The following was asked on stackexchange but I think it also belongs here:

https://math.stackexchange.com/questions/1513446/transitive-models-and-ch

Suppose $M, N$ are two countable transitive models of ZFC which have same ordinals, cofinalities and reals (but not necessarily same sets of reals!). Suppose $M$ models the continuum hypothesis. Can we conclude that $N$ also models continuum hypothesis? I can see that if this fails, then neither one of $M, N$ is included in the other. But what if $M, N$ are incomparable.

  • 2
    CW by request of Ashutosh, because he was only relaying the question. – Todd Trimble Nov 5 '15 at 2:13
  • I should clarify that Monroe had an entire machinery (anonymous collapse and he spoke about it in GSCL 2014) to get much stronger results around April 2014. – Ashutosh Nov 5 '15 at 3:33
  • I actually answered the question on Math.SE. – Asaf Karagila Nov 5 '15 at 7:25
  • (There are other ways to relay questions from MSE to MO, by the way. For example: meta.mathoverflow.net/questions/1967/…) – Asaf Karagila Nov 5 '15 at 7:28
  • @Asaf, if I am not mistaken, the question was answered here before your answer there, although the link to Monroe's paper over on MSE by hot_queen came even earlier. – Joel David Hamkins Nov 5 '15 at 11:54
up vote 12 down vote accepted

This strange situation can indeed happen.

Start with a countable transitive model of set theory $W$, satisfying CH, and let $G$ be $W$-generic for the forcing $\newcommand\Add{\text{Add}}\Add(\omega,\omega_1)^W$. So the model $M=W[G]$ continues to satisfy CH. Now, let $g$ be $W[G]$-generic for the collapse of $\omega_2^W$ to $\omega_1^W$, using the collapse forcing as it is defined in $W$, and temporarily consider $W[G][g]$. Notice that the collapse forcing was countably closed in $W$, and so it remains at least countably distributive in $W[G]$. So we have added no new countable sequences of ordinals in going from $W[G]$ to $W[G][g]$. Inside $W[G][g]$, the ordinals $\omega_1^W$ and $\omega_2^W$ are now in bijection, and so there is an isomorphism $\pi:\Add(\omega,\omega_1)^W\cong \Add(\omega,\omega_2)^W$ between these two forcing notions as posets. Furthermore, since by design the bijection of the ordinals has the property that every countable piece of it is in $W$, the same property is true for this isomorphism $\pi$. Let $H=\pi[G]\subset\Add(\omega,\omega_2)^W$ be the isomorphic copy of $G$ induced by the isomorphism $\pi$. I claim that $H$ is $W$-generic for $\Add(\omega,\omega_2)^W$, because if $A$ is any maximal antichain in this forcing in $W$, then by the c.c.c. it follows that $A$ is countable, and so $\pi^{-1}A\subset\Add(\omega,\omega_1)^W$ is a maximal antichain in $W$ for the first forcing, since this much of $\pi$ is in $W$. And so, since $G$ must meet $\pi^{-1}A$, it follows that $H$ meets $A$; so $H$ is $W$-generic, and we may let $N=W[H]$. So $N$ is a model of $\neg\text{CH}$, since we've added $\omega_2$ many Cohen reals. Both $M$ and $N$ are c.c.c. extensions of $W$, and so they have the same cardinals and cofinalities.

Let's now argue that they have the same reals. First off, every real of $N=W[H]$ is certainly in $W[G][g]$, where $H$ is constructed, and the reals of $W[G][g]$ are the same as the reals of $W[G]=M$, since the $g$ forcing is countably distributive. So every reals of $N$ is in $M$. Conversely, if $x$ is a real in $M=W[G]$, then $x$ is in $W[G\upharpoonright\alpha]$ for some $\alpha<\omega_1$. And since $\pi\upharpoonright\Add(\omega,\alpha)$ is in $W$, it follows that $x$ is in $W[H\upharpoonright\pi[\alpha]]$, which is contained in $W[H]=N$. So they have the same reals.

In conclusion, $M$ and $N$ have the same reals, the same cardinals, the same cofinalities, but one has CH and one does not, as desired.

One may cast the argument in terms of forcing over $V$; there is no need to go to countable transitive models. Namely, start in $V$, with CH, and then force to add $\omega_1$ many Cohen reals to form $V[G]$. Now collapse $\omega_2$ to $\omega_1$ using the ground model collapse, and in $V[G][g]$ define $H$ as the copy of $G$ induced by that isomorphism. It now follows by the argument above that $V[G]$ and $V[H]$ have the same reals, the same cardinals and cofinalities, but one has CH and the other does not.

And of course, there is nothing special about $\omega_2$ in these arguments, we could have used $\omega_3$ or any other regular cardinal in $W$ just as easily.

  • 1
    The situation is possible. See my answer on math.SE – Asaf Karagila Nov 5 '15 at 7:25
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    But haven't I also given a proof here that it happens? I find my argument to be more direct, since it doesn't use symmetric models. There is no need to consider $L(\mathbb{R})$ explicitly, although of course the models will have the same $L(\mathbb{R})$ since they have the same reals. – Joel David Hamkins Nov 5 '15 at 11:44
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    Well, sure. Your argument, however, can be reduced to passing through $L(\Bbb R)$, which can be clearer and easier to see why these things work nicely (after forcing over $L(\Bbb R)$ to have say, $2^{\aleph_0}=\aleph_2$, you didn't add any new reals, so every real is in $L$ or Cohen over $L$, so this is again a Cohen extension). – Asaf Karagila Nov 5 '15 at 11:54
  • I suppose one can view that as a reduction, but I'm inclined to say that the symmetric model argument can be reduced to a direct forcing argument. – Joel David Hamkins Nov 5 '15 at 12:01

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