Single Accumulation point How to show that the origin
is the only accumulation point of the set $$z, = \frac{i}{n} (n = 1, 2, ...)$$(I was able to show it's a accumulation point,but was not able to show it's the only one)
def of accumulation point:A point $z$ is said to be an accumulation point of a set $S$ if each deleted neighborhood of $z$ contains at least one point of $S$.
Thanks for your help
 A: For all $z\in\mathbb C$, $|z-z_n|\geq |z|-|z_n|$ as a consequence of the triangle inequality.  With your $z_n=\dfrac{i}{n}$ this implies $|z-z_n|\geq |z|-\frac1n$.  If $z\neq 0$, then there exists $N$ such that $\frac{1}{N}<\frac12|z|$, from which it follows that for all $n\geq N$, $|z-z_n|>\frac12|z|$.  Thus the disk of radius $\frac12|z|$ centered at $z$ contains none of the points $\{z_N, z_{N+1},z_{N+2},\ldots\}$.  Of the remaining finitely many points $\{z_1,z_2,\ldots,z_{N-1}\}$, consider all of the distances $|z-z_k|$.  There is a smallest positive one of these because the set is finite*.  Call that smallest positive distance $d$.  Then the open disk centered at $z$ with radius $\min\{d,\frac12|z|\}$, with $z$ removed, contains no $z_n$.  This shows that $z$ is not an accumulation point.
*(It could be that $N=1$, in which case there is nothing further to show; rather than taking the minimum of the empty set you just wouldn't need $d$ at all.)
More generally, when you have a convergent sequence $\{w_n\}$ with $w=\lim\limits_{n\to\infty}w_n$, then for all $\varepsilon>0$, the disk of radius $\varepsilon$ centered at $w$ contains all but finitely many of the $w_n$s.  Hence if $z\neq w$, taking $\varepsilon=\frac12|z-w|$, the open disk centered at $z$ with radius $\frac12|z-w|$ will contain only finitely many of the $z_n$s, and you can take a smaller radius if necessary to avoid the remaining finitely many points.
