Is an isolated singularity of a complex-valued function an isolated point of the domain of the function? Let $\Omega\subseteq\mathbb C$ be open and $f:\Omega\to\mathbb C$. We say that $z_0\in\mathbb C$ is an isolated singularity of $f$ if $f$ is holomorphic on a punctured neighborhood $N$ of $z_0$.
On the other hand, there is the following paragraph in the Wikipedia article of this notion:

Formally, and within the general scope of functional analysis, an isolated singularity for a function $f$ is any topologically isolated point within an open set where the function is defined.

I don't get this. In a general topological space $E$, we say that $x\in E$ is an isolated point of $B\subseteq E$ if there is an open subset $U\subseteq E$ with $B\cap U=\{x\}$.
But why is $z_0$ an isolated point of $N$?
 A: In general, one should remember that anybody can edit a Wikipedia page, regardless of their qualification. In this case, by looking through the history of edits, this particular offending sentence (or, rather, some version of it) was initially written by an engineer, who clearly did not understand the definition; then another engineer tried to correct it but made it even worse. The correct sentence should be something like:
"In the language of topology, an isolated singularity of $f$ is an isolated point of the set $\partial \Omega$."
or
"In the language of topology, an isolated singularity of $f$ is an isolated point of the set ${\mathbb C}\setminus \Omega$."
(There is some disagreement if $\infty$ is allowed as an isolated singularity; if it is, then $\partial \Omega$ is understood as the boundary of $\Omega$ in the Riemann sphere.)
A: Do remember that anyone can do any changes in wikipedia so don't completely rely on wikipedia.
An isolated singularity of f is an isolated point of the set ∂Ω.
