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i'm struggling with the following problem:

Let $f:\mathbb{C}\setminus\{0\}\to\mathbb{R}_+$, $(x,y)\mapsto\sqrt{x^2+y^2}$. Show that $f$ is open for saturated sets $A$ under the assumption, that $\mathbb{C}\setminus\{0\}$ and $\mathbb{R}_+$ are equipped with the subspace topology of the euclidean topology.

i know a set $A$ is saturated iff $A=f^{-1}(f(A))$, but i dont know where this comes to use.

my attempt:

i want to show that for $z\in A\subseteq\mathbb{C}\setminus\{0\}$ the set $f(A)$ is a neighborhood of $f(z)$, but how do i compute $f(A)$? since the euclidean topology is just intervals $(x,y)$ i guess i could write $A$ as the product of such intervals, but then i'm stuck.

HannahBloom
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  • Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun May 05 '20 at 15:38
  • sorry, i didnt want to make it too specific – HannahBloom May 05 '20 at 15:44

2 Answers2

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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 $f[A]$.

Henno Brandsma
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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.

Matematleta
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