Can such a Dedekind cardinal exist? Motivated by idle curiosity and this question about characterizing countable sets I ask:

Is it consistent with ZF that there is an uncountable set $S$ such that, for every infinite set $X\subseteq S,$ there is a surjection from $X$ to $S$?

My thoughts: If $\kappa$ is the cardinality of such a set $S,$ then $\kappa$ is Dedekind-finite, $2^\kappa$ is Dedekind-infinite, and $2^\kappa=2^\lambda$ for every infinite cardinal $\lambda\le\kappa.$
 A: Yes, this holds in the second Cohen model. This model has a certain sequence of sets $P_0,P_1,\dots$ of order two. The headline property of this model is that there is no choice function for this sequence, illustrating Bertrand Russell’s famous boots and socks analogy:

“Among boots we can distinguish right and left, and therefore we can make a selection of one out of each pair, namely, we can choose all the right boots or all the left boots; but with socks no such principle of selection suggests itself, and we cannot be sure, unless we assume the multiplicative axiom, that there is any class consisting of one sock out of each pair.”

We'll need a slightly stronger property. For each $n\geq 0$ let $S_n$ denote the set of choice functions for $P$ restricted to $n$ i.e. the Cartesian product $\prod_{i=0}^{n-1} P_i$ where by convention, the empty product is the singleton set $\{\emptyset\}.$ Let $S=\bigcup_{n=0}^\infty S_n.$
Let $\pi_n$ denote the symmetry of $S$ that acts by transposing the two elements of $P_n$ - so it flips the value of $x_n$ of each element $(x_0,\dots,x_n,\dots,x_{m-1})\in S_m$ for $m>n,$ and acts trivially on $S_m$ for $m\leq n.$ Note that $S$ is infinite, even in ZF. The property I need is:

(*) for any subset $X\subseteq S$ there exists $k$ such that $X$ is invariant under $\pi_n$ for all $n\geq k$

A model of ZFA (a weaker theory than ZF) where this holds is “The second Fraenkel model” as described in Jech’s book Set Theory, 3rd Millenium ed, Example 15.50. To get a model of ZF, use Theorem 15.53 (Jech-Sochor). “The second Cohen model” is basically the same thing but specifically implements the elements of $P_n$ as sets of reals.
Let $X\subseteq S$ be infinite. We will show that there is a surjection from $X$ to $S.$
Let $k$ be the integer given by (*). Since $X$ is infinite, it intersects infinitely many of the sets $S_n.$ Enumerating these indices gives a strictly ascending sequence $n(0)<n(1)<\dots$ such that $X\cap S_{n_j}\neq\emptyset$ for each $j.$
Let $a_0,\dots,a_{2^k-1}$ be an enumeration of $S_k.$
Let $S_{i,j}$ denote the set of elements of $S_i$ whose first $\min(i,k)$ elements agree with $a_j.$
For all $i\geq 0$ and all $0\leq j<2^k$ define $f_{i,j}$ to be the function from $X\cap S_{n(k+i2^k+j)}$ to $S_{i,j}$ defined by the following procedure:

*

*Replace the first $k$ elements by $a_j\in S_k.$

*Truncate to the first $i$ components.

Step 2 is possible because $i\leq i2^k+j\leq n(k+i2^k+j).$ This $f_{i,j}$ is surjective because $\pi_k,\dots,\pi_{i-1}$ generate a transitive group action on $S_{i,j}.$ So the union of all $f_{i,j}$ is a surjection $f$ from a subset of $X$ to $S.$ Of course we can extend the domain to $X$ by sending elements of $X\setminus\mathrm{dom}(f)$ to any fixed element of $S.$
