Transitive models and CH 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.
 A: 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. 
