Let \( f \) be a meromorphic function on a (connected) Riemann Surface \( X \). Show that the zeros and the poles of \( f \) are isolated points.
Solution
Let \( X \) be a Riemann surface and let \( f \) be a meromorphic function on \( X \). Then \( f \) can be considered as a holomorphic map \( f \ \colon X \longrightarrow \hat{\mathbb{C}} \) which is not identically equal to \( \infty \). But for holomorphic maps between Riemann Surfaces the Identity Theorem holds.
Remark: Using the Identity Theorem we can also prove that the set of ramification points of a proper, non-constant, holomorphic map between Riemann Surfaces consists only of isolated points.
The exercise can also be found at mathimatikoi.org
Solution
Let \( X \) be a Riemann surface and let \( f \) be a meromorphic function on \( X \). Then \( f \) can be considered as a holomorphic map \( f \ \colon X \longrightarrow \hat{\mathbb{C}} \) which is not identically equal to \( \infty \). But for holomorphic maps between Riemann Surfaces the Identity Theorem holds.
- If the set of zeros of \( f \) contained a limit point, then, by the Identity Theorem, \( f \) should be equal to \( 0 \). But we have assumed that \( f \) is not constant.
- If the set of poles of \( f \) contained a limit point, then, by the Identity Theorem, \( f \) should be equal to \( \infty \). But we have excluded that case by definition of a meromorphic function.
Remark: Using the Identity Theorem we can also prove that the set of ramification points of a proper, non-constant, holomorphic map between Riemann Surfaces consists only of isolated points.
The exercise can also be found at mathimatikoi.org
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