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Suppose $f : G \longrightarrow \mathbb{C}$ is a holomorphic function on a connected open set $G$. Then the set $N := \{w \in G: f(w)= 0\}$ is closed in $G$. This fact is used in Reinhold Remmert's Complex Analysis to prove the identity theorem. It seems trivial, but I cannot see why it is true...

Marcel S
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    It's true for every continuous function. – Daniel Fischer May 27 '17 at 18:43
  • Perhaps more important is that if $f$ is not identically zero, then the set of zeroes is in fact discrete; see here: https://math.stackexchange.com/questions/1483249/show-that-the-set-of-zeros-of-f-is-discrete – TomGrubb May 27 '17 at 18:44

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A holomorphic function is continuous, notice that $N=f^{-1}(\{0\})$ with $\{0\}$ being closed.

C. Falcon
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