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Inspired by Question # 2420627 "Prove that there is a bijection $f:\mathbb R\times \mathbb R\to \mathbb R$ in the form of $f(x,y)=a(x)+b(y)$" and the answer and comments by Thomas Andrews.

Proposition. There exist subsets $A, B$ of $\mathbb R,$ each a bijective image of $\mathbb R,$ such that the function $g:A\times B\to \mathbb R,$ where $g(a,b)=a+b,$ is a bijection. That is, $\forall x\in \mathbb R\; \exists! (a,b)\in A\times B\;(x=a+b).$

In $ZFC$ this is easily proven. I suspect that in $ZF$ it may be undecidable (Assuming $Con (ZF)$ of course).

Any thoughts or references on this? Perhaps an example like $L(S),$ for some $S$ in a Forcing extension of $V,$ where $L(S)$ doe not satisfy the Proposition.

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  • $\begingroup$ Note that forcing extensions preserve the axiom of choice, so that's not where to look. Symmetric submodels of forcing extensions can lack choice. $\endgroup$
    – Noah Schweber
    Commented Sep 8, 2017 at 1:38
  • $\begingroup$ @Noah Schweber . If $S$ is not in the ground model then $L[S]$ may fail to satisfy $AC$. For example there are forcing extensions $V[G_1], V[G_2]$ where $\mathbb R^{V[G_1]}=\mathbb R^{V[G_2]}$ but $|\mathbb R|^{[V[G_1]}\ne |\mathbb R|^{V[G_1]}$ so $L[\mathbb R^{V[G_1]} \Vdash \neg AC.$ $\endgroup$
    – DanielWainfleet
    Commented Sep 8, 2017 at 1:47
  • $\begingroup$ @NoahSchweber.Thanks for the edit $\endgroup$
    – DanielWainfleet
    Commented Sep 8, 2017 at 1:51
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    $\begingroup$ I do not know what notation you are using. Under the standard interpretation, $L[X]$ is always a model of choice. $\endgroup$
    – Andrés E. Caicedo
    Commented Sep 8, 2017 at 3:08
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    $\begingroup$ @Andrés: If I recall correctly, Kunen actually uses the opposite notation in his new book. Which is just awful. I recall having made a similar comment to someone some time ago, and they whipped open the book and showed me that "they are only following Kunen". Maybe it was only about $L(\Bbb R)$, which was denoted by $L[R]$ (like Shelah sometimes does in his papers). But nonetheless, that was the notation used. $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 8, 2017 at 9:34

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