What choice principles does "every set is in bijection with a transitive set" imply?

Question

A set $X$ is transitive if $x\in y\in X\implies x\in X$. We shall say $\mathsf{TC}$ to mean the axiom "For all $X$ there is transitive $Y$ such that $|X|=|Y|$", where $|X|=|Y|$ means that there is a bijection $X\to Y$. I am interested in how much choice we obtain by assuming $\mathsf{TC}$.

Question. Does $\mathsf{TC}$ imply any classical choice principles? E.g. $\mathsf{AC}_X(Y)$, $\mathsf{DC}_\lambda$, comparability principles, etc.

Assuming the Axiom of Choice, $\mathsf{TC}$ holds because every set is in bijection with an ordinal. On the other hand, $\mathsf{TC}$ implies that every infinite set surjects onto $\omega$ (if $X$ is transitive and infinite then the rank function will have infinite image), so $\mathsf{TC}$ is not a consequence of $\mathsf{ZF}$.

My guess would be that $\mathsf{TC}$ has a stronger interaction with Hartogs and Lindenbaum numbers than just $(\forall X)\aleph^\ast(X)\neq\aleph_0$. Maybe $\mathsf{TC}$, or perhaps $\mathsf{TC}+\mathsf{KWP}_1$, could imply full $\mathsf{AC}_\mathsf{WO}$.

Answer 1

Here is one instance, although not with a "classical" choice principle.

Namely, your principle TC implies the rigid relation principle RR, a weak choice principle introduced by Justin Palumbo and myself in this paper:

• Joel David Hamkins and Justin Palumbo, The rigid relation principle, a new weak choice principle, Math Logic Quarterly, 58 (6):394-398 (2012), DOI:10.1002/malq.201100081, arXiv:1106.4635.

The rigid relation principle is the assertion that every set admits a rigid binary relation. This is true under TC, since $\langle X,\in\rangle$ is rigid whenever $X$ is a transitive set, and so every set that is bijective with a transitive set admits a rigid binary relation.

Meanwhile, Justin and I proved that RR is strictly intermediate between ZF and ZFC.

Answer 2

You can say a bit more than just every infinite set surjects onto $\omega$. Every infinite set is Dedekind infinite.

The reason is simple. If $T$ is an infinite transitive set, then $T\cap V_\omega$ is an infinite transitive set, and it is countable. To see this, note that if $T\cap V_\omega$ is finite, then, since $T$ is infinite, there is a gap in the finite ranks, but this is impossible as the rank function on a transitive set has a downwards closed image (i.e., its image is an ordinal).

This can probably be lifted to get slightly more out of this argument, but the "inspect locally, conclude globally" is bound to fail here. If every infinite set of reals is Dedekind-infinite, then every set of reals is equipotent to a transitive set. Simply consider the reals as $V_{\omega+1}$ and then every infinite set $T\subseteq V_{\omega+1}$ is equipotent with $T\cup V_\omega$, which is transitive. Equally, use $\mathcal P(\omega)$ and take $T\cup\omega$.

But, remember that we can have sets of reals which are very eccentric, in the technical sense, while still being Dedekind-infinite. So there is no reason to expect that the above argument can be used for anything significantly stronger.

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Here's a nice thing to ponder about while walking in a park.

Assume $\sf KWP_1$ and for a set $x$ let $\kappa_x$ be the least such that $x$ maps into $\mathcal P(\kappa_x)$. Assume the following holds: $\kappa_x<\aleph(x)$ for all $x$. Then $\sf TC$ holds.

To see that, simply replace $x$ with a copy in $\mathcal P(\kappa_x)$, and since $\kappa_x+x=x$, as it is Dedekind-infinite, we can assume that said copy also contains $\kappa_x$ itself.

So, firstly, this seems to be flexible enough to extend to $\sf KWP_\alpha$, with some obvious caveats that we need to have more conditions on how the copy behaves with respect to adding subsets of a lower Kinna–Wagner rank. This might end up a bit too messy.

Secondly, is this consistent without choice at all? It seems likely, presumably if we violate choice on the reals in a way that preserves $\sf KWP_1$, this should work out. But I don't have a proof off the top of my head.

Answer 3

After some thought the version of Löwenheim-Skolem used below appears to imply AC. So this upper bound is trivial.

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An upper bound is the downward Löwenheim-Skolem Theorem plus every infinite set is Dedekind infinite.

Given an infinite set $X$. Let $\mathcal{L}_X$ be the language containing $=$, $\in$, and one constant symbol for each element of $X$. Let $M$ be the transitive closure of $X$ viewed as a $\mathcal{L}_X$ model where each constant is interpreted by the corresponding element of $X$. Since $|X| + \omega = |X|$, the downward Löwenheim-Skolem Theorem says that $M$ has an elementary submodel $N$ with $|N| \leq |X|$. We must actually have $|N| = |X|$ since the constants are distinct. Since $N \vDash$ extensionality, $N$ is isomorphic to its Mostowski collapse $T$. This $T$ is a transitive set with $|T| = |X|$.

Answer 4

I had previously asked this question in relation to a definition of cardinality that I gave then as: The cardinality of a set $x$ is the set of all sets equinumerous with $x$ having all elements in their transitive closures being strictly subnumerous to $x$.

$\sf ZF$ proves that for every set $x$ there is a set $\overset x \nabla$ of all sets strictly subnumerous to $x$ having every element in their transitive closures being strictly subnumerous to $x$. But, it doesn't prove the existence of an injection from $x$ to $\overset x \nabla$ . But, with your $\sf TC$ principle clearly this works. But, I'm not sure if it is equivalent with it.