Equivalence classes of "$x \sim y \Longleftrightarrow x -y $ is rational".

Question

Given the equivalence relation $x \sim y \Longleftrightarrow x -y $ is rational on the interval $[0,1)$.

How do we reason* that there are uncountably infinite number of equivalence classes?

*A rigorous proof is not required but still welcome.

Edit

The equivalence relation on $[0,1)$ given by $x∼y⇔x−y$ is rational.

If the set of equivalence classes is countable, then $(0,1)$ is the countable union of countable sets and therefore is countable, which is a contradiction. — roger, Mar 30 '13 at 17:50

score 5 · Accepted Answer · answered Mar 30 '13 at 17:50

Let $x\in[0,1)$, show that $x/\sim$ is countable (because every element is of the form $x+q$ for some unique rational $q$).

If there would be countably many equivalence classes then $[0,1)$ would be the countable union of countable sets, which is not the case.

score 2 · Answer 2 · answered Mar 30 '13 at 17:51

Let's say we have $\kappa$ many equivalence classes, each classes having $\omega$ number of elements (countably infinite), so we have $$c=|[0,1)|=\kappa\cdot\omega$$ and, as among cardinals, $\kappa\cdot\omega=\max(\kappa,\omega)$, we get that $\kappa=c=$ continuum.

Equivalence classes of "$x \sim y \Longleftrightarrow x -y $ is rational".

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