Permutation model in which infinite sets are weakly Dedekind-infinite but not Dedekind-infinite

Question

I’m trying to create a permutation model $N$ in which every infinite set is weakly Dedekind-infinite (i.e. for every infinite set $A\in N$ there is a surjective map $f:A\rightarrow\omega$ in $N$), but the model contains an infinite $B\in N$ which is not Dedekind-infinite (i.e. there is no injective function $f:\omega\rightarrow B$ in $N$). I’ve been struggling, so if anyone can help me out that would be greatly appreciated.

Answer 1

The second Fraenkel model (in the terminology of Jech's AOC monagraph, see section 4.4) is a permutation model that has these properties.

The model is constructed as follows: We start with countably-many atoms grouped like $A = \bigcup_n \{a_n,b_n\}$, consider the group of permutations $\pi$ such that for all $n,$ either $\pi a_n= a_n$ and $\pi b_n = b_n,$ or $\pi a_n= b_n$ and $\pi b_n = a_n$, and then take the permutation model $\mathcal V$ of this group with finite supports.

As usual for finite supports, $A$ is infinite and Dedekind-finite in $\mathcal V$: For any $f:\omega \to A$ with infinite range and any finite $E\subseteq A,$ there is an $n$ such that $a_n,b_n\notin E$ and either $a_n$ or $b_n$ in the range of $f.$ Then the $\pi$ that just swaps $a_n$ and $b_n$ fixes $E$ and has $\pi f \ne f,$ and thus $f\notin \mathcal V.$

But unlike the other two permutation models Jech studies in chapter $4,$ note that $A$ is not weakly Dedekind-finite: The surjection $A\to \omega$ given by $a_n\mapsto n$ and $b_n\mapsto n$ is clearly symmetric with support $\emptyset$.

To show that every infinite set is weakly Dedekind-infinite in $\mathcal V,$ we can use the fact that $\mathcal V$ satisfies the axiom of multiple choice (i.e. for any collection $X$ of nonempty sets, there is a function $f$ on $X$ such that for all $x\in X,$ $f(x)$ is a finite subset of $x$), which is proven as part of Jech's theorem 9.2. The idea of the proof is that for any nonempty $X\in \mathcal V$ and $x\in X,$ the set $G(X,x) = \{\pi x: \pi X = X\}$ is finite, since $\pi x$ only determined by what $\pi$ does on the support of $x$, and is clearly unchanged by any $\pi$ with $\pi X= X.$ By choosing a canonical $X$ and $x$ in each orbit of the full group and defining $F(\rho X) = \rho G(X,x)$, we can see $F$ is a symmetric multiple choice function on $\mathcal V.$

So with that result, it just remains to show multiple choice implies every infinite set is weakly Dedekind-infinite. Let $X$ be infinite and for $n\in \omega,$ let $X_n = \{Y\subseteq X: |Y|=n\}.$ Let $Y_n$ be a multiple choice sequence from $X_n$ and let $Y'_n = \cup Y_n,$ so each $Y_n'$ is a finite subset of $X$ with size $\ge n.$ Then by finding the next $Y_n$ with new elements (which always will exist since the size keeps growing) and taking the new elements, we can get a sequence $Z_n$ of disjoint nonempty subsets of $X.$ Then for $x\in X,$ define $g(x)$ be the unique $n$ such that $x\in Z_n$ if $x\in \bigcup_n Z_n,$ and otherwise let $g(x) =0$, and $g$ is a surjection $X\to \omega.$

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It should be noted that this only establishes independence of the two propositions over ZFA, and the argument seems unpromising to transfer to ZF since multiple choice is equivalent to choice over ZF. (Also, according to Howard and Rubin, which may be outdated, it's unknown if every infinite set is weakly Dedekind infinite in the second Cohen model.) However, as noted in the comments, the basic Cohen model witnesses independence over ZF, and exercise 5.20 in Jech sketches the argument.