Talk:Removable singularity

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[edit] Incorrect statement?

I believe that your statement of Reimann's theorem is incorrect. While I can't seem to find a good description of the theorem online anywhere I am pretty sure that:

a) It is defined as the limit at a of the function existing (i.e. the limit is the same as you approach from all sides, but the function is defined as something else when it is actually at a. (sorry I don't feel like inserting the latex right now).

b) In the example of the Heaviside functin where f(x)=0 x<0 and f(x)=1 x>=1 we have a function that is obviously bounded at 0, but this is not a removable singularity. No redefinition of the Heaviside function will make this continuous.

Perhaps this definition of Reimann's theorem makes more sense in the complex analysis sense, but if so you should be more specific. --anon

I believe Riemann's theorem applies only to holomorphic functions. Perhaps a different wording could make this clearer. —Caesura^(t) 22:07, 8 December 2005 (UTC)

Yes, this phenomenon is unique to complex analysis. If the function is bounded at the singularity, that's enough to guarantee not only its continuity, but even infinite differentiability. I clarified this in the article. Oleg Alexandrov (talk) 01:02, 9 December 2005 (UTC)