How is the “neighbourhood” part of ‘holomorphic’ defined?

Question

According to my understanding, we say a function on the complex numbers is holomorphic if and only if it is complex differentiable in a neighbourhood of every point in its domain—not just that point.

This is a very vague qualification in my opinion. How is the neighbourhood defined?

In my mind, I’m thinking that the “neighbourhood” part is redundant. Say $f:D\to\Bbb C$, where $D\subseteq\Bbb C$, and say $f$ is complex differentiable for all $z\in D$. If $N(z,r)=\{w\in D:\lvert z-w\rvert<r\}$ is a neighbourhood of $z$, then we already know every point in the neighbourhood is complex differentiable because all $w\in D$ are complex differentiable.

Logically it’s unreasonable for this universally used definition to include a redundancy, so I conclude I must be misunderstanding what is meant by “neighbourhood.”

Answer 1

Here is the definition which one finds in standard textbooks (Ahlfors, Rudin, etc.)

Definition. Let $\Omega$ be an open subset of the complex plane. Then a function $f: \Omega\to {\mathbb C}$ is said to be holomorphic in $\Omega$ if it is complex-differentiable at every point of $\Omega$.

Logically speaking, it is equivalent to the definition from Wikipedia, but the latter obscures (a bit) the assumption that $\Omega$ is open. In fairness to the rest of the Wikipedia article, later on (in the "Definition" part) it also gives the standard definition of a holomorphic function.

Answer 2

What this means is clarified by naming all the quantities involved, and stating the condition with precise logical quantifiers and logical connectives, like this:

For every point $z$ in the domain there exists a number $r > 0$ such that for every point $w$ in the open ball centered on $z$ of radius $r$, the function is differentiable at $w$.

"Neighborhood" here refers to the ball centered on $z$ of radius $r$.