Talk:Identity theorem

The analytic continuation argument is circular, though. Charles Matthews 21:06, 30 December 2005 (UTC)

It seems to me that saying that a set contains an open set does not prove that the set is itself open. Something appears to be missing in this proof. Something needs to be said about the w in S I think to resolve this. (In fact the proof doesn't even show that S is non-empty) Msg555 (talk) 21:57, 13 May 2011 (UTC)


 * Looks alright to me. It shows that for every w in S there exists an open set in S containing w - that proves openness. And the fact that S is nonempty follows from the hypothesis in the very first line of the article: if f=g in some neighbourhood of z, then z is in S. Jutreras (talk) 10:30, 17 June 2011 (UTC)

The improved version is IMHO wrong. For example if the set S has just one element there is no contradiction to h(z) != 0 in B\{c}. — Preceding unsigned comment added by 87.152.246.140 (talk) 22:04, 4 December 2012 (UTC)

The statement of the theorem is wrong. You require something more than equality on any non-empty subset. Otherwise, all holomorphic functions are constant functions. 50.67.69.151 (talk) 05:35, 1 February 2014 (UTC)

Full theorem
The claims and proofs in this article a decade ago were a bit messy. After having looked at the more thorough German version of the same article, and worked through the details myself I have now added a ‘the following are equivalent‘ type full characterisation of when two holomorphic functions are equal (equivalent: when a holomorphic function vanishes). Imo we should keep this new section, delete the old sections, and add an applications section (e.g. the thing about fibres being discrete or the entire domain). drusus null 23:49, 19 April 2021 (UTC)

Accumulation Point MUST be assumed inside the set
If the set where the functions agree accumulated but to a point on the boundary (which is outside the open set) then there is no contradiciton. The function sin(1/z) is analytic on the open domain C-{0}, it has a set of zeros that accumulate, yet the function is not zero. It is not a contradiciton bc 0, the accumulation point is not in the domain of definition. 130.234.20.37 (talk) 13:53, 26 January 2023 (UTC)