Strong limit cardinals in AD

Question

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?

Answer 1

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.

Answer 2

EDIT: My claim that $\Theta$ is a high limit ordinal is of course nonsense, as Asaf pointed out.

Meanwhile, nothing interesting happens when you drop choice: ZF certainly does prove that high limit ordinals exist. Consider the sequence $(\lambda_\alpha)_{\alpha\in Ord}$ given by

• $\lambda_0=\omega$.

• $\lambda_\eta=\sup\{\lambda_\alpha: \alpha<\eta\}$ for $\eta$ a limit.

• $\lambda_{\alpha+1}=\sup\{\beta\in Ord:$ there is a surjection from $2^{\lambda_\alpha}$ to $\beta\}$ (this is just $h(\lambda_\alpha)$, in your notation).

The existence of $\lambda_\alpha$ for each $\alpha\in Ord$ follows from Replacement, once we know that $\lambda_{\alpha+1}$ exists whenever $\lambda_\alpha$ does. This latter fact might look like it requires choice, but it is in fact provable in ZF (originally I believe by Lindenbaum) using Hartog's theorem: note that if a set $A$ surjects onto an ordinal $\delta$, then $\delta$ injects into the powerset $\mathcal{P}(\alpha)$.

Note that indeed this is exactly the proof that strong limit ordinals exist: the only place we use choice in that argument is in showing that $2^\alpha$ is an ordinal when $\alpha$ is, and that's not necessary here since your definition of $h$ already ensures that the output is an ordinal.

Then for any limit $\eta>0$, $\lambda_\eta$ is by definition a high limit ordinal. And it's easy to see that $\lambda_\omega$ is the least high limit ordinal.