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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.

DanielWainfleet
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  • Note that forcing extensions preserve the axiom of choice, so that's not where to look. Symmetric submodels of forcing extensions can lack choice. – Noah Schweber Sep 08 '17 at 01:38
  • @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 $L[\mathbb R^{V[G_1]} \Vdash \neg AC.$ – DanielWainfleet Sep 08 '17 at 01:47
  • @NoahSchweber.Thanks for the edit – DanielWainfleet Sep 08 '17 at 01:51
  • @bof. That was a typo of mine. I edited. – DanielWainfleet Sep 08 '17 at 03:03
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    I do not know what notation you are using. Under the standard interpretation, $L[X]$ is always a model of choice. – Andrés E. Caicedo Sep 08 '17 at 03:08
  • @AndrésE.Caicedo I meant $L(X)$ as in Kunen's textbook . I always get that notation mixed up with $L[X].$ – DanielWainfleet Sep 08 '17 at 03:56
  • @NoahSchweber. I made a notation error : $L(S)$, not $L[S]$. – DanielWainfleet Sep 08 '17 at 03:58
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    @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. – Asaf Karagila Sep 08 '17 at 09:34
  • @AsafKaragila. I am using Kunen's "old" book, first edition. Thank you. – DanielWainfleet Sep 08 '17 at 18:50

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