# Adding Cohen reals one at a time

We know that if we start with a ctm $\mathbb{B}$ and force with the poset of finite functions from $\omega$ to $2$, we add a single Cohen real. We also know that if we force with the poset $\mathbb{P} = Fn(\kappa \