Is there a uniform way to define an upper bound on $|X|$ union of $|X|$-sized sets in ZF?

Question

Working in ZF (without choice); Let $A \times B$ denote, as usual, the cartesian product of sets $A$ and $B$. Let $|X|$ denote the cardinality of $X$ defined after Scott's. Lets define: $$|X|^2= | X \times X |$$

Does ZF prove that any set union of $|X|$ many sets each of size $|X|$, is subnumerous to $|X|^2$?

If that fails then is there a definable function on $X$ such that every union of $|X|$ many |X|-sized sets is subnumerous to? Could that be for example $|\mathcal P(X)|^2$?

Answer 1

No, not even remotely, not even ever.

$\sf ZF$ does not prove that a countable union of countable sets is countable, so for $X=\omega$ your question is already proved false. Moreover, the results of Morris and many others show that $\sf ZF$ cannot even place a bound on cardinalities of countable unions of countable sets.

In other words, it is consistent with $\sf ZF$ that for every set $X$, there is a set $Y$ which is a countable union of countable sets and $|Y|\nleq|X|$. Specifically, Morris proved that it is consistent:

For every ordinal $\alpha$, there is a set $A_\alpha$ which is a countable union of countable sets, but $\mathcal P(A_\alpha)$ can be mapped onto $\omega_\alpha$.

Recently, I've extended said result to replace $\omega_\alpha$ by $V_\alpha$.

This style of failure of choice can be extended in many different ways, all of which are very terrible.