# Cardinality of power sets decides all of cardinal arithmetic?

Assuming ZFC, is it possible to have two models which agree on the cardinality of all the power sets, but disagree on the cardinality of some other cardinal exponentiation (meaning that they agree on the function $F$ such that $F(\alpha) = \beta$ iff $2^{\aleph_\alpha} = \aleph_\beta$, but in one model ${\aleph_\alpha} ^ {\aleph_\beta} = {\aleph_\gamma}$ whereas in the second model ${\aleph_\alpha} ^ {\aleph_\beta} = {\aleph_\delta}$ with $\gamma \neq \delta$)?

Put another way: If we decide (for example by forcing) the cardinality of every power set ($2^{\aleph_\alpha}$ for all $\alpha$), does it automatically decide the result of every possible cardinal exponentiation (${\aleph_\alpha} ^ {\aleph_\beta}$ for all $\alpha,\beta$)?

For example, when we assume GCH, we have an immediate formula for the cardinality of every cardinal exponentiation (see Jech 5.15), and we have no "freedom" to choose alternative values for them.

• By agree on cardinality of the power set, do you mean agree on the actual cardinality of the actual power sets, or just agree on the values of the function $F(\alpha)=\beta$ such that $2^{\aleph_\alpha}=\aleph_\beta$? (In the latter case, for example, any two models of GCH agree but this is not necessarily the situation in the first, stricter, definition.) – Asaf Karagila Dec 20 '15 at 7:24
• I meant the latter and corrected the statement of the question above. – Alon Navon Dec 20 '15 at 12:50

The answer is no assuming the existence of a model of $\mathsf{ZFC+GCH}$ having a supercompact cardinal.

First, using Silver's forcing, there is a generic extension $V[K]$ where $\kappa$ is still measurable but $2^{\kappa}=\kappa^{++}$. Then we can use Prikry's forcing to obtain a generic extension $V[K][H]$ of $V[K]$ such that all bounded subsets of $\kappa$ are in $V[K]$, all cardinals are preserved, $\kappa$ is still a strong limit and $\operatorname{cf}\kappa=\omega$. Let $G=K\ast H$.

As we assumed $V\models\mathsf{GCH}$, it's not hard to prove that in $V[G]$ we have $2^\lambda=\lambda^+$ for all $\lambda>\kappa$; since the poset yielding $V[K]$ has size $\kappa^{++}$ in $V$ and the poset giving the extension $V[K][H]$ has size $\kappa^{++}$ in $V[K]$.

Now let us work in $V[G]$. We have $$\kappa^{\aleph_0}=\kappa^{\operatorname{cf}\kappa}=2^{\kappa}=\kappa^{++},$$ and $$(\kappa^{+3})^{\omega_1}=(2^{\kappa^{++}})^{\omega_1}=2^{\kappa^{++}}=\kappa^{+3},$$ thus we can force with $Add(\omega_1,\kappa^{+3})$ to obtain a generic extension $V[G][H']$ where $2^{\omega_1}=\kappa^{+3}$, and $2^{\lambda}=\kappa^{+3}$ for all $\omega_1\leq\lambda\leq \kappa^{++}$. In $V[K]$, $\kappa$ is measurable, thus in there $\kappa=\aleph_\kappa$, so as cardinals are preserved in $V[G]$ we get that $\kappa=\aleph_\kappa$ is also true in $V[G][H']$.

Let $\beta_0$ be such that $2^{\aleph_0}=\aleph_{\beta_0}$ in $V[G]$. Then $\beta_0<\kappa$; as all bounded subsets of $\kappa$ in $V[G]$ are in $V[K]$. We also have $2^{\aleph_0}=\aleph_{\beta_0}$ in $V[G][H']$.

Thus if we consider the following function $$F(\alpha)=\begin{cases} \beta_0 & \text{if}&\alpha=0 \\\kappa+3 & \text{if}& 1\leq\alpha\leq\kappa+2\\\alpha+1 &\text{if}&\alpha\geq\kappa+3 \end{cases},$$ it follows that for all ordinals $\alpha$, $$V[G][H']\models 2^{\aleph_\alpha}=\aleph_{F(\alpha)},$$ and as the poset we used in $V[G]$ is $<\omega_1$-closed there, we get that $\kappa^{\aleph_0}=\kappa^{++}$ in this extension too.

Now, let $\mathbb P\in L$ be a poset, cardinal preserving, such that if $K'$ is $L$-generic over $\mathbb P$, we have for all ordinals $\alpha$, $$L[K']\models 2^{\aleph_\alpha}=\aleph_{F(\alpha)}.$$ In $L$, $\kappa$ is inaccesible, and thus in this model $\aleph_\kappa=\kappa$, so as $\mathbb P$ preserves cardinals we get that $\aleph_\kappa=\kappa$ in $L[K']$ too.

Let us work in $L[K']$. The singular cardinals hypothesis is true; since $0^\sharp$ does not exist, so we get that as $2^{\aleph_0}<\kappa$ and $\kappa$ is regular, $\kappa^{\aleph_0}=\kappa$.

Therefore we have that in both models $V[G][H']$ and $L[K']$, $2^{\aleph_\alpha}=\aleph_{F(\alpha)}$ for all ordinals $\alpha$, but $$V[G][H']\models \aleph_\kappa^{\aleph_0}=\aleph_{\kappa+2}\text{ and }L[K']\models \aleph_\kappa^{\aleph_0}=\aleph_\kappa.$$

Note: This argument should go through with no problem using just a measurable cardinal $\kappa$ of Mitchell order $\kappa^{++}$, working in Mitchell's model for such $\kappa$, using Gitik and Woodin's forcing. However as I'm not that familiar with this method, I used Silver's instead.

• I was planning to write such an answer, but you wrote it in far better details. So I'm glad that I wasn't able to! – Asaf Karagila Dec 20 '15 at 18:24
• @AsafKaragila, I knew I had to rush :^). – Camilo Arosemena-Serrato Dec 20 '15 at 18:26
• Probably less in the coming month or so. I've got so much work, so I will have far less time to write good answers like this. Therefore I'll try to limit them into areas where I know that I have more to say (e.g. foundational questions or AC related things), rather than questions I have less to say (e.g. large cardinals and inner models). – Asaf Karagila Dec 20 '15 at 18:27
• I'm confused why such a complicated solution is necessary. Let $M_1=L[U],$ let $\mathbb{P}$ be Prikry forcing, and $M_2=L[U][G].$ Then $M_i \models GCH$ but $M_1 \models \aleph_{\kappa}^{\aleph_0}=\aleph_{\kappa}$ while $M_2 \models \aleph_{\kappa}^{\aleph_0}=\aleph_{\kappa+1}.$ – Elliot Glazer Sep 18 '18 at 5:29
• @ElliotGlazer why don't you turn your comment into an answer? – Camilo Arosemena-Serrato Sep 18 '18 at 19:36