Show that a function maps open saturated sets into open sets

A set $A$ is saturated for $f$ if $A$ contains $x$ and $f(x)=f(x')$ then $x' \in A$ as well. So with any point $a$ of $A$, $A$ must also contain the whole circle $\{z \in \Bbb C\setminus \{0\}: |z|= a\}$.

An open saturated set contains besides this $a$ also a whole ball around $a$ an this then implies we have a whole "band" of circles inside $A$, and so an interval around $f(a)$ inside $f[A]$.


Use the fact that a function maps open saturated sets into open sets if and only if it is a quotient map.

Now, if $f^{-1}(S)$ is open in $\mathbb C\setminus \{(0,0)\}$ for an arbitrary $S\in \mathbb R^+$, since translations and contractions/dilations are homeomorphsms, we may assume without loss of generality $f^{-1}(S)=U\setminus \{(0,0)\},\ $ the open punctured unit disk. Then, $f(U\setminus \{(0,0)\})=(0,1)$ is open in $\mathbb R^+$ so $f$ is a quotient map.

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Analysis

General Topology

Complex Analysis

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