Uncountable collection of cardinals, without the axiom of choice

Question

I think that it is possible to have a countable collection of sets, in $\sf ZF$, such that they all have different cardinalities, namely $\{E_n \mid n ≥ 1\}$ with $E_0 = \Bbb N, E_{n+1}=\mathcal P(E_n)$. However, I was wondering:

Is it possible to construct, in $\sf ZF$, an uncountable collection of sets such that they all have different cardinalities?

(This is possible with the axiom of choice, we just take $\{\aleph_{\alpha} \mid \alpha < \omega_1\}$). I could take the union of all the $E_n$'s, which has cardinality greater than any of the $E_n$, and apply $\mathcal P(\cdot)$ again and again, but I always stay with countably many sets, in my opinion.

Thank you for your help!

Answer 1

The axiom of choice is not needed at all: $\{\aleph_\alpha: \alpha<\omega_1\}$ provably exists in ZF alone. Recall that $\aleph_\alpha$ is defined as follows:

• $\aleph_0=\omega$,

• $\aleph_\lambda=\sup\{\aleph_\alpha: \alpha<\lambda\}$ for $\lambda$ a limit, and

• $\aleph_{\alpha+1}$ is the least ordinal $>\aleph_\alpha$ which is not in bijection with $\aleph_\alpha$.

The proof that $\aleph_\alpha$ exists for each ordinal $\alpha$ (and that $\{\aleph_\alpha: \alpha<\beta\}$ exists for each ordinal $\beta$, and that $\aleph_\alpha\equiv\aleph_\beta\iff\alpha=\beta$) doesn't use anything outside of ZF (indeed, it uses much less than ZF).

Answer 2

The definition of the aleph numbers does not hinge on the axiom of choice. Rather it hinges on Hartogs theorem and Replacement. Both follow from $\sf ZF$, of course.

We define $\aleph_0$ to be $\omega$, and at successor steps we take the least ordinal which has a strictly larger cardinality than the previous step---such ordinal exists due to Hartogs theorem and Replacement. At limits we take the limits, of course.

But you can also just define by transfinite recursion a sequence of power sets, and it's fine. You don't need choice to prove $\omega_1$ exists. (At limits, take unions, of course.)