Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I would like some help with the proof of the preservation of measurable cardinals in mild extensions, as I am a bit new with forcing.

By mild extensions, I mean the generic extension produced from a forcing notion of size less than $\kappa$, where $\kappa$ is a large cardinal (in my question measurable). I am reading the proof from Levy-Solovay's article "Measurable Cardinals and the Continuum Hypothesis" (1967). The idea of the proof is that if $U$ is a non-principal, $\kappa$-complete ultrafilter on $\kappa$ (in $V$), then $W=\{X\subseteq \kappa:\exists Y\in U(Y\subseteq X)\}$ is a non-principal $\kappa$-complete ultrafilter on $\kappa$ in $V[G]$. Levy & Solovary use ideals instead of filters, which requires only slight modifications.

I am confused at the part where we try to prove that $W$ is an ultrafilter. Here is what I understand. If $X\subseteq \kappa$ in $V[G]$, let $p$ be such that $p\Vdash \dot{X}\subseteq \kappa$; we define the set of potential values of $X$ in $V$: $$T=\{\alpha<\kappa:\exists q\leq p(q\Vdash \alpha\in \dot{X})\}. $$ $T$ is a subset of $\kappa$ in $V$, so either $T\in U$ or $T\not\in U$. Now somehow we have to show that $T\in U$ implies $X\in W$ and $T\not\in U$ implies $X\not\in W$ (and this is the points where I am stuck). Can you please help me with this argument?

share|improve this question
    
While this question would be on topic on MSE, I disagree with its closing here. I think this question is on topic here. –  Asaf Karagila Jul 20 '13 at 23:36
    
I agree with Asaf. I find any question concerning the interaction of forcing and large cardinals to be a graduate-level question that is on-topic for MO. –  Joel David Hamkins Jul 20 '13 at 23:45
2  
Joel has completely answered the question, but I hope the following two remarks might be useful for St.D. or for other readers. (1) Contrary to the implicit assumption in the question, if $T\in U$, it need not follow that $X\in W$. As Joel showed, it does follow that some $q\leq p$ forces $X$ to be in $W$, but $G$ might happen not to contain such a $q$. (2) Since the notion of forcing $P$ has size $<\kappa$, you can assume, replacing it with an isomorphic copy, that it is in $V_\kappa$. That will simplify Joel's $j$ argument slightly as $j$ will fix $P$ and all its members. –  Andreas Blass Jul 21 '13 at 4:13

2 Answers 2

up vote 5 down vote accepted

If $T$ is not in $U$, then the complement of $T$ is in $U$, which means that $U$ concentrates on ordinals that $p$ forces are not in $\dot X$, which means that $p$ forces that the complement of $\dot X$ is in $W$.

If $T$ is in $U$, then let $T_q=\{\alpha\mid q\vdash\check\alpha\in \dot X\}$ be the ordinals that $q$ forces to be in $\dot X$, for $q\leq p$. Since $T=\bigcup_{q\leq p}T_q$ and this is a small union, it follows from the completeness of $U$ and $T\in U$ that some $T_q\in U$. In this case, $q$ forces $\dot X\in W$ since clearly $q$ forces $T_q\subset \dot X$.

So either $p$ forces the complement of $\dot X$ into $W$, or some $q\leq p$ forces $\dot X\in W$. So no condition can force that $\dot X$ is not decided by $W$, and so it is in ultrafilter.

By the way, a second way to prove the theorem is to start with an elementary embedding $j:V\to M$ with critical point $\kappa$ and suppose that $G\subset\mathbb{P}$ is $V$-generic. Since the forcing is small, it follows that $j(D)=j''D$ for any $D\subset\mathbb{P}$. Let $H=j''G$. It follows easily that $H=j''G\subset j(\mathbb{P})$ is $M$-generic. We may therefore lift the embedding by defining $j^\ast(\tau_G)=j(\tau)_H$. This is well-defined and elementary since $j''G\subset H$. It is easy to see that $j=j^\ast\upharpoonright V$, and so this really is a lift. In particular, $\kappa$ is still measurable in $V[G]$, because it is the critical point of the embedding $j^\ast:V[G]\to M[H]$.

Finally, one can prove that the two approaches are equivalant, in the sense that if $j$ is the ultrapower by $U$ in $V$, then $j^\ast$ is the ultrapower by $W$ in $V[G]$.

share|improve this answer
    
A nice reference to add here is Joel's PhD thesis, Lifting and extending measures by forcing; fragile measurability, UC Berkeley, 1994. –  Andres Caicedo Jul 21 '13 at 0:22
    
Thanks for the answer. @Andres Caicedo: Is there a link for the thesis? In Dr. Hamkin's website, the text is not available. –  St. D. Jul 21 '13 at 0:46
    
You're welcome, St. D. And thanks for the reference, Andres! –  Joel David Hamkins Jul 21 '13 at 0:48
    
@St.D. You may email Joel about this. Or, if he doesn't have an online copy to send you, and he is fine with this, I could email you a pdf of a copy I scanned years ago. –  Andres Caicedo Jul 21 '13 at 0:48
    
Andres---that is fine. Could you kindly email me a copy also? (sigh...) Also, St.D., if you send me an email, I can also send you another relevant reference. –  Joel David Hamkins Jul 21 '13 at 0:54

This is an instance of a more general phenomenon. Let $\kappa$ be a regular cardinal, and $I$ be a $\kappa$-complete ideal on some set $Z$. Let $\mathbb{P}$ be a forcing of size less than $\kappa$, and let $G \subseteq \mathbb{P}$ be generic over $V$. In $V[G]$, let $\bar{I} = \{ X \subseteq Z : (\exists Y \in I) X \subseteq Y \}$.

It is easy to show that in $V[G]$, $\bar{I}$ is $\kappa$-complete, and that every $\bar{I}$-positive $X$ contains a $I$-positive set $Y$ from $V$. So the boolean algebra $\mathcal{P}^V(Z)/I$ is dense in $\mathcal{P}^{V[G]}(Z)/\bar{I}$. To apply this to the present situation, note that in general $\mathcal{P}(A)/J$ is isomorphic to the two-element boolean algebra iff the dual of $J$ is an ultrafilter.

This is all subsumed by a more general fact, Foreman's Duality Theorem. See:

Foreman, Matthew. Calculating quotient algebras of generic embeddings. Israel J. Math. 193 (2013), no. 1, 309–341. http://www.ams.org/mathscinet/search/publdoc.html?pg1=MR&s1=3038554

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.