In "Lectures on Riemann surfaces" by Otto Forster, before the proof of Rado's theorem which asserts that every Riemann surface has a countable topology, the author commented "Clearly this is trivial for compact Riemann surfaces". This is not clear at all for me. Could someone help me understand this ?
Why is it trivial for compact Riemann surfaces to have a countable topology?
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complex-analysis
riemann-surfaces
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0Thank you @ChrisEagle and ZevChonoles for quick answers. The above is the direct quotation from the book. I understand it as second countable or "has a countable base for its topology" as you suspect. – 2011-12-23
1 Answers
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Such a surface is a union of finitely many open disks. Each disk has a countable basis consisting of balls of rational radius around points with rational coordinates. The union of the bases for each disk is a countable basis for the whole space.
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0Thank you ! Now it is clear for me too. – 2011-12-23