# Forcing with product vs. box product

If we want to add one real number the simplest way to do it is to use Cohen forcing. The poset is $\lbrace p\colon n\to 2\mid n\in\omega\rbrace$ which is a countable set. We can think of this as approximating a new set by finite sets.

If we want to add $\omega$-many pairwise generic real numbers we can do it by taking the poset $\lbrace p\colon n\times m\to 2\mid n,m\in\omega\rbrace$. This forcing approximates infinitely many new sets by finite parts of finitely many of the new sets. Interestingly enough both these sets are countable and therefore forcing with one is the same as forcing with the other.

But we can also take $\lbrace p\colon\omega\times n\to 2\mid n\in\omega\rbrace$ and approximate all new sets at each stage. This poset is not countable anymore.

If we look at these things as topological spaces then the original Cohen forcing is really just a countable dense subset of the Cantor space, and we are adding a generic point to the Cantor set. The second forcing is the product of countably many Cantor sets, it is homeomorphic to the original Cantor set.

The third notion of forcing, though, is the box product of countably many Cantor sets. So at least topologically it is different. I also have to admit that I never saw anyone using this approach.

Is there anything wrong it? Is it at all different from the first/second approach? What do we gain/lose when we use the box-product over the usual product?

## 2 Answers

Your third notion of forcing, with conditions $p:\omega\times n\to 2$ ordered by extension as $n$ increases, is the same as the forcing to add a function $\omega$ to the reals of the ground model. This forcing is equivalent to the forcing $\text{Coll}(\omega,\mathbb{R})$, to collapse the ground model continuum to be countable.

To see this, observe that each condition $p$ is essentially the same as a finite list of reals, corresponding to the slices of $p$. Since we may extend any such condition so as to mention any given ground model real on a slice, the generic function will be a surjection from $\omega$ to the ground model reals $\mathbb{R}$. So this forcing collapses $2^\omega$ to $\omega$, and since it has size continuum, it must be forcing equivalent to $\text{Coll}(\omega,\mathbb{R})$.

In the spirit of your question, let me mention a fourth way of adding $\omega$ many Cohen reals, namely, the full support product $\Pi_n \text{Add}(\omega,1)$. Thus, conditions are simply countable sequences of finite binary sequences, ordered by extension in the natural way. This forcing is something like your third way, except that we do not insist on having uniformly constant $n$. That is, the conditions do not need to have the same size on each coordinate. Thus, we may think of a condition $p$ as a function from the region in the plane $\mathbb{N}\times\mathbb{N}$ to $2$, where the domain consists of pairs $(n,m)$ with $m\leq f(n)$ for some function $f:\mathbb{N}\to\mathbb{N}$. (Your third poset has only constant functions $f$.) Meanwhile, this version of the poset also collapses the ground model continuum to $\omega$. To see this, consider the following operation: consider the generic function, which fills out the entire matrix with $0$s and $1$s. Start at any height $k$ in the first column, and look at the number $n_0$ of $0$s directly above it (which is finite by genericity); then go to the $n_0$th digit of the second column, get that bit $a_0$, and let $n_1$ be the number of zeros following it; and so on. This defines a certain binary sequence $a_n$, which is determined by $k$. The point now is that any given binary sequence will arise as such a sequence for some $k$ via the generic filter, since for any condition $p$, I can extend it so as to ensure that my given binary sequence is coded. Thus, the forcing adds a surjection from $\omega$ to the ground model reals, and so the forcing is equivalent to $\text{Coll}(\omega,\mathbb{R})$.

• Thank you for the wonderful answer. I have to admit that when I was writing the question I had in mind your fourth addition rather than my third forcing, and until this very moment I didn't even notice the dissimilarities. However correct me if I am wrong the third notion is dense in your fourth notion, so the fact that the third collapses the continuum immediately implies the fourth does (and vice versa). – Asaf Karagila Nov 27 '12 at 7:57
• Asaf, the third is not dense in the fourth, since in the full-support product forcing, the support needn't be uniformly bounded. For example, consider the condition which is "all zeros below the diagonal". This is a valid condition in my forcing, but no condition in your forcing can ensure it, and in fact no generic filter for your forcing can be all zero below the diagonal (since we may extend any of your conditions to a condition with a $1$ below the diagonal). But nevertheless, the two forcing notions are equivalent, since they are both $\text{Coll}(\omega,\mathbb{R})$. – Joel David Hamkins Nov 27 '12 at 10:55
• Ah, yes. You are very correct about the diagonal argument. They are equivalent nonetheless which is the important thing! – Asaf Karagila Nov 27 '12 at 12:13
• Not only was Asaf thinking of the full-support product when he wrote the question, that's also what I was thinking of when I read it. In particular, the "Somewhat surprisingly" at the start of my answer refers to the rather clever "indirect addressing" argument for the fact that the fourth notion collapses the continuum. My (rather poor) excuse for misreading the question is that the fourth notion corresponds more closely than the third to the topological concept of box product, mentioned in the title. (Another excuse might be telepathic communication to me of Asaf's intention.) – Andreas Blass Nov 27 '12 at 18:18
• Ah, now I understand your "surprising" remark a bit better, since it does seem surprising that the full-support product collapses the continuum, and I think the indirect addressing argument is a little tricky---I find it to be a good exercise for those learning forcing. – Joel David Hamkins Nov 27 '12 at 18:25

Somewhat surprisingly, your third forcing collapses the cardinal of the continuum to $\aleph_0$. That is, it adjoins a surjection from $\omega$ onto the ground model's set of reals.

• Somewhat surprisingly... and perhaps the initial seed for the study of countable support iteration which eventually led to proper forcing! – François G. Dorais Nov 26 '12 at 22:49
• Thank you Andreas, and Francois for an interesting piece of historical trivia. – Asaf Karagila Nov 27 '12 at 7:58