# Question regarding singular points of a complex function

Definition 1: The point $$z_0$$ is called a singular point of a function $$f$$ if $$f$$ fails to be analytic at $$z_0$$ but is analytic at some point in every neighborhood of $$z_0$$

Definition 2: A singular point $$z_0$$ is said to be isolated if, in addition,there is a deleted neighborhood of $$z_0$$ throughout which $$f$$ is analytic

Example: The function $$\frac{1}{\sin(\frac{\pi}{z})}$$

has the singular points $$z=0$$ and $$z=\frac{1}{n} (n=\pm 1,\pm2,\cdots)$$, all lying on the segment of the real axis from $$z=-1$$ to $$z=1$$. $$\underline{Each\; singular\; point \;except\; z=0 \;is\; isolated.}$$

I'm confusing this underlined term, because i know clearly $$z=0$$ is not an isolated singular point, since every neighborhood of $$0$$ contains another singular points. This is ok.

But how the numbers of the form $$\frac{1}{n}$$ is isolated ? If i take, $$n=1000000000000000000000$$, then how we guarantee the neighborhood of this $$\frac{1}{n}$$ does not contains some remaining singular points ?

I'm sorry, this is very elementary for others, but i'm confusing

Any help is appreciated !

Given $n,$ the ball of radius $r=\min\{|\frac{1}{n}-\frac{1}{n+1}|,|\frac{1}{n-1}-\frac{1}{n}|\}$ centered at $\frac{1}{n}$ has no other singular points.