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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.”

gen-ℤ ready to perish
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    It means open ball around your point. Since its on the complex plane, its just an open disk. – Sergio Escobar May 05 '20 at 01:22
  • Where did you find this definition? Are you sure this is holomorphic rather than analytic? For the latter it makes sense to require a neighborhood. – Moishe Kohan May 05 '20 at 01:46
  • @MoisheKohan Wikipedia and MSE concur on the definition and seem confirm that a function that a function is analytic if and only if it is holomorphic. Are you saying the definition is wrong? – gen-ℤ ready to perish May 05 '20 at 01:48
  • @gen-zreadytoperish: Yes these are equivalent, but it is a theorem. Wikipedia and MSE are not always reliable sources. Did you see this definition in any textbook? It is not wrong, just cumbersome. One usually defines analyticity on open subsets, then there is no need to invoke neighborhoods. – Moishe Kohan May 05 '20 at 01:53
  • @MoisheKohan No, I did not see this in a textbook. It would be nice if I could check a book out of the library to read deeper into the matter, but all libraries in my region (metroplex of Dallas, Texas) are closed. – gen-ℤ ready to perish May 05 '20 at 01:54
  • Oh, I see. I just checked Rudin's book, he has the correct definition; same for Ahlfors. No idea where the Wikipedia definition came from, just ignore it. – Moishe Kohan May 05 '20 at 02:01
  • @MoisheKohan Well, what’s the correct definition then? – gen-ℤ ready to perish May 05 '20 at 02:04
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    I suspect the Wikipedia definition is a laboured way of saying "a holomorphic function must be defined on an open subset of $\Bbb C$". – Angina Seng May 05 '20 at 02:41
  • @AnginaSeng That sounds like the best explanation. I’d give that a $\color{green}{\checkmark}$, but it would be best with a mention of why we need this extra qualification is needed – gen-ℤ ready to perish May 05 '20 at 05:40
  • It’s very rude to not explain why you’re downvoting, especially when several people liked the answers to the question, and the answers are predicated on the existence of the question. – gen-ℤ ready to perish May 09 '20 at 20:06

2 Answers2

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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.

Moishe Kohan
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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 $r$.

Lee Mosher
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  • This is a redundancy, is it not? – gen-ℤ ready to perish May 05 '20 at 01:24
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    @gen-zreadytoperish No, why do you think so? – saulspatz May 05 '20 at 01:28
  • 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. – gen-ℤ ready to perish May 05 '20 at 01:32
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    @gen-zreadytoperish only if $D$ is an open set, which is the whole point of this statement. You can reword the definition of holomorphicity to only be defined on open sets, but that is exactly the concept of a neighborhood. Nothing here is redundant. – Brevan Ellefsen May 05 '20 at 02:00