I heard the following fact a while back from Joel Hamkins, who told me at the time that he learned it from Hugh Woodin:

Observation: Let $M$ be a countable transitive model of $\mathsf{ZFC}$. Then there are forcing extensions $M[c]$ and $M[d]$, each by adding a Cohen real, which are non-amalgable. By this, I mean that if $N \supseteq M$ is a model of $\mathsf{ZFC}$ has the same ordinals as $M$ then $N$ cannot contain both $M[c]$ and $M[d]$.

(We don't actually need that $M$ is transitive, but let's throw it in to make things easier.)

This is not hard to prove so I'll include a sketch of the argument.

Proof sketch: Since $M$ is countable it's coded by some real $m$. We will construct $c$ and $d$ so that together they allow us to build $m$. Hence, any $N \supseteq M[c] \cup M[d]$ with the same ordinals as $M$ will see that all its ordinals are all countable, and hence cannot be a model of $\mathsf{ZFC}$.

Enumerate the countably many dense sets for Cohen forcing over $M$ as $D_0, D_1, \ldots$. We build $c$ and $d$ in $\omega$ many stages. Start with $c_0$ which meets $D_0$. Take $d_0$ to be $0$s up to the length of $c_0$ followed by a $1$ followed by $m(0)$, then extend to meet $D_0$. We can then keep going: $c_{n+1}$ extends $c_n$ by $0$s until it reaches the length of $d_n$ followed by a $1$ then extend to meet $D_{n+1}$. Next, $d_{n+1}$ extends $d_n$ with $0$s up to the length of $c_{n+1}$ followed by a $1$ followed by $m(n+1)$. At the end, $c$ and $d$ are the unions of, respectively, the $c_n$s and the $d_n$s.

They are generic because they meet every dense set. If we have both $c$ and $d$ in our model of $\mathsf{ZFC}$ then we can reconstruct $m$ by looking at the blocks of $0$s alternating between the two reals and use that to find $m(0)$, then $m(1)$, and so on. ∎

This construction relies upon being able to choose both $c$ and $d$. My question is whether we can get this sort of non-amalgability if we're only allowed to choose one of them.

Question: Let $M \models \mathsf{ZFC}$ be countable (and transitive, if we like) and let $c$ be a Cohel real generic over $M$. Is there a Cohen real $d$ generic over $M$ so that $M[c]$ and $M[d]$ are non-amalgable?

Phrased in the negative, this question is:

Question: Let $M \models \mathsf{ZFC}$ be countable (and transitive). Is there a Cohen real $c$ generic over $M$ so that if $d$ is any Cohen real generic over $M$ then $M[c]$ and $M[d]$ are amalgable?

  • $\begingroup$ Nice question! My article on upward closure has further remarks on amalgamation and non-amalgamation: jdh.hamkins.org/…. $\endgroup$ Sep 15, 2016 at 21:19

1 Answer 1


Let $c$ be Cohen over $M$. Construct $d$ as follows. Let $X$ be an infinite set of integers all of whose infinite subsets compute the height of $M$. Inductively construct $d$ such that $d$ lies in every open dense set coded in $M$ and the set of positions at which $c, d$ disagree is an infinite subset of $X$.

  • $\begingroup$ Very nice! This is great. $\endgroup$ Sep 15, 2016 at 21:45

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