# Decomposition space of $\mathbb{C}$ by concentric circles [closed]

What are the topological properties of the quotient space $X$ obtained from $\mathbb{C}$ by identifying points of the same modulus? I.e., the space $X=\mathbb{C}/E$ where $E$ is the equivalence relation $x\sim y$ iff $|x|=|y|$. Is this space homeomorphic to any commonly known topological space? What if we remove a countable dense subset from $\mathbb{C}$, obtaining a space we call $\mathbb{C}_0$ and ask the same question for $X_0=\mathbb{C}_0/E_0$, where $E_0$ is $E$ restricted to $\mathbb{C}_0$?

[Added April 14, 2015] This question is a particular instance of a more general problem which I am trying to understand. Suppose $Y$ is a Polish space (i.e., a complete separable metric space) and $E$ is an equivalence relation on $Y$ such that every equivalence class is homeomorphic to the unit circle. Do we know that the number of equivalence classes is either finite, countably infinite, or $\ \mathfrak{c}$? My illustrative example turns out to be trivial and not very helpful. I suspect the answer in general may depend on CH. In ZFC we can get a YES if we assume more about the equivalence relation, e.g., it is a closed (or even coanalytic) subset of the product space $Y\times Y$. I suspect the stated assumptions on $E$ do not imply this condition.

## closed as off-topic by Anton Petrunin, abx, Joonas Ilmavirta, YCor, Alex DegtyarevApr 13 '15 at 5:45

This question appears to be off-topic. The users who voted to close gave these specific reasons:

• "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – Joonas Ilmavirta, YCor, Alex Degtyarev
• "This question does not appear to be about research level mathematics within the scope defined in the help center." – Anton Petrunin, abx
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• I must be missing something: why isn't the answer to your first question just the usual half-axis $[0,\infty)$ with its usual topology? – Yemon Choi Apr 13 '15 at 0:42
• Oh, of course! This is obvious. – Fred Dashiell Apr 13 '15 at 0:49
• The second question does look more interesting, though, so I don't think the question needs to be closed just yet – Yemon Choi Apr 13 '15 at 0:58
• The second question is not any more difficult than the first. If $0 \in \mathbb{C}_0$, then the quotient is $[0,\infty)$ again, because equivalence classes in $\mathbb{C} \smallsetminus \{0\}$ are uncountable so removing a countable set cannot have any significant effect. If $0 \notin \mathbb{C}_0$, then the quotient is $(0,\infty)$. – Dave Witte Morris Apr 13 '15 at 1:43
• I think the later, general question raised by Fred Dashiell makes this post worth reopening – Yemon Choi Apr 14 '15 at 16:16

The number zero has no other element in its equivalence class and so let us set it aside. The non-zero complex numbers, as a group under multiplication is the direct product of its two subgroups: (i) numbers of unit modulus, and (ii) positive real numbers. That is $z=re^{i\theta}$, the polar decomposition. So the quotient by one component would be the other. Now attach the number set aside we see $[0,\infty)$