Necessary and sufficient condition for $\left \lvert \mathbb{R} \big / \! \! \equiv \right \rvert = |\mathbb N|$

Question

I was thinking about the following: Let $\equiv$ be an equivalence relation on $\mathbb R$. What is the necessary and sufficient condition that $\equiv$ must satisfy in order for $\left \lvert \mathbb{R} \big / \! \! \equiv \right \rvert=|\mathbb N|$ to be true? My intuition tells me that, if $\left \lvert \mathbb{R} \big / \! \! \equiv \right \rvert=|\mathbb N|$ is true, then each equivalence class of $\mathbb{R} \big / \! \! \equiv$ has an uncountable number of elements, and if each equivalence class has a countable or finite number of elements, then $\left \lvert \mathbb{R} \big / \! \! \equiv \right \rvert>|\mathbb N|$. Is there any theorem/result answering this kind of question?

Answer 1

Your intuition is close to correct. The criterion you need is that there be a countably infinite number of equivalence classes.

It follows that at least one of the classes must have uncountably many elements, but that's all you can assert. Consider the example where each integer is in its own class and everything else is a single uncountable class.

Answer 2

Suppose that, in $\Bbb R$, you define$$x\sim y\iff(x\in\Bbb N\wedge y=x)\vee(x,y)\notin\Bbb N^2.$$Then the equivalence classes are $\Bbb R\setminus\Bbb N$ and $\{1\}$, $\{2\}$, $\{3\}$, and so on. So, $\Bbb R/\sim$ is countable. But there is one and only one equivalence class with an uncountable number of elements.

And, in general, if $\Bbb R/\sim$ is countable, then at least one equivalence class has to have an uncountable number of elements.

Answer 3

Sometimes it's more comfortable to think of equivalences classes as surjective functions. The relationship $|\mathbb R / \equiv|=|\mathbb N|$ is the same as saying there exists a surjective function $$f:\mathbb R\mapsto\mathbb N,\quad f(a)=f(b)\iff a\equiv b$$ As pointed out by other answers, the only conclusion you can draw is that at least one of the equivalence classes has $2^{\aleph_0}$ elements, since $$n\cdot2^{\aleph_0}=2^{\aleph_0} \quad\forall n\in \mathbb N_+\\\aleph_0\cdot2^{\aleph_0}=2^{\aleph_0}\\$$