Splitting infinite sets into disjoint sets of equal cardinality

Question

When given an infinite set $X$, it seems to me very reasonable that one can 'split' it up into two disjoint subsets $A$ and $B$ such that all three have the same cardinality. For countable sets, this is rather easy, for then we can index the elements in $X$ by $(x_n)_{n \in \mathbb{N}}$, and we can take $A$ as the elements indexed by even $n$, and $B$ those by odd ones. Of course, countability isn't needed per se; with a set indexed by $\mathbb{R}$, it can be split up rather easily as well. Where I run into difficulty is a general infinite set, where this indexing trick doesn't seem to work, because the indexing set isn't 'known' well enough (in the sense that I can't make an easy choice, as with $\mathbb{N}$ or $\mathbb{R}$).

My question is if this is possible for any infinite set $X$.

Answer 1

In general, if $X$ is infinite, $\vert X \vert = \vert X \vert+\vert X \vert$, ie. there exists a bijection between $X$ and $X\sqcup X$, where $X\sqcup X$. denotes the disjoint union. There's an obvious injection from $X$ to $X\sqcup X$, and there's an injection from $X\sqcup X$ to $X\times X$. Since $X$ is infinite, we have $\vert X \vert = \vert X\times X\vert$ (as a well-known consequence of AC), and so $\vert X \vert = \vert X\sqcup X \vert$. Thus, there exists a bijection $f: X \to X\sqcup X$. Now let $A = f^{-1}[\{0\}\times X], B = f^{-1}[\{1\}\times X]$. Then $A,B$ are disjoint and have the same cardinality as $X$, and $X=A\cup B$.