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 Degtyarev Apr 13 '15 at 5:45

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

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    $\begingroup$ 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? $\endgroup$ – Yemon Choi Apr 13 '15 at 0:42
  • $\begingroup$ Oh, of course! This is obvious. $\endgroup$ – Fred Dashiell Apr 13 '15 at 0:49
  • $\begingroup$ The second question does look more interesting, though, so I don't think the question needs to be closed just yet $\endgroup$ – Yemon Choi Apr 13 '15 at 0:58
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    $\begingroup$ 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 $(0,\infty)$. $\endgroup$ – Dave Witte Morris Apr 13 '15 at 1:43
  • $\begingroup$ I think the later, general question raised by Fred Dashiell makes this post worth reopening $\endgroup$ – 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 $[0,\infty)$

To see this pictorially think of the complex plane as an infinite table cloth. Now bring a magic wand with some gum at its tip and place it at zero and pull up the cloth with the wand in the third dimension so that the table cloth starts hanging like a thread from the tip of the wand. (you can squeeze it!) So we this cloth looks like a hanging rope with points at same distance from origin merged, the equivalence relation you mentioned. this rope is the real line from zero upwards.

  • $\begingroup$ Isn't this just a lengthier version of what I said in my first comment? $\endgroup$ – Yemon Choi Apr 14 '15 at 16:15
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    $\begingroup$ Thats right; you gave the answer first; I gave an explanation justifyuing; I'd encourage you to rewrite your comment as answer, then I'll upvote that and delete my answer. OK? $\endgroup$ – P Vanchinathan Apr 14 '15 at 16:48
  • $\begingroup$ Thanks. Firstly, my comment was trivial (as the OP acknowledges). Secondly, my comment is now no longer an answer because the OP has written a new and more substantial question. Thirdly, I can't add an answer even if I wanted to, because the question is currently closed. $\endgroup$ – Yemon Choi Apr 15 '15 at 14:47

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