Convergent sequences and accumulation points

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.

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 $a_n \neq a$.