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The relation $\sim$ is as follows: $x \sim y$ if $x-y \in \mathbb{Z}$ and I can't really figure out what are the classes of equivalence from that relation. Help please.

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    $\begingroup$ Hint: try winding a piece of string about your finger while you think about this question. $\endgroup$ – Rob Arthan Oct 30 '16 at 23:22
  • $\begingroup$ For the future: It's generally preferable if you put the entire question in the post, instead of just the title. $\endgroup$ – AJY Oct 30 '16 at 23:28
  • $\begingroup$ See also this post: Show: Quotient space is homeomorphic to unit sphere $\endgroup$ – Martin Sleziak Oct 31 '16 at 9:16
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The class of $x$ is simply $\xi (x) = [x]=x+\mathbb Z = \{ x+n : n\in\mathbb Z\}$. The map $\xi : {\Bbb R} \mapsto {\Bbb R}/{\Bbb Z}$ associates to $x$ its class. Perhaps more important, a set $A$ in the quotient is open iff $\xi^{-1}(A)$ is open in ${\Bbb R}$.

You may then show that $\cos$ and $\sin$ (of the reals) gives rise to a well-defined continuous bijective function $$\phi\in {\Bbb R}/{\Bbb Z}\mapsto (\cos 2\pi \phi,\sin 2\pi \phi) \in S^1\subset {\Bbb R}^2$$ which provides the wanted homeomorphism.

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Hint: Remember $S^1 = \{\exp{i 2\pi s} : s \in [0, 1)\}$

Define $\Theta : S^1 \rightarrow \mathbb{R}/\sim$ as $\Theta(\exp{i2\pi s}) = \{s + k : k \in \mathbb{Z}\}$ and show that $\Theta$ is an homeomorphism.

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Since my original approach was flawed, I feel somewhat obligated to not only correct myself, but also think through more of the details.

First of all, for every $[x] \in \mathbb{R}/ \sim$ there is a unique $\tilde{x} \in [0,1)$ such that $x \sim \tilde{x}$, namely $\tilde{x} := x - \lfloor x \rfloor$, where $\lfloor x \rfloor := \max \{ z \in \mathbb Z \mid z \le x \}$.

Now consider the map $$ j: \mathbb R / \sim \to S^{1}, [x] \mapsto e^{i \tilde{x} 2 \pi}. $$ Note that $i(x) = i(y)$ iff $x-y \in \mathbb Z$ iff $x \sim y$ iff $\tilde{x} = \tilde{y}$, so the above map can also be expressed as $$ j: \mathbb R / \sim \to S^{1}, [x] \mapsto e^{i x 2 \pi} $$ if for every $a \in \mathbb R / \sim$ we fix some (any) $x \in \mathbb R$ with $a = [x].

By this argument, $j$ is injective and clearly surjective.

To see that $j$ is a homeomorphism, let us further consider the projection $$ p \colon \mathbb R \to \mathbb R / \sim, x \mapsto [x]. $$ and $$ c: \mathbb R \to S^{1}, x \mapsto e^{ix \pi}. $$

$c$ is continous and hence, for any open $O \subseteq S^{1}$, the set $c^{-1}"[O] \subseteq \mathbb R$ is open. It follows, from the definition of the quotient space topology that $j$ is continuous.

On the other hand, if $O \subseteq \mathbb{R} / \sim$ is open, then $U := p^{-1}"O \subseteq \mathbb R$ is open and in fact $j"[O] = c"[U]$. Since $c$ is an open map, it follows that $j$ is open as well.

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  • $\begingroup$ Uh...? No, as the OP noted the quotient is the circle. What you might be trying to say is that every real number has a unique representative in $[0,1)$ under this equivalence relation. $\endgroup$ – Pedro Tamaroff Oct 30 '16 at 23:22
  • $\begingroup$ You're writing that $[0,1)\simeq S^1$. This is false. $\endgroup$ – Pedro Tamaroff Oct 30 '16 at 23:25
  • $\begingroup$ @PedroTamaroff Shoot, you're right. $[0,1)$ isn't compact. $\endgroup$ – Stefan Mesken Oct 30 '16 at 23:26

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