# Why are holomorphic functions defined in reference to an open set?

I presume that it is because one needs to take a limit to define the derivative, but is there another reason?

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No, the only thing is that each point should be in a disc that is part of the domain of definition. E.g. your guess is the reason. –  AD. Nov 3 '10 at 5:36
Because power series automatically converge on discs (assuming the radius of conv. is positive), so if you want to meaningfully say your function is locally represented by power series it makes sense that the domain is an open set. (Note that it is common to speak about a holomorphic function on, say, a closed disc, but that is understood to mean it is defined on some open set containing that closed disc.) –  KCd Nov 3 '10 at 9:35

You may want to take a look at When is a Function that Satisfies the Cauchy-Riemann Equations Analytic? by J. D. Gray and S. A. Morris, The American Mathematical Monthly, Vol. 85, No. 4, (Apr., 1978), pp. 246-256.

Yes, in a sense, we want open domains so limits can be computed, but we have a notion of directional limit, so this is not much of a reason; what I mean is, some justification is needed that we actually lose something otherwise.

The paper above gives examples showing how the Cauchy-Riemann equations alone do not suffice for analyticity, even in the presence of continuity, and proceeds to look for weak additions to these two requirements.

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