I'm slightly confused at the proof of this theorem, hopefully someone can help.(adsbygoogle = window.adsbygoogle || []).push({});

Identity theorem: Suppose X and Y are Riemann surfaces, and [itex]f_1,f_2:X \to Y[/itex] are holomorphic mappings which coincide on a set [itex]A \subseteq X[/itex] having a limit point [itex]a \in X[/itex]. Then [itex]f_1[/itex] and [itex]f_2[/itex] are identically equal.

The proof starts out with: Let G be the set of all points [itex]x \in X[/itex] having an open neighbourhood W such that [itex]f_1|_W = f_2|_W[/itex].

Now, it seems like they assume without argument that G is non-empty. Why is G non-empty?

Also, maybe I'm just confused about this, what does it mean that A has a limit pointain X? As I understand it,ais simply a point which cannot be separated from A by two open sets. Couldn't it be the case that A is a one-point set, doesn't every set have a limit point? In that case I don't see how G could be non-empty.

EDIT: Ok, upon some thought I think I recall that a limit point [itex]a \in X[/itex] of A is a point which cannot be separated from [itex]A / \{a\}[/itex]. Is this right? In that case A must be infinite. Still, this doesn't resolve the question of the existence of such a W as described above.

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# Identity theorem

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