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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?

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    $\begingroup$ 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. $\endgroup$ – Wojowu Jun 10 at 15:07
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    $\begingroup$ 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. $\endgroup$ – Elliot Glazer Jun 10 at 16:05
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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.
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  • $\begingroup$ This is more or less the same as the model described by Farmer, if the parameters are chosen correctly. $\endgroup$ – Asaf Karagila Jun 10 at 18:58
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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.

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    $\begingroup$ The way you explain this model makes me sad. $\endgroup$ – Asaf Karagila Jun 10 at 17:21
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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)?

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    $\begingroup$ 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." $\endgroup$ – Elliot Glazer Jun 10 at 17:51
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    $\begingroup$ @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. $\endgroup$ – Wojowu Jun 10 at 18:02

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