In $\mathsf{ZF}$, for each set $A$ exists an ordinal $\alpha$ such that no surjection $A\to \alpha$ exists

Question

Some observations.

• The statement can easily be proven in $\mathsf{ZFC}$, but we do not assume that $\mathsf{AC}$ holds.

• If one proves that $S=\{\beta \mid \text{there is a surjection $A\to \beta$}\}$ is a set, the statement follows by taking $\alpha=\sup S$.

• An obvious candidate would be $\alpha=h(A)$, the Hartogs number of $A$.

All my attempts to prove the statement required some form of $\mathsf{AC}$. I'm starting to wonder if the statement can be proven in $\mathsf{ZF}$.

Answer 1

You can really just repeat the proof of Hartogs theorem, but use surjections instead. An easy proof is simply consider the fact that if there exists a surjection $f\colon X\to Y$, then there is an injection from $Y$ into $\mathcal P(X)$. So the Hartogs number of $\mathcal P(X)$ is an upper bound of this ordinal you're looking for.

Interestingly enough, it is not necessarily the Hartogs number, and it may not be equal to the Hartogs of the power set either.

(And one last piece of trivia, the name of the least non-zero ordinal that $X$ does not map onto is the Lindenbaum number.)