Q:

What is the structure of $S$?

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.

A:

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.

