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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 $X$.

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  • $\begingroup$ You should first decide/specify what you mean by "same cardinality". Cardinality for finite sets is pretty obvious, for infinite sets they all end up being "equally infinite". To me it feels like you want them to be of equal measure. $\endgroup$ – N.Bach May 15 '17 at 12:30
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    $\begingroup$ @N.Bach "equal cardinality" is a strictly defined mathematical term. Two sets have the same cardinality if and only if there exists a bijection between them. $\endgroup$ – 5xum May 15 '17 at 12:31
  • $\begingroup$ Not an expert: assuming Choice, you can well-order your set, i.e. it has the same cardinality as some ordinal $\alpha$. I guess the ordinal $\alpha+\alpha$ does have the same cardinality too. A bijection between your set $X$ and $\alpha+\alpha$ will give you such a partition. But this is just a sketch. Maybe the difficulty is now to construct the bijection $f:\alpha\leftrightarrow \alpha+\alpha$. $\endgroup$ – M. Winter May 15 '17 at 12:33
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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 $X=A\cup B$.

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    $\begingroup$ In fact, every infinite set $X$ being in bijection with $X\times X$ is equivelant to the axiom of choice. $\endgroup$ – Jason May 15 '17 at 15:48

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