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Let $ f : U \subset \mathbb C \to \mathbb C $ a continuous function where $ U $ is an open and connected set.

Prove that the image of $ f$ cannot be a curve on compex plane.

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False without further conditions on $f$. Is $f$ analytic? Injective? Either would be sufficient. – Erick Wong Jul 1 '12 at 20:46
@ErickWong: I found this exercise exactly as I post it. Can you give your solution with the further conditions ? – passenger Jul 1 '12 at 20:50 (this requires analyticity of $f$) – Giuseppe Negro Jul 1 '12 at 20:54
up vote 2 down vote accepted

Are you sure there are no additional constraints, e.g. the Cauchy-Riemann equations are satisfied (continuity + C.R. -> analytic). One could simply take $f(z) = \Re(z)$, which is continuous but not analytic -- the image of an open set (say the ball of radius 1 around the origin) is a curve (the interval $[-1,1]$).

The reason this cannot hold for an analytic function is because of the open mapping theorem. Analytic functions are open mappings, so an open set $U$ has to map to an open set $f(U)$, which cannot be a curve.

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O.K I see thank you very much! Can we find an example of a non-analytic function for which the claim fails ? – passenger Jul 1 '12 at 21:05
Yes, $f(z) = \Re(z)$, as stated. The important thing to understand is that a curve is not open in the topology of $\mathbb{R}^2$. – snarski Jul 1 '12 at 21:12

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