# Dedekind-“finiteness” for arbitrary limit cardinals

In $$\mathbf{ZF}$$, it is possible for a set $$A$$ to be infinite but not to admit a countable set. In other words, for any $$\alpha\in\omega$$, there is an injection from $$\alpha$$ into $$A$$, but there is no injection from $$\omega$$ into $$A$$. If we replace $$\omega$$ by a successor cardinal $$\kappa^+$$ in the above statement, any $$A$$ of cardinality $$\kappa$$ works, so there exists an example in $$\mathbf{ZFC}$$. However, for limit cardinals, the question seems unclear to me.

Therefore: If $$\kappa>\omega$$ is a limit cardinal and $$A$$ is a set such that for any $$\alpha\in\kappa$$, there is an injection from $$\alpha$$ into $$A$$, does it follow in $$\mathbf{ZF}$$ that there is an injection from $$\kappa$$ into $$A$$? Additionally, does the existence of such an $$A$$ imply the existence of a Dedekind-finite set?

• Equivalently, the question is whether limit cardinals can occur as Hartogs number. I suspect the answer is yes, and that it need not imply existence of Dedekind-finite sets, but have failed to find any references. – Wojowu Jun 10 at 15:07
• It's consistent with ZF that $\aleph(\mathbb{R})=\aleph_{\omega}.$ Ref: "A model of Z-F set theory with $\aleph(2^{\omega}) = \aleph_{\omega}$" by Derrick and Drake. – Elliot Glazer Jun 10 at 16:05

## 3 Answers

Start with your favourite model of $$\sf ZFC$$, your favourite regular cardinal $$\mu$$, and your favourite limit cardinal $$\lambda>2^\mu$$.

Now consider the $${<}\mu$$-support product $$\prod_{\alpha<\lambda}\operatorname{Add}(\mu,\alpha)$$. With automorphism groups that act on each individual component in the product, and with a filter of subgroups generated by fixing a bounded1 number of component pointwise, we get a symmetric system. It is easy to see that in the symmetric extension $$2^\mu$$ has an injection from any $$\alpha<\mu$$, since that is given by the $$\alpha$$th component.

But if there is an injection from $$\lambda$$ into $$2^\mu$$, then this injection codes a set of ordinals, and by a standard homogeneity argument we can show that any set of ordinals added to the model must have been added by a bounded part of the product, and so that is impossible.

Now, since this forcing is $$\mu$$-closed, has $$\mu^+$$-cc and the filter of groups has $$\operatorname{cf}(\lambda)$$-completeness we instantly get $$\sf DC_{<\operatorname{cf}(\lambda)}$$ to hold in the extension as well.

Picking $$\mu=\omega$$ and $$\lambda=\beth_{\omega_1}$$, for example, provides us with $$\sf DC$$, and with $$\aleph(\Bbb R)=\beth_{\omega_1}$$.

1. We can actually replace bounded by any other support, as long as it does not contain any cofinal subsets. This will affect how much $$\sf DC$$ holds in your model, though.
• This is more or less the same as the model described by Farmer, if the parameters are chosen correctly. – Asaf Karagila Jun 10 at 18:58

Yes, it is possible; one can see this by a variant of a standard proof of the consistency of ZF + $$\neg$$AC, as witnessed in symmetric submodels of forcing extensions. Start with universe $$V=L$$. Let $$\lambda$$ be a limit cardinal. Force, adding a $$\lambda$$-sequence $$G$$ of Cohen reals with the finite support product. Then $$L$$ and $$L[G]$$ have the same cardinals. Let $$H^*$$ be the "symmetric $$\mathcal{H}_\lambda$$ of the extension"; that is, $$H^*=\bigcup_{\alpha<\lambda}\mathcal{H}_\lambda^{L[G\upharpoonright\alpha]},$$ where $$\mathcal{H}_\gamma$$ denotes the set of all sets hereditarily of size $$<\gamma$$. Let $$M=L(H^*)$$ (or one could use $$M=\mathrm{HOD}_{H^*\cup\{H^*\}}^{L[G]}$$). Then $$M\models\mathrm{ZF}$$, $$L\subseteq M\subseteq L[G]$$ (so $$L,M,L[G]$$ have the same ordinal cardinals, so $$\lambda$$ is a limit cardinal in $$M$$), $$\mathcal{H}_\lambda^M=H^*$$ (so for each $$\alpha<\lambda$$ there is an $$\alpha$$-sequence of distinct reals in $$M$$), but there is no $$\lambda$$-sequence of distinct reals in $$M$$. (These facts can be proven much like in the proofs of failure of AC mentioned above, using the homogeneity of the forcing etc.) So the set $$A$$ of reals of $$M$$ witnesses the property.

