Talk:Hartogs' lemma

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[edit] Is this stated correctly?

The page claims:

More precisely, on Cⁿ for n ≥ 2, any analytic function F defined on the complement of a compact set K extends (necessarily uniquely) to an analytic function on Cⁿ.

Suppose we take F(z)=1/z, defined on C\{0}, and let K={0}, which is certainly compact. Then F doesn't extend to an analytic function on C, unless the meaning of "analytic" is somehow extended beyond what I'm used to. --Trovatore 19:42, 23 October 2005 (UTC)

Sorry, I missed the condition n≥2. --Trovatore 19:44, 23 October 2005 (UTC)