How do I find the equivalence classes of the relation $\mathcal{R} = \{(x, y) \in \mathbb{C} \mid x - y \in \mathbb{R}\}$?

Question

I'm currently trying to solve the following tasks:

1) Prove that the relation $\mathcal{R} = \{(x, y) \in \mathbb{C} \mid x - y \in \mathbb{R}\}$ is an equivalence relation.

2) Find all equivalence classes of $\mathcal{R}$.

I already solved the first one, but I'm having "trouble" with the second one. If I'm not mistaken, there is an infinite amount of equivalence classes, right?

• $x \sim 0 = \{0\}$, because for "$x - 0 \in \mathbb{R}$" to be true, $x$ has to be $0$.

And you can continue doing this, for an infinite amount of equivalence classes:

• $x \sim 1 = \{1\}$

• $x \sim 2 = \{2\}$

• $x \sim 3 = \{3\}$

• $\cdots$

If this is actually correct, how do I correctly write it down?

Can I just write something like "$x \sim i = \{i\}$ for $i \in [0, \infty[$"?

Answer 1

Hint:

Try to show that given a complex number $x\in\Bbb C$ where we write $x$ in the form $x=a+bi$ with $a,b\in\Bbb R$ and another complex number $y$ written as $y=c+di$ with $c,d\in\Bbb R$...

...that $x$ is related to $y$ if and only if $b=d$.

So... $2+i$ is related to $5+i$ since $(2+i)-(5+i)=-3$ is real for example and $5$ is related to $0$ since $(5)-(0)=5$ is real and so on...

As to the question of the equivalence classes... if we were to picture the complex numbers as appearing on the complex plane, what does it mean for $b=d$?

So, the set of all numbers related to some complex number $a+bi$ will be the set of all complex numbers whose imaginary part is equal to $b$... how might you write that? How might you write the collection of all such equivalence classes?