Can there be a proper class of Dedekind-finite cardinals?

Question

To formalise the question: Does $\mathsf{ZF}$ prove the existence of an ordinal $\alpha$ such that, whenever $X$ is Dedekind-finite, there is an injection $X\to V_\alpha$? (Equivalently the existence of some set $Y$ such that every Dedekind-finite set embeds into $Y$).

A set $X$ is Dedekind-finite if there is no injection $\omega\to X$.

It is consistent that such an $\alpha$ exists. The obvious case is $\mathsf{ZFC}$, or any model of "Dedekind-finite if and only if finite", as in such models every Dedekind-finite set embeds into $V_\omega$.

Similarly, the answer is yes in any model of $\mathsf{SVC}$: If there is $A$ such that, for all $X$, there is an ordinal $\eta$ and a surjection $\eta\times A\to X$, then for all $X$ there is an $\eta$ such that $X$ embeds into $\mathscr{P}(A)\times\eta$. Taking $\eta$ minimally, if $X$ is Dedekind-finite then $\eta\leq\omega$, and thus every Dedekind-finite set embeds into $\mathscr{P}(A)\times\omega$.

Answer 1

The answer to the question in the body is no (and to that in the title, yes) - in Theorem 4.5 of this paper it is shown that there may be a proper class $K$ which is Dedekind-finite (which, for our purposes, we can take to mean all of its subsets are Dedekind-finite). Any proper class surjects onto the class of ordinals (by considering $V$-ranks), so for all large enough $\beta$, $K\cap V_\beta$ maps onto the Hartogs number of $|V_\alpha|$, and thus cannot embed into $V_\alpha$.

Theorem 4.6 shows that we can have way more - it is possible for the class $K$ to itself definably map onto $V$, which implies that any set is surjected onto by some Dedekind-finite set.

Answer 2

Not only there can be a proper class of them, every set can be the image of a Dedekind-finite set.

This is embedded in the proof of the Morris model, and in https://arxiv.org/abs/1911.09285 it is made more explicit.