulation point is necessarily the limit.
A: Here's a somewhat more general proof of this:
Let $X$ be a Hausdorff topological space, and let $\left\{x_n\right\}_{n\in\mathbb N}$ be a sequence in $X$ that converges to $x\in X$. Suppose that $y\in X$ is an accumulation point of $\left\{x_n\right\}_{n\in\mathbb N}$ such that $x\neq y$. Then, since $X$ is Hausdorff, there are disjoint open neighborhoods $U$ and $V$ of $x$ and $y$, respectively. Since $\left\{x_n\right\}_{n\in\mathbb N}$ converges to $x$, there is an $N\in\mathbb N$ such that $x_n\in U$ whenever $n\geq N$. However, this implies that there are at most $N-1$ elements of $\left\{x_n\right\}_{n\in\mathbb N}$ in $V$ different from $y$. Denote this set of finite elements by $\left\{y_n\right\}_{n=1}^m$. Again, since $X$ is Hausdorff, there are disjoint open neighborhoods $U_n$ and $V_n$ of $y_n$ and $y$, respectively. Then $\bigcap_{n=1}^m V_n$ is an open neighborhood of $y$ containing no elements of $\left\{x_n\right\}_{n\in\mathbb N}$, which is a contradiction.
A: Yes, you are right. If $x_n$ is a convergent sequence, then


*

*$\{x_n\}$ has an accumulations point if and only if $\{x_n\}$ is infinite.

*$\{x_n\}$ has at most one accumulation point and if it has one, then it coincides with the limit.


None of these statements is true for non-convergent squences.
