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Let $S^1$ be the circle of radius 1 viewed as a subset of the complex plane. Then it is a subgroup of $\mathbb{C}^*$. Let $\mathbb{Q}, \mathbb{Z}$ be the additive groups of rational and integer numbers respectively. Then $\mathbb{Q}/\mathbb{Z} \cong S^1$ as groups via $[r] \mapsto e^{2\pi i r}$. In particular, they have the same number of elements. Now $S^1$ is uncountable. Intuitively $\mathbb{Q}/\mathbb{Z}$ should be countable. What am I missing? Is there a general set theoretic argument behind this?

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    $\begingroup$ It's $\Bbb R / \Bbb Z \cong S^1$, not $\Bbb Q / \Bbb Z \cong S^1$. Also, we can map $S^1$ bijectively to $[0,2\pi)$, which is uncountable because it's a nondegenerate interval of real numbers. $\endgroup$ – tilper Jun 2 '17 at 14:47
  • $\begingroup$ Use the stereographic projection. $\endgroup$ – Wuestenfux Jun 2 '17 at 14:47
  • $\begingroup$ $\Bbb Q/\Bbb Z\cong S^1$ is simply not true. It is $\Bbb R/\Bbb Z$. But answering your question: No. $\Bbb Q/\Bbb Z$ is not uncountable but nicely countable. $\endgroup$ – M. Winter Jun 2 '17 at 14:50
  • $\begingroup$ @tilper The map given is a group isomorphism. Sorry, the intended question was about the cardinality $\mathbb{Q}/\mathbb{Z}$. I edited the title. Clearly $S^1$ is uncountable. $\endgroup$ – user 1987 Jun 2 '17 at 14:52
  • $\begingroup$ Ok, but the isomorphism is flawed. See my answer for why. $\endgroup$ – tilper Jun 2 '17 at 14:57
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Left as a comment but I'll expand on it here:

It's $\Bbb R / \Bbb Z \cong S^1$, not $\Bbb Q / \Bbb Z \cong S^1$. Also, we can map $S^1$ bijectively to $[0,2\pi)$, which is uncountable because it's a nondegenerate interval of real numbers.

With your map from $\Bbb Q / \Bbb Z$ to $S^1$ where you send $r$ to $e^{2\pi i r}$, what element gets sent to $e^{\sqrt2 \pi i}$? $e^{\sqrt2 \pi i}$ is certainly in $S^1$ but under the map $r \mapsto e^{2 \pi ir}$, $r$ would have to be $\dfrac1{\sqrt 2}$, which is definitely not in $\Bbb Q / \Bbb Z$. So $\Bbb Q / \Bbb Z \ncong S^1$. What we have in fact is $\Bbb R / \Bbb Z \cong S^1$.


Apparently this isn't part of your intended question, but I'll leave it here anyway for posterity.

More simply, any element of $S^1$ can be described as $e^{i\theta}$, where $\theta \in [0, 2\pi)$. Thus the map $e^{i\theta} \mapsto \theta$ is a bijection between $S^1$ and $[0,2\pi)$. Any nondegenerate interval of real numbers can be bijectively mapped to all of $\Bbb R$. Thus $S^1$ can be bijectively mapped to $\Bbb R$, and so $S^1$ is uncountable.

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    $\begingroup$ I see my mistake now. We simply have $\mathbb{Q}/\mathbb{Z} \cong \{z \in \mathbb{C}^* \mid z^n = 1 \text{ for some } n\in \mathbb{N}\}$. $\endgroup$ – user 1987 Jun 2 '17 at 14:59
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$\Bbb Q$ is countable. $\Bbb Q/\Bbb Z$ consists of equivalence classes of rational numbers. So there should be an embedding $$\iota:\Bbb Q/\Bbb Z\hookrightarrow\Bbb Q.$$ For example map $[q]\in\Bbb Q/\Bbb Z$ onto the unique rational number $p$ in $[0,1)$ with $p\in[q]$. So it is an injection, hence the cardinality of $\Bbb Q/\Bbb Z$ is at most countable. Since it is infinite, it is countable as there are no infinities below $\aleph_0$.

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Notice that $\mathbb{Q}/\mathbb{Z} = \{q + \mathbb{Z}:q \in \mathbb{Q}, \ 0 \leq q < 1\}$. This is precisely all the rationals in [0,1). Of course it is countable.

As @tilper and @M. Winter have mentioned, it is not true that $\mathbb{Q}/\mathbb{Z} \cong S^1$. Instead, $\mathbb{R}/\mathbb{Z} \cong S^{1}$. Nevertheless, with some work, you can show that $\mathbb{Q}/\mathbb{Z}$ is the torsion subgroup of $\mathbb{R}/\mathbb{Z}$. Therefore $\mathbb{Q}/\mathbb{Z}$ can be realized as the torsion subgroup of $S^1$, which is the multiplicative group of root of unity in $\mathbb{C}^{\times}$.

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