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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 ?

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    Riemann surfaces do not have countable topologies. Do you mean second countable?2011-12-23
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    @Aki: You mean "has a **countable base for its topology**", a.k.a. is second-countable. This is a much weaker condition than "has a countable topology".2011-12-23
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    What is "countable topology"? Does it mean that it has a topology which is first or second countable?2011-12-23
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    Thank 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

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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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    Thank you ! Now it is clear for me too.2011-12-23