Does there exist an equivalence relation on a set such that the set of equivalence classes and all individual equivalence classes are all uncountable?

Question

I have thought about many examples, usually involving equivalence relations on $\Bbb{R}$, and so far none of them fulfill the criteria. I have tried to think about the various definitions of equivalence relation, equivalence classes, countability and so on but I have yet to prove or disprove it for sure. At this point I feel like there exists no such equivalence relation, but I would like to know for sure and how you would go about proving this.

On $\mathbb R^2$, let $(x,y)\sim(x’,y’)$ iff $(y=y’)$. — fantasie, Oct 31 '21 at 12:19

score 0 · Accepted Answer · answered Oct 31 '21 at 16:57

For a general approach to this, note that if we are assuming the axiom of choice, then every infinite set $S$ has the same cardinality as $S\times S$ (proving this is a little more complicated, you can find more info here). Thus there exists a bijection $f:S\rightarrow S\times S$, and we can construct an equivalence relation such that $x\sim y$ if and only the first part of $f(x)$ equals the first part of $f(y)$. If $S$ is uncountable, this equivalence relation has the properties you want.

A similar result without choice seems possible, but is probably more complex if it exists.

Does there exist an equivalence relation on a set such that the set of equivalence classes and all individual equivalence classes are all uncountable?

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