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I'm trying to understand the quotient space ${\mathbb R}/{\mathbb Z}$ (with the quotient topology) and I am stuck with the following question:

How can I show that ${\mathbb R}/{\mathbb Z}$ is compact?

One can either establish a homeomorphism between ${\mathbb R}/{\mathbb Z}$ and the set $\{(x,y):x^2+y^2=1\}$ or directly show by definition of compactness. In either way I don't know how to go on. Or is there a handy theorem that one can use here?

[Added:] Thanks to David's comment and Daniel's elaboration, one should note that in this post $\mathbb{Z}$ should be understood as a group acting on $\mathbb{R}$. More explicitly,

Consider the set $X=\mathbb {R}$ of all real numbers with the ordinary topology, and write $x \sim y$ if and only if $x − y$ is an integer. Then the quotient space $X/\sim$ is homeomorphic to the unit circle $S^1$ via the homeomorphism which sends the equivalence class of $x$ to $\exp(2\pi ix)$. More details are in the Wikipedia article.

  • Consider the map $t\mapsto (\cos(2\pi t),\sin(2\pi t))$. – Adam Hughes Mar 07 '17 at 02:42
  • @AdamHughes: I would be interested in seeing more details, thanks. By considering the cosets of $\mathbb{R}/\mathbb{Z}$, one can easily give a bijection between $\mathbb{R}/\mathbb{Z}$ and one of the fundamental domains $[0,1)$ and thus give a bijection between $\mathbb{R}/\mathbb{Z}$ and the unit circle $S^1$. Right now I don't see how the homeomorphism is given. –  Mar 07 '17 at 02:50
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    Hey Jack, this is just a hint/comment, not an answer. Best of luck! – Adam Hughes Mar 07 '17 at 02:51
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    You can't, since it is not compact. – David Hartley Mar 07 '17 at 20:16
  • @DavidHartley: What are you talking about? –  Mar 07 '17 at 20:34
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    At the end of the "Examples" section on the wikipedia page, there is the "Note: The notation R/Z is somewhat ambiguous. If Z is understood to be a group acting on R then the quotient is the circle. However, if Z is thought of as a subspace of R, then the quotient is a countably infinite bouquet of circles joined at a single point."; If $\mathbb{R}/\mathbb{Z}$ denotes the space obtained by identifying the subspace $\mathbb{Z}\subset \mathbb{R}$ to a single point, the resulting quotient space is not compact. – Daniel Fischer Mar 07 '17 at 20:39
  • @DanielFischer: Stupid me. I didn't notice that at all. Let me edit the post. (And my apology to David.) –  Mar 07 '17 at 21:00

2 Answers2

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$\mathbb{R}/\mathbb{Z}$ has a (minor) inconvenience, which is that it doesn't come from a compact space, so we can't guarantee compactness straightforwardly, and neither can we use the quotient map directly to exhibit a homeomorphism (although this is easily manageable in this special case), since we only have a continuous bijection a priori. But in order to avoid that inconvenience, one can do the following:

Prove that $[0,1]/\sim $, where $0 \sim 1$, is homeomorphic to $S^1$. For that, consider $f:[0,1] \to S^1 $ given by $f(t)=e^{2\pi i t}$, then pass to the quotient using the universal property. It is easily verified to be a bijection in the quotient. Having compact domain (since $[0,1]/ \sim=\pi([0,1])$ and $\pi$ is continuous), it is a homeomorphism (since $S^1$ is Hausdorff).

Done that, now prove that $[0,1]/ \sim$ is homeomorphic to $\mathbb{R}/\mathbb{Z}$. For that, consider $f:[0,1]\to \mathbb{R}$ the inclusion, compose with the quotient to yield a function $[0,1]\to \mathbb{R}/\mathbb{Z}$, then use the universal property to get a function $[0,1]/\sim \to \mathbb{R}/\mathbb{Z}$, which is a continuous bijection from a compact set to a Hausdorff set again.

Aloizio Macedo
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The map $$ f\colon [0,1]\to \mathbb{R}/\mathbb{Z}, \qquad f(x)=x+\mathbb{Z} $$ is continuous and surjective.

egreg
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  • I wanted to know originally how to establish a homeomorphism between $\mathbb{R}/\mathbb{Z}$ and $S^1$ but I somehow ended up with posting a related question regarding compactness of $\mathbb{R}/\mathbb{Z}$, for which this should be the simplest way. Thank you! –  Mar 07 '17 at 13:55
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    @Jack Then consider $\varphi\colon\mathbb{R}\to S^1$, $\varphi(x)=e^{2i\pi x}$, which is a surjective continuous homomorphism (seeing $S^1$ as a subgroup of the multiplicative group of $\mathbb{C}$), with kernel $\mathbb{Z}$. – egreg Mar 07 '17 at 14:38
  • That gives a group isomorphism between $\mathbb{R}/\mathbb{Z}$ and $S^1:={z\in\mathbb{Z}:|z|=1}$. How does one get that this is also a homeomorphism? –  Mar 07 '17 at 15:13
  • @Jack A continuous bijection between compact Hausdorff spaces is a homeomorphism. – egreg Mar 07 '17 at 15:40
  • Thanks for your elaboration. Let me summarize as follows: The continuous surjection $\varphi:\mathbb{R}\to S^1$ induces a continuous bijection $g:\mathbb{R}/\mathbb{Z}\to S^1$ (with $\varphi=g\circ\pi$ where $\pi$ is the canonical projection); since the compactness of $\mathbb{R}/\mathbb{Z}$ is done and it is known that $S^1$ is Hausdorff, the map $g$ is a homeomorphism. –  Mar 07 '17 at 15:55
  • @Jack Yes, that's correct. – egreg Mar 07 '17 at 15:57