• The way you explain this model makes me sad. – Asaf Karagila Jun 10 at 17:21

Yes, it is possible. Below I will prove two specific results in this direction: 1. If there is an infinite Dedekind-finite set, then there is such a set $$A$$ for $$\kappa=\aleph_\omega$$, 2. It is consistent with ZF (EDIT: at least assuming consistency of an inaccessible) that there is such a set $$A$$ for $$\kappa=\aleph_{\omega_1}$$, but there are no infinite Dedekind finite sets.

Both rely on the following general result: suppose there is a regular cardinal $$\lambda$$ and a set $$B$$ such that $$B$$ surjects onto $$\lambda$$, but $$\lambda$$ doesn't inject into $$B$$. Then for any cardinal $$\kappa$$ of cofinality $$\lambda$$ there is a set $$A$$ such that all cardinals $$<\kappa$$ embed into $$A$$, but $$\kappa$$ doesn't.

Indeed, let $$f:B\to\lambda$$ be a surjection. Let $$\kappa_\alpha,\alpha<\lambda$$ be an increasing sequence converging to $$\kappa$$. Let $$A$$ be the union of sets $$\{b\}\times\kappa_{f(b)}$$ for all $$b\in B$$. Clearly all $$\kappa_\alpha$$, and hence all cardinals below $$\kappa$$, embed into $$A$$. On the other hand, suppose we had an injection $$\kappa\to A$$. Composing with the obvious projection $$A\to B$$ gives us a well-orderable subset of $$B$$, which by assumption has size smaller than $$\lambda$$. But then we see the image of $$\kappa$$ is a union of less than $$\lambda$$ sets of size strictly smaller than $$\kappa$$ (since $$\{b\}\times\kappa_{f(b)}$$ are all strictly smaller than $$\kappa$$), contradicting the assumption that $$\kappa$$ has cofinality $$\lambda$$. This concludes the proof.

Now let me justify the claims from the start. If there is an infinite Dedekind-finite set $$D$$, then the conditions are satisfied for $$\lambda=\omega$$ if we take $$B$$ to be the set of all finite sequences of elements of $$D$$. It is not hard to see $$B$$ is infinite and Dedekind-finite if $$D$$ is, and $$f:B\to\omega$$ taking any sequence to its length is surjective. Therefore we may take $$\kappa=\aleph_\omega$$.

For an example not using Dedekind-finite sets, we recall that, assuming consistency of an inaccessible, it is consistent with ZF that DC holds and $$\omega_1$$ doesn't embed inside $$\mathbb R$$ (for instance, models of ZF+DC+"all sets of reals are Lebesgue measurable" satisfy this). DC (or just countable choice) implies there are no infinite Dedekind-finite sets and that $$\omega_1$$ is a regular cardinal. Since $$\mathbb R$$ surjects onto $$\omega_1$$ (interpret a real as a binary sequence coding some relation on $$\omega$$; if it codes well-order map it to its length, otherwise map to $$0$$), we can apply the above construction with $$\lambda=\omega_1,B=\mathbb R$$ and $$\kappa=\aleph_{\omega_1}$$.

Some interesting questions still remain, for instance: is it possible that the above holds for some $$\kappa$$ of countable cofinality, but there is no infinite Dedekind-finite sets? Is it possible for this to hold for all limit cardinals $$\kappa>\omega$$ (with or without an infinite Dedekind-finite set)?

• ZF + DC + "$\omega_1$ doesn't inject into $\mathbb{R}$" has consistency strength an inaccessible since it implies $\omega_1$ is inaccessible in $L.$ In particular, it's not a theorem of "all sets of reals have property of Baire." – Elliot Glazer Jun 10 at 17:51
• @ElliotGlazer You seem to be right, I misremember some results due to Rasonnier. I thought an uncountable well-orderable subset of $\mathbb R$ implies existence of a set without property of Baire, but it only seems to imply existence of a nonmeasurable set. I will edit accordingly. – Wojowu Jun 10 at 18:02