In this question, I will be working in ZF.


Let $h(\kappa)$ for a cardinal $\kappa$ (not necessarily an ordinal) be the smallest ordinal $\alpha$ such that there is no surjection from a set of size $\kappa$ onto $\alpha$.


$h(\mathfrak{c})$ is defined as $\Theta$.


Let an uncountable ordinal $\alpha$ be a **high limit ordinal** if and only if for any cardinal $\kappa$ with no surjection onto $\alpha$, there is no surjection of $2^\kappa$ onto $\alpha$.


Equivalently, for any cardinal $\kappa$ such that $h(\kappa)\leq\alpha$, $h(2^\kappa)\leq\alpha$.




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Under AC, the high limit ordinals are precisely those which are strong limit cardinals.


**Questions: Under AD, what properties does the smallest high limit ordinal have? Is it possible in ZF that there exists no high limit ordinal?**



Strong limit cardinals in AD
First, let me point out that the notion of "high limit cardinal" is misleading, and perhaps not quite what you want it to be.


The reason is that strong limit cardinals are defined, in the absence of choice, as $\aleph$ numbers $\alpha$, such that for all $\beta<\alpha$, there is no surjection from $V\_\beta$ onto $\alpha$.


It is easy to see why a strong limit cardinal is a high limit cardinal. But the converse need not be true. Clearly if there is no map from $\omega$ onto $\alpha$, then there is no map from $V\_\omega$ onto $\alpha$; and then by induction we get all the way to $V\_{\omega+\omega}$. But there is no reasonable way without choosing surjections to prove that in that case there is no surjection from $V\_{\omega+\omega}$ onto $\alpha$ either.




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But regardless, $\Theta$ is not a high limit cardinal. Note that there is a surjection from $\mathcal P(\Bbb R)$ onto $\Theta$, simply by mapping $A$ to a pre-well-order it codes, or to $0$ if it doesn't code one.


Therefore it is certainly not a high limit ordinal. 


And finally, as to the question about ZF, even without choice there are arbitrarily large *strong* limit cardinals, and they are all high limit cardinals. The proof is practically the same proof as the usual one. Start with $\alpha\_0$, let $\alpha\_{n+1}$ be what you denote as $h(V\_{\alpha\_n})$, and let $\alpha=\sup\alpha\_n$. It is easy to see that $\alpha$ is a strong limit cardinal, and therefore a high limit cardinal.


