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?

  • $\begingroup$ While this question would be on topic on MSE, I disagree with its closing here. I think this question is on topic here. $\endgroup$
    – Asaf Karagila
    Jul 20, 2013 at 23:36
  • $\begingroup$ 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. $\endgroup$ Jul 20, 2013 at 23:45
  • 2
    $\begingroup$ 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. $\endgroup$ Jul 21, 2013 at 4:13

2 Answers 2


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 $V[G]$.

  • $\begingroup$ A nice reference to add here is Joel's PhD thesis, Lifting and extending measures by forcing; fragile measurability, UC Berkeley, 1994. $\endgroup$ Jul 21, 2013 at 0:22
  • $\begingroup$ Thanks for the answer. @Andres Caicedo: Is there a link for the thesis? In Dr. Hamkin's website, the text is not available. $\endgroup$ Jul 21, 2013 at 0:46
  • $\begingroup$ You're welcome, St. D. And thanks for the reference, Andres! $\endgroup$ Jul 21, 2013 at 0:48
  • $\begingroup$ @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. $\endgroup$ Jul 21, 2013 at 0:48
  • $\begingroup$ 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. $\endgroup$ Jul 21, 2013 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


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