Can every ordinal be reachable from $\aleph_0$, in ZF?

Question

Define: $ x \text{ is reachable from } y \iff \\ \exists z (|z|\leq |y| \land \forall m \in z (|m| \leq |y|) \land |x|\leq|\bigcup (z)|)$

Given that ZF is consistent with only $\aleph_0$ being regular*, and so any ordinal supernumerous to $\aleph_0$ would have $\aleph_0$ as its confinality; then would this mean that ZF is consistent with saying that every ordinal is reachable from $\aleph_0$?

Answer 1

It's provable in ZF (without choice) that the ordinal $\omega_2$ is not the union of countably many countable sets. Proof: Suppose the contrary: $\omega_2=\bigcup_{n\in\omega}A_n$ with each $A_n$ countable. Without loss of generality, the $A_n$'s are disjoint (just replace each $A_n$ with $A_n-\bigcup_{k<n}A_k$). For each $n$, let $f_n$ be the unique order-isomorphism from $A_n$ to its order-type $\alpha_n$, which, being a countable ordinal, is $<\omega_1$. Then you can map $\omega_2$ one-to-one into $\omega_1\cdot\omega$ by sending each $x\in A_n$ to $(\omega_1\cdot n)+f_n(x)$. Since $\omega_1\cdot\omega<\omega_2$, this is a contradiction.

If I remember correctly, this result is due to Tom Jech.

Edit: The result is indeed due to Jech:

MR0644752 Jech, Thomas

On hereditarily countable sets.

J. Symbolic Logic 47 (1982), no. 1, 43–47

Answer 2

(My previous answer was of course nonsense as Andreas Blass pointed out.)

In natural language your definition is: "$x$ is reachable from $y$ iff $x$ can be covered by $y$-many sets each of which has $y$-many elements." Of course in the absence of choice this is not the same as admitting a surjection from $y^2$.

So your question is asking whether it is consistent that every ordinal is a countable union of countable sets. In fact, already we can show in $\mathsf{ZF}$ that $\omega_2$ is not a countable union of countable sets (although per Gitik it is consistent that it has countable cofinality). See e.g. here.