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I want to show that every infinite genus Riemann surface $M$ has a proper closed subset such that, $M^*\setminus E$ ($M^*$ is the one-point compactification of $M$) is connected and locally connected, and $E$ is not of essentially of a finite genus.

Am I right to say that let's take a closed path start at infinity and back to infinity including all genus?

Lukas Geyer
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user191988
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    Can you clarify what "and $E$ is not of essentially of a finite genus" is supposed to mean? And what is a path "including all genus"? – Lukas Geyer Nov 12 '14 at 19:12
  • If every infinite genus Riemann surface compactifies to something homeomorphic to the countable-genus torus, then this looks fine-but I don't actually know that that's true. – Kevin Carlson Nov 12 '14 at 19:17

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