# $M\models 2^\alpha=\alpha^+\ \forall \alpha<\kappa\longrightarrow 2^\kappa=\kappa^+$ if $M$ is a model of ZFC given by a measurable cardinal

$M\models 2^\alpha=\alpha^+\ \forall \alpha<\kappa\longrightarrow 2^\kappa=\kappa^+$ if $M$ is a model of ZFC given by a measurable cardinal

In Jech 1978, page 319 (lemma 28.13), we read:

Let $\kappa$ be a measurable cardinal. If $2^{\kappa} > \kappa^+$, then the set $\{\alpha < \kappa : 2^α > α^+\}$ has measure one for every normal measure on κ. Consequently, if $2^α = α^+$ for all cardinals $α < κ$, then $2^κ = κ^+$.

Proof. Let $D$ be a normal measure on $κ$, and let $M = Ult_D(V )$. If $2^α = α^+$ for almost all $α$, then, since $[d]_D = κ$, we have $M \models 2^κ = κ^+$.

Now why is the latter true? I proved that $[d]^+=[d^+]$, where $d$ is the identity function of $\kappa$ seen in the ultrapower and $d^+$ is the function $d(\alpha)=\alpha^+$ for all cardinals $\alpha<\kappa$ (and arbitrarily defined on the other ordinals less than $\kappa$, since they have measure $0$). But now I don't know how to use this to conclude.

Thank you in advance for any suggestions.

EDIT: Sorry for my previous question,the comouter saved it wrong.

• So, what's your question? (Also: Please improve the formatting of your question. The MathJax code in your quote is broken in many places.) – Stefan Mesken Oct 22 '17 at 16:12
• Now your quote is missing part of the argument, e.g. that $M$ and $\operatorname{Ult}_D(V)$ have the same powerset of $\kappa$. – Stefan Mesken Oct 22 '17 at 16:21
• Yes: the latter part was clear, so I omitted it from the quote: my question is about the last sentence of the actual quote. Thank you. – W. Rether Oct 22 '17 at 16:24
• The latter part depends on the argument that you now omitted from the proof. – Stefan Mesken Oct 22 '17 at 16:26
• Are you sure? I thought that the last part showed only that, assuming the statement true in $M$, then it is true in $V$. – W. Rether Oct 22 '17 at 16:33

The fact that $M \models 2^{\kappa} = \kappa^{+}$ follows immediately from Łoś's theorem. In fact $$\{ \alpha < \kappa \mid 2^{\alpha} = \alpha^+ \} = \{ \alpha < \kappa \mid \operatorname{GCH}_{d(\alpha)} \} \in D$$ and thus $$M \models \operatorname{GCH}_{[d]},$$ i.e. $$M \models 2^{\kappa} = \kappa^{+}.$$

However, we also get:

Claim. $V \models 2^{\kappa} = \kappa^{+}$.

Proof. Since $M \subseteq V$, we have that $(\kappa^{+})^{M} \ge (\kappa^{+})^{V}$ and, since $V$ and $M$ have the same powerset of $\kappa$, we get that $V \models 2^{\kappa} \le (\kappa^{+})^{M}$. It hence suffices to show that $(\kappa^{+})^{V} = (\kappa^{+})^{M}$.

Suppose otherwise. Then $(\kappa^{+})^{V} < (\kappa^{+})^{M}$ and thus, in $M$, there is a well-order $R \subseteq \kappa \times \kappa$ of order type $(\kappa^{+})^{V}$. This set exists in $V$ which is absurd. Q.E.D.

• It took you six minutes. You lied in the comments! – Asaf Karagila Oct 22 '17 at 17:45
• @AsafKaragila Yes, I'm painfully aware. Turns out that - in my draft - I considered $M$ as my model and $\operatorname{Ult}_{D}(M)$ as its ultrapower. It took a bit extra time to fix that. – Stefan Mesken Oct 22 '17 at 17:58
• C'mon, either $M$ is a set model, or an ultrapower. What next? $\beta$ for Woodin cardinals??? :P – Asaf Karagila Oct 22 '17 at 17:59
• @AsafKaragila You are the one who called $\kappa$ a Woodin cardinal! Plus, in my part of the universe, $M$ is an inner model ;-) – Stefan Mesken Oct 22 '17 at 18:05
• So... supercompact cardinals cannot be called $\kappa$ anymore? Or are you claiming that supercompact cardinals are not Woodin cardinals? ;-) – Asaf Karagila Oct 22 '17 at 19:14