What is the structure of $S$?

Question

Suppose we define an equivalence relation on $\mathbb R$ by $aRb$ iff $\{a\}=\{b\}$ for $a,b\in\mathbb R$. Here $\{.\}$ defines the fractional part. In other words, $aRb$ iff $a-b\in\mathbb Z$. Suppose $S$ is the collection of all equivalence classes defined by this equivalence relation. What is the structure of $S$?

So the way I proceeded is:

Suppose $\bar{a}$ is the equivalence class for $a\in\mathbb R$. Then, $\bar{a}$ is the set of all real numbers located at integer distances from $a$. So $\bar{a}$ is a set of equidistant points (if we try to represent $\bar{a}$ graphically).

The same is true for $\bar{b}$ for any other $b\in \mathbb R$. Every such equivalence class consists of a sequence of equidistant points.

So the collection $S$ of these equivalence classes will be the entire real line!

But our professor drew $S$ as a circle containing all elements from $0$ to $1$ on the circumference, with $0$ and $1$ being the identical point on the circumference. He further hinted that we need to consider the function $z\mapsto e^{2\pi zi}$ for $z\in\mathbb R$. I don't understand this.

Answer 1

Hint.

Name $T$ the equivalence relation. And define $$ \begin{array}{l|rcl} \varphi : & \mathbb R & \longrightarrow & \mathbb U\\ & t & \longmapsto & e^{2i\pi t} \end{array}$$

You know that $\varphi$ is surjective. Now you will be able to prove that the inverse image of $z \in \mathbb U$ by $\varphi$ is an equivalence class of the equivalence relation $T$.

This proves that $\varphi$ induces a bijection $\overline{\varphi}$ between $S$ and $\mathbb U$. Which is exactly what your professor mentionned.

Answer 2

This equivalence relation makes $0.1\equiv1.1\equiv2.1\equiv3.1$ as you have stated. So each equivalence class has a representative on the unit interval $[0,1)$. Since $1\equiv0$ it makes sense to tie the two ends of this interval together, making a circle.

Answer 3

Hint: Show that $\psi : (\mathbb R, +) \to (S^1, \cdot)$ defined by $\psi (x) = e^{2\pi i x}$ is a homomorphism. Notice that $\ker \psi = \mathbb Z$. Then show that is isomorphic to $S^1$, by using the Isomorphism Theorem.

Where $S^1$ is the unitary circle and $S = \frac{\mathbb R}{\mathbb Z}$.

Answer 4

Your reasoning shows that every $x\in\mathbb{R}$ is contained in some equivalence class, which is what you'd expect from an equivalence relation (an equivalence relation always partitions a set in disjoint equivalence classes). However, the elements of the collection of equivalence classes are (of course), equivalence classes, not individual element of $\mathbb{R}$.

So indeed $\bigcup_{\bar{a}\in S}\bar{a}=\mathbb{R}$, but this doesn't say anything about the structure of $S$. The equivalence class $\bar{a}$ can be identified with some element of $[0,1)$ (namely the element {a}). And since $\bar{0}=\bar{1}$ we can also identify the end point of this interval to form something looking a lot like a circle. Now it's up to you to maybe make this more concrete by using the hint.

---

Edit: If you are asked to 'draw' $S$ without any other hints or notes, then that question is just very vague. You are right to be slightly confused then. This is the same as asking one to 'draw' $S=\{\{1\},\{2,3\},\emptyset\}$; There is no clear 'best'/'standard'/'obvious' way to draw this:

1. are we supposed to draw for each element in the untion of $S$ a single point on the real line?
2. Are we supposed to draw $1$ point for each subset of $S$ at the value of the average of that subset?
3. ... etc

So the answer is: if you are asked to make a drawing that 'shows' the structure of $S$, it is up to you to decide how to make this structure most clear using a drawing. In your original example the mapping to the unit circle seems like one of the best choices but, again, there is no $1$ unique right/best answer here.