Prove that if $f$ is an analytic function on an open subset $U$ of $\mathbb{C}$ then the non empty preimage of $\mathbb{R}$ under $f$ cannot be compact. I have been trying this problem for a long time now but to no avail. I was trying to prove this by assuming the contrapositive and then showing that $f$ is constant. Can anyone please help me with this? Thanks for any help.

  • $\begingroup$ The preimage of the real line can be the empty set, and that is compact. $\endgroup$ – Martin R Apr 11 '18 at 14:58
  • $\begingroup$ Thanks for the comment. I have edited. $\endgroup$ – Ester Apr 11 '18 at 15:00
  • $\begingroup$ Is $f$ defined only on $U$? $\endgroup$ – Antoine Apr 11 '18 at 15:10
  • $\begingroup$ Yes, only on U which is not even connected, not on the entire complex plane. $\endgroup$ – Ester Apr 11 '18 at 15:20

Suppose otherwise. Then $f\bigl(f^{-1}(\mathbb{R})\bigr)$ is a non-empty compact subset of $\mathbb R$. Therefore, there is a $x\in f\bigl(f^{-1}(\mathbb{R})\bigr)$ such that $x=\max f\bigl(f^{-1}(\mathbb{R})\bigr)$. So, $x\in\mathbb R$ but the image of $f$ contains no real number greater than $x$. That's impossible, by the open mapping theorem. Unless of course, $f$ is constant, in which case $f^{-1}(\mathbb{R})=f^{-1}\bigl(\{x\}\bigr)$, which is not compact.

  • $\begingroup$ Clear and concise! +1 $\endgroup$ – Fimpellizieri Apr 11 '18 at 15:12
  • $\begingroup$ $f^{-1}(\mathbb{R})$ is a set of complex numbers, so what is $\max f^{-1}(\mathbb{R})$ ? $\endgroup$ – Martin R Apr 11 '18 at 15:14
  • $\begingroup$ If $f(z) = i z$, what is $\max f^{-1}(\mathbb{R})$? $\endgroup$ – Antoine Apr 11 '18 at 15:14
  • $\begingroup$ @MartinR I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Apr 11 '18 at 15:19

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