Choice Set is Uncountable

Show that any choice set for the rational equivalence relation on a set of positive outer measure must be uncountably infinite

I have been thinking about this problem on and off for a few days, but I haven't been able to come with much. Let $E$ be a set with positive outer measure, and $C_E$ some choice set of $E$. I know that $E$ must be uncountable, otherwise it will have a outer measure of zero. This means that $E$ must contain an uncountable number of irrational numbers. Initially I thought that the equivalence class of an irrational would be a singleton, but this isn't so since $(1-\pi) + \pi = 1$. I could use a hint.

EDIT:

Let $[x] = \{y \in E ~|~ x-y \in \Bbb{Q} \}$ denote the rational equivalence class of the on $E$ of the element $x \in E$. Then $[x] = \{x + q ~|~ \mbox{ for some } q \in \Bbb{Q}\}$, which suggests that the elements of $[x]$ can be put into a one-to-one correspondence with some subset $\Bbb{Q}$, making the cardinality of $[x]$ at most countably infinite. As I noted above, a set with positive outer measure must be uncountably infinite. Now if there were only a countably infinite number of equivalence classes, then these equivalence classes, which partition $E$, couldn't add up to make the uncountable set $E$ (i.e., it would be impossible for their union to be $E$). Hence, there must be an uncountable number of equivalence classes. Since $C_E$ is formed by taking a single element from each equivalence class, it's clear that $C_E$ must also be uncountably infinite.

How does that sound?

• What if the equivalence classes are not disjoint? – Bijayan Ray May 15 at 10:21
• @BijayanRay Equivalence classes are always disjoint, provided that the relation is in fact an equivalence relation (i.e., a relation that is reflexive, symmetric, and transitive). – user193319 May 15 at 10:27