I’d like to describe a certain interesting and surprising situation that can happen with models of set theory.

Theorem. If $\newcommand\ZFC{\text{ZFC}}\ZFC$ set theory is consistent, then there are two models of $\ZFC$ set theory $M$ and $N$ for which

• $M$ and $N$ have the same real numbers $$\newcommand\R{\mathbb{R}}\R^M=\R^N.$$
• $M$ and $N$ have the ordinals and the same cardinals $$\forall\alpha\qquad \aleph_\alpha^M=\aleph_\alpha^N$$
• But $M$ thinks that the continuum hypothesis $\newcommand\CH{\text{CH}}\CH$ is true, while $N$ thinks that $\CH$ is false.

This is a little strange, since the two models have the set $\R$ in common and they agree on the cardinal numbers, but $M$ thinks that $\R$ has size $\aleph_1$ and $N$ will think that $\R$ has size $\aleph_2$.  In particular, $M$ can well-order the reals in order type $\omega_1$ and $N$ can do so in order-type $\omega_2$, even though the two models have the same reals and they agree that these order types have different cardinalities.

Another abstract way to describe what is going on is that even if two models of set theory, even transitive models, agree on which ordinals are cardinals, they needn’t agree on which sets are equinumerous, for sets they have in common, even for the reals.

Let me emphasize that it is the requirement that the models have the same cardinals that makes the problem both subtle and surprising. If you drop that requirement, then the problem is an elementary exercise in forcing: start with any model $V$, and first force $\CH$ to fail in $V[H]$ by adding a lot of Cohen reals, then force to $V[G]$ by collapsing the continuum to $\aleph_1$. This second step adds no new reals and forces $\CH$, and so $V[G]$ and $V[H]$ will have the same reals, while $V[H]$ thinks $\CH$ is true and $V[G]$ thinks $\CH$ is false. The problem becomes nontrivial and interesting mainly when you insist that cardinals are not collapsed.

In fact, the situation described in the theorem can be forced over any given model of $\ZFC$.

Theorem. Every model of set theory $V\models\ZFC$ has two set-forcing extensions $V[G]$ and $V[H]$ for which

• $V[G]$ and $V[H]$ have the same real numbers $$\newcommand\R{\mathbb{R}}\R^{V[G]}=\R^{V[H]}.$$
• $V[G]$ and $V[H]$ have the same cardinals $$\forall\alpha\qquad \aleph_\alpha^{V[G]}=\aleph_\alpha^{V[H]}$$
• But $V[G]$ thinks that the continuum hypothesis $\CH$ is true, while $V[H]$ thinks that $\CH$ is false.

Proof. Start in any model $V\models\ZFC$, and by forcing if necessary, let’s assume $\CH$ holds in $V$. Let $H\subset\text{Add}(\omega,\omega_2)$ be $V$-generic for the forcing to add $\omega_2$ many Cohen reals. So $V[H]$ satisfies $\neg\CH$ and has the same ordinals and cardinals as $V$.

Next, force over $V[H]$ using the forcing from $V$ to collapse $\omega_2$ to $\omega_1$, forming the extension $V[H][g]$, where $g$ is the generic bijection between those ordinals. Since we used the forcing in $V$, which is countably closed there, it makes sense to consider $V[g]$.  In this extension, the forcing $\text{Add}(\omega,\omega_1^V)$ and $\text{Add}(\omega,\omega_2^V)$ are isomorphic. Since $H$ is $V[g]$-generic for the latter, let $G=g\mathrel{“}H$ be the image of this filter in $\text{Add}(\omega,\omega_1)$, which is therefore $V[g]$-generic for the former. So $V[g][G]=V[g][H]$. Since the forcing $\text{Add}(\omega,\omega_1)$ is c.c.c., it follows that $V[G]$ also has the same cardinals as $V$ and hence also the same as in $V[H]$.

If we now view these extensions as $V[G][g]=V[H][g]$ and note that the coutable closure of $g$ in $V$ implies that $g$ adds no new reals over either $V[G]$ or $V[H]$, it follows that $\R^{V[G]}=\R^{V[H]}$. So the two models have the same reals and the same cardinals. But $V[G]$ has $\CH$ and $V[H]$ has $\neg\CH$, in light of the forcing, and so the proof is complete. QED

Let me prove the following surprising generalization.

Theorem. If $V$ is any model of $\ZFC$ and $V[G]$ is the forcing extension obtained by adding $\kappa$ many Cohen reals, for some uncountable $\kappa$, then for any other uncountable cardinal $\lambda$, there is another forcing extension $V[H]$ where $H$ is $V$-generic for the forcing to add $\lambda$ many Cohen reals, yet $\R^{V[G]}=\R^{V[H]}$.

Proof. Start in $V[G]$, and let $g$ be $V[G]$-generic to collapse $\lambda$ to $\kappa$, using the collapse forcing of the ground model $V$. This forcing is countably closed in $V$ and therefore does not add reals over $V[G]$. In $V[g]$, the two forcing notions $\text{Add}(\omega,\kappa)$ and $\text{Add}(\omega,\lambda)$ are isomorphic. Thus, since $G$ is $V[g]$-generic for the former poset, it follows that the image $H=g\mathrel{“}G$ is $V[g]$-generic for the latter poset. So $V[H]$ is generic over $V$ for adding $\lambda$ many Cohen reals. By construction, we have $V[G][g]=V[H][g]$, and since $g$ doesn’t add reals, it follows that $\R^{V[G]}=\R^{V[H]}$, as desired. QED

I have a vague recollection of having first heard of this problem many years ago, perhaps as a graduate student, although I don’t quite recall where it was or indeed what the construction was — the argument above is my reconstruction (which I have updated and extended from my initial post). If someone could provide a reference in the comments for due credit, I’d be appreciative.  The problem appeared a few years ago on MathOverflow.