Re: Every set x equinumerous with a set y disjoint from x?

["MoeBlee"]

Rupert wrote:
> That looks good to me. David's proof looks good to me as well.

So would I be correct to take it that the consensus is that the theorem
almost surely needs choice?

If that is the case, I find it interesting that this is not mentioned
much (if at all) in basic textbooks. It seems so intuitive that for any
set there is an equinumerous set disjoint from the original set, so the
need for the axiom of choice would be another in the list of "reasons"
for the axiom.

MoeBlee