If $|X|<|Y|$ then $|Y|=|Y-X|$ (with $Y$ infinite)

Question

Like the title says, I would like to prove that if $|X|<|Y|$ then $|Y|=|Y-X|$. (with $Y$ infinite)

I know I have to use the axiom of choice, but I've no idea about how to proceed.

Any help is welcome :-)

Answer 1

Assuming the axiom of choice, one can show that $|X|+|Y|=\max\{|X|,|Y|\}$. This suffices in order to prove that $|Y|=|Y-X|$ using the following [trivial] claim: $|Y|=|Y\setminus X|+|X|$.

It is necessary to use the axiom of choice, and in fact this is equivalent to the axiom of choice. If the axiom of choice fails then there is some $X$ which cannot be well-ordered and $A$ which can be well-ordered such that $|A|\nleq|X|$. Take $Y=A\cup X$ (and assume this is a disjoint union).

Then $|X|<|Y|$, since $|A|\leq|Y|$ but $|A|\nleq|X|$; and at the same time $|Y-X|<|Y|$, since $|Y-X|=|A|$, which is well-orderable, whereas $|Y|$ itself is not well-orderable.