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Work in a reasonable theory of determinacy such as $\mathsf{ZF+DC+AD}$. Which of the following identities are true for arbitrary infinite sets?

  1. $|A^2|=|A^3|$ (motivated by an MSE question that asks if this identity implies choice; a well known theorem of Tarski says $|A|=|A^2|$ implies choice)
  2. $|A|=|A\sqcup A|$
  3. $|A\times\omega|=|A|$

More generally, is there a procedure to determine the validity of any cardinality identity involving $A$, $\omega$, $\times$ and $\sqcup$?

Second try: As pointed out by Asaf Karagila, the answer is negative for a somewhat stupid reason...Here are some attempts to make the question more interesting:

  • Work in $\mathsf{ZF+AD}+V=L(\mathbb{R})$, or determine the truth in $L(\mathbb{R})$ under large cardinals (are these the same?)
  • Only consider "small" infinite sets $A$ (such as images of $\mathbb{R}$).
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    $\begingroup$ In general, 2 is equivalent to 3. $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 2 at 18:11
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    $\begingroup$ @AsafKaragila I always got points off for saying 2=3 in math classes. $\endgroup$
    – Noah Schweber
    Commented Aug 2 at 18:33
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    $\begingroup$ @NoahSchweber: Equivalent isn't the same as equal in my foundational framework. $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 2 at 19:30
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    $\begingroup$ As a first approximation, you want to assume $\mathsf{AD}^++V=L(\mathcal P(\mathbb R))$, what I call a "natural model" of determinacy. (Working in $L(\mathbb R)$ is a particular case of this.) $\endgroup$
    – Andrés E. Caicedo
    Commented Aug 2 at 20:27
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    $\begingroup$ @Noah Yes, that's correct. The cardinal structure of natural models of $\mathsf{AD}^++\lnot \mathsf{AD}_{\mathbb R}$ is significantly different from that of natural models of $\mathsf{AD}_{\mathbb R}$, though. $\endgroup$
    – Andrés E. Caicedo
    Commented Aug 2 at 20:41

2 Answers 2

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Simon Thomas "Superrigidity and countable Borel equivalence relations" Corollary 4.9 gives a countable Borel equivalence relation E shows E and 2E are not Borel bireducible. (The examples involve probability measures and group actions.) It is quite possible that under $\mathsf{AD}^+$, $|\mathbb{R} / E| < |\mathbb{R} / E \sqcup \mathbb{R} / E|$ for this equivalence relation (as Woodin claims).

Woodin "The cardinals below $|[\omega_1]^{<\omega_1}|$" Corollary 71 gives a set $X_0$ so that $|X_0| < |X_0 \sqcup X_0|$ under $\mathsf{AD}_\mathbb{R} + \mathsf{DC}$.

By Hjorth's $E_0$-dichotomy ("A Dichotomy for the Definable Universe"), under $\mathsf{AD}^+$, every set which is a surjective image of $\mathbb{R}$ either injects into the power set of an ordinal (is linearly orderable) or $\mathbb{R} / E_0$ injects into it (is not linearly orderable). Woodin's example is a set of the former type and Thomas's example is a set of the latter type.

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  • $\begingroup$ Thanks! These papers look very relevant. But a priori Borel bireducibility may not be the same as having the same cardinality, am I correct? And from what I understand from the comments by Noah and Andrés above, $\mathsf{AD}_{\mathbb{R}}$ contradicts $V=L(\mathbb{R})$ so there is still hope that the cardinality structure in $L(\mathbb{R})$ is nicer? $\endgroup$
    – n901
    Commented Aug 7 at 19:27
  • $\begingroup$ It is very unlikely that Simon Thomas' proof cannot be adapted to prove a genuine cardinality result under $\mathsf{AD}^+$. Typically in this area, the relevancy of Borelness is to use the Baire property, Lebesgue measurability, and the Lusin-Novikov countable section uniformzation which all hold in full generality under $\mathsf{AD}^+$ (which holds in $L(\mathbb{R})$). $\mathsf{AD}_R$ does fails in $L(\mathbb{R})$ because Uniformization fails in $L(\mathbb{R})$. It is possible that Woodin's example could be a theorem of $\mathsf{AD}^+$ (but the proof are quite complicated). $\endgroup$
    – user534667
    Commented Aug 7 at 22:07
  • $\begingroup$ There is a perspective that Simon Thomas proof involving descriptive set theory (and Lie groups) gives evidence that this phenomenon is the correct behavior of cardinalities in a definable setting (like determinacy) albeit it is not the most simple behavior. $\endgroup$
    – user534667
    Commented Aug 7 at 22:19
  • $\begingroup$ Borel cardinality results of the sort discussed here transfer to the $\mathsf{AD}^+$ setting, and we have a good understanding of how to verify claims of this kind. Richard and I wrote a paper a lifetime ago illustrating this with Glimm–Effros-type dichotomies. We never published it, but see our published trichotomy paper. The Borel version of such a result holds in small $\mathsf{ZFC}$ models within the determinacy model, and an ultraproduct argument and Vopěnka forcing allows us to lift it. $\endgroup$
    – Andrés E. Caicedo
    Commented Aug 8 at 0:17
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    $\begingroup$ @AndrésE.Caicedo Many dichotomies in Woodin paper like the $S_1$-dichotomy are not true in $L(\mathbb{R})$. However $S_1 = \{f \in [\omega_1]^{<\omega_1} : \sup(f) = \omega_1^{L[f]}\}$ can be defined in any model of $\mathsf{AD}$ and its basic properties can be proved in $\mathsf{AD}$. The set $X_0$ is very complicated defined in terms of $\mathbb{P}_\mathrm{max}$ but seems to me that it could be defined in any model of $\mathsf{AD}$. Potentially some of its basic property like $|X_0| < |X_0 \sqcup X_0|$ could be proved in $\mathsf{AD}^+$. $\endgroup$
    – user534667
    Commented Aug 8 at 0:45
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Starting from $L(\Bbb R)$, we can take a symmetric extension which preserves $\sf DC$ and adds an $\omega_1$-amorphous set, just somewhere far above $\Theta$.

So we cannot prove that (1) or (2) hold for all sets. This is to be expected, since $\sf AD$ only had power up to the power set of the reals, and the universe is much much larger than that.

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  • $\begingroup$ Thank you, I have editted the question. Are there still "loopholes" remaining? $\endgroup$
    – n901
    Commented Aug 2 at 18:33
  • $\begingroup$ Probably? Let's wait for someone who's more familiar with the structure of the universe around the reals... $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 2 at 19:30
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    $\begingroup$ @AsafKaragila At least assuming large cardinals there can't be too simple loopholes since then the theory of $L(\mathbb{R})$ is generically absolute (I don't remember if this follows from determinacy in $L(\mathbb{R})$ alone). $\endgroup$
    – Noah Schweber
    Commented Aug 2 at 19:37
  • $\begingroup$ @Noah: Not sure how that'd be relevant? $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 2 at 19:49
  • $\begingroup$ @AsafKaragila I assumed a "loophole" in this context meant an easy way to consistently-but-not-provably get bad behavior high in the cumulative hierarchy. Since forcing doesn't work for the modified question, to the best of my knowledge we don't currently have any tools for that sort of thing. $\endgroup$
    – Noah Schweber
    Commented Aug 2 at 20:00

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