In the complex analysis text book "Complex Variables and Applications 8th edition", it states the function $1/sin (\pi/z)$ has singular points $z = 0$ and $z = 1/m; (m = \pm 1,2,3,4,\dots.) $
I sort of understand why z=0 is not considered an isolated singular point. However why are all the other singular points isolated? Could you explain this to a math undergrad? Thank-you
On
For any positive integer $m$, $1/m$ is a finite distance away from its closed neighbours in the set $A =\{ 1/n : n \in \{1,2,3,\dotsc \}\}$. (You can convince yourself of this easily enough: they are obviously $1/(m \pm 1)$, and you can easily subtract these from $1/m$ and take the modulus.)
Hence you can stick an open ball around $1/m$ for any positive integer $m$ so that $1/m$ is the only point of $A$ inside it. This is precisely what isolated means.
Isolated means we can draw a ball around the point and have no other singularities be in that ball. given a number of the form $z=1/m$, we can draw a sufficiently small ball such that no other number $z=1/n$ is in that ball, neither is $0$, so they are isolated.
On the other hand, there is no ball around $0$ that can avoid numbers of the form $1/m$, since we can always get as close to 0 as we want with numbers of the form $1/m$, so 0 is not isolated.