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

The Lévy-Solovay theorem says that small forcings do not create measures. J.D. Hamkins has generalized this to a larger class of forcings called gap forcings. I would assume this cannot be generalized to all forcings, but I cannot think of a counterexample. Is there a forcing notion that creates a $\kappa$-complete (or even countably complete) measure $\mu$ on some uncountable cardinal $\kappa$ such that $\mu \cap V$ is not in $V$?

share|improve this question
1  
Trevor, I think you might want to change the conclusion of your question to "$\kappa$ is not already measurable in $V$." Otherwise, I think you could use a sufficiently generic permutation of $\kappa$ (or a similar device) to twist an existing measure into one such that $\mu \cap V \notin V$. Of course, this kind of answer completely misses the point. Another variant of your question could ask for a new measure on $\kappa$, one that is not a "trivial variation" of a measure on $\kappa$ that comes from the ground model, but I don't think that's what you want either. –  François G. Dorais Oct 28 '12 at 10:57

2 Answers 2

up vote 12 down vote accepted

Your title question is asking whether the measurability of a measurable cardinal is downwards absolute to ground models: if $\kappa$ is measurable in a forcing extension $V[G]$, must it be measurable in $V$? This is a question that makes sense for any of the large cardinal notions.

The answer is that, although the smaller large cardinal notions such as inaccessibility and Mahlo-ness are downward absolute, this phenomenon does not generally hold for the much larger large cardinals. Specifically, a non-measurable cardinal $\kappa$ can become measurable after forcing with $\text{Add}(\kappa,1)$.

Here is one way to see it. Suppose $\kappa$ is a measurable cardinal in $V$. We may force $2^\kappa=\kappa^+$ while preserving the measurability of $\kappa$, since this adds no new subsets to $\kappa$; so suppose $2^\kappa=\kappa^+$ already in $V$. Let $\mathbb{P}$ be the Easton support forcing iteration of length $\kappa$, which forces with $\mathbb{Q}_\gamma=\text{Add}(\gamma,1)$ to add a Cohen subset at every inaccessible cardinal $\gamma$. Let $\mathbb{Q}_\kappa=\text{Add}(\kappa,1)$ be the stage $\kappa$ forcing, to do so at the top. Suppose $G\ast g\subset\mathbb{P}\ast\mathbb{Q}_\kappa$ is $V$-generic. First, I claim that $\kappa$ is measurable in $V[G][g]$. This follows from the usual lifting arguments, which appear in many of my papers. Start with $j:V\to M$, an ultrapower embedding by a measure on $\kappa$. The forcing $j(\mathbb{P})$ is isomorphic to $\mathbb{P}\ast\mathbb{Q}\ast\mathbb{P}_{tail}$, where the tail forcing is $\leq\kappa$-closed in $M[G][g]$. Since $|j(\kappa^+)|^V=\kappa^+$, we may enumerate the dense subsets of $\kappa$ in $M[G][g]$ in a $\kappa^+$ sequence in $V[G][g]$. And since $M[G][g]^\kappa\subset M[G][g]$ in $V[G][g]$, we may thereby diagonalize to produce an $M[G][g]$-generic filter $G_{tail}\subset\mathbb{P}_{tail}$, and thus lift $j$ to $j:V[G]\to M[j(G)]$, where $j(G)=G\ast g\ast G_{tail}$. Similarly, the object $g$ is essentially a condition in $j(\mathbb{Q}_\kappa)$, and so we may diagonalize again to produce an $M[j(G)]$-generic filter $h\subset j(\mathbb{Q}_\kappa)$ containing it, and thus lift $j$ fully to $j:V[G][g]\to M[j(G)][j(g)]$, with $j(g)=h$. Thus, $\kappa$ is measurable in $V[G][g]$.

Meanwhile, and this is the main point, I claim that $\kappa$ is not measurable in $V[G]$. Suppose it were, with embedding $j:V[G]\to \bar M$ in $V[G]$. By elementarity, since $V[G]$ thinks it is a forcing extension by $\mathbb{P}$, it follows that $\bar M=M[j(G)]$ for some inner model $M$. (My theorems on approximation and covering show that in fact $M\subset V$ and $j\upharpoonright V:V\to M$ is definable in $V$, but we don't need that here.) We may factor the forcing as $M[j(G)]=M[G\ast \bar g\ast\bar G_{\rm tail}]$, where $\bar g\subset\text{Add}(\kappa,1)^{M[G]}$ is $M[G]$-generic. But note that $P(\kappa)^V\subset M$ and so also $P(\kappa)^{V[G]}\subset M[G]$. Thus, $\bar g$ will have to be really $V[G]$-generic for this forcing, which is a contradiction if $\bar g\in V[G]$. So there can be no such embedding in $V[G]$, and so $\kappa$ is not measurable there. This kind of argument is used many times in my paper Destruction or preservation as you like it.

Thus, we have a model $\bar V$, namely $\bar V=V[G]$, where $\kappa$ is not measurable, but it becomes measurable after forcing with $\text{Add}(\kappa,1)$.

The same essential argument works with all the stronger large cardinals. If we had started with $\kappa$ supercompact, for example, then we could make a model where it is not measurable, but becomes supercompact after forcing with $\text{Add}(\kappa,1)$.

Kunen observed that one can make a more extreme example as follows (Saturated ideals, Journal of Symbolic Logic, 43(1):65--76, March 1978). He relies on the fact that if one first adds a homogeneous Souslin $\kappa$-tree, and then forces with that tree, the combined forcing is equivalent to $\text{Add}(\kappa,1)$. But the first step kills the weak compactness of $\kappa$. So combining this observation with the previous, one arrives at the conclusion:

If $\kappa$ is a measurable cardinal, then there is a forcing extension $\bar V$ in which $\kappa$ is no longer weakly compact, but forcing with a $\kappa$-Souslin tree in $\bar V$ makes $\kappa$ suddenly measurable in the forcing extension $\bar V[g]$. Furthermore, if $\kappa$ was tall, strong or supercompact in th original ground model, then it will also retain those stronger properties in $\bar V[g]$.

So a non-weakly compact cardinal can become supercompact by forcing.

Addition. Consider forcing $\mathbb{P}$ which is an Easton support iteration of length $\kappa$, which at inaccessible $\gamma$ forces with the lottery sum of either doing nothing, or using $\text{Add}(\gamma,1)$. The argument above shows that if $\kappa$ is measurable and $G$ is $V$-generic for $\mathbb{P}$, then $\kappa$ is measurable in $V[G]$. We don't need the stage $\kappa$ forcing, now, since we may opt for trivial forcing in the stage $\kappa$ lottery. But also, if we force to add $V[G]$-generic $g\subset\kappa$, then $\kappa$ remains measurable in $V[G][g]$, since we may instead opt for the nontrivial forcing at stage $\kappa$. But the key argument above show that every normal measure in $V[G]$ must concentrate on $\gamma$ for which we did trivial forcing, and every normal measure $\mu$ on $\kappa$ in $V[G][g]$ concentrates on the $\gamma$ for which we did nontrivial forcing. It follows that such a $\mu$ in $V[G][g]$ must have $\mu\notin V[G]$. So we can preserve measurability, while preventing normal measures from extending ground model measures.

share|improve this answer
    
It is obvious that any notion incompatible with $L$ is not downwards absolute. No? –  Asaf Karagila Oct 28 '12 at 12:16
    
Asaf, that would be for downwards absoluteness generally, rather than downwards absoluteness to ground models. –  Joel David Hamkins Oct 28 '12 at 12:21
    
Of course. I blame my hasty reading on the faulty LaTeX :-) –  Asaf Karagila Oct 28 '12 at 12:28
    
By the way, François's variations of the question are also very interesting, and I can explain further examples of that, e.g. where a measurable cardinal is preserved, but no new measure lifts an old measure, etc. –  Joel David Hamkins Oct 28 '12 at 12:35
    
Thanks, Andreas! –  Joel David Hamkins Oct 28 '12 at 12:49

Just some comments to complement Joel's answer:

That forcing can destroy and then recreate measurability is due to Kunen:

Kenneth Kunen. Saturated ideals, The Journal of Symbolic Logic 43 (1) (1978), 65–76. MR0495118 (80a:03068)

The same is true for real valued measurability, this is due to Gitik:

Moti Gitik, Saharon Shelah. More on Real-valued measurable cardinals and forcing with ideals, Israel Journal of Mathematics 124 (2001), 221–242 ([GiSh 582]). MR1856516 (2002g:03110)

For a while it was open whether we can destroy and then reconstruct measurability while still preserving weak compactness. In all arguments I was aware of, it is essential that weak compactness fails in the intermediate model, as what we accomplish is a (destructible) instance of failure of stationary set reflection. Joel points out that his argument actually preserves weak compactness, and can be carried out in other settings to preserve measurability or stronger properties.

More generally, one can ask whether measurability can "appear spontaneously" by forcing, through some other means, just as we obtain saturation on the non-stationary ideal on $\omega_1$ by forcing MM, which cannot be traced back to some measure in some inner model that the forcing is reconstructing.

For another proof of the specific result you are asking (but one that destroys weak compactness in the intermediate extension), see section 4 of my paper on RVM cardinals:

Real-valued measurable cardinals and well-orderings of the reals. In Set theory. Centre de Recerca Matemàtica, Barcelona, 2003–2004, Joan Bagaria, and Stevo Todorcevic, eds.; Trends in Mathematics, Birkhäuser Verlag, Basel, 2006, pp 83–120, MR2267147 (2007g:03064)

share|improve this answer
    
Andres, I don't think the big open question you mention is open, because in the argument I give in my answer, $\kappa$ is weakly compact in $V[G]$. This can be shown by a weak compactness lifting argument, working with weak compactness embeddings. The obstacle concerning $\bar g$ does not arise with weak compactness embeddings, since the model has only $\kappa$ many dense sets, instead of all of them. –  Joel David Hamkins Oct 28 '12 at 15:21
    
And higher up, the main theorems in my "as you like it" paper show that one can make cardinals $\kappa$ that, for example, lose their $\kappa^+$-supercompactness while keeping measurability, but then can gain their supercompactness again. And the same for $\kappa^{++}$ and so on higher up. –  Joel David Hamkins Oct 28 '12 at 15:26
    
Ah, I see. This is nice! It is a different way of destroying measurability than by killing stationary reflection. –  Andres Caicedo Oct 28 '12 at 15:48

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.