Talk:Identity theorem

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

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)[reply]

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)[reply]

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)[reply]

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)[reply]

Full theorem[edit]

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[edit]

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)[reply]