Existence of a partition of a set $X$ which contains at least 2 subsets of $X$ with the same cardinality

Question

If $X$ is a set s.t $|X|\geq \aleph _0$ then I want to prove that there exist $A,B\subset X$ s.t $A\cap B=\emptyset$, $A\cup B=X$ and $|A|=|B|=|X|$. How can I construct such subsets?

Answer 1

This is easy to do for $\aleph_0$: take even and odd numbers. You can use this to prove the statement for a more powerful $X$; the trick is to split $X$ into disjoint sets of cardinality $\aleph_0$ (you will need the axiom of choice, naturally). To this end, well order $X$ (which can be done by the Zermelo theorem) and put elements $a\prec b$ into one set if they are within a finite distance, i.e. there exist some $n\ge 1$ and $x_1,\dots,x_n=b$ such that $a$ precedes $x_1$, $x_1$ precedes $x_2$, $\dots$, $x_{n-1}$ precedes $x_n=b$. Now you can "halve" each of your disjoint sets, which are of cardinality $\aleph_0$ (this I leave for you to prove).

You can use the same idea for other things, e.g. to split $X$ into a countable number of equipotent sets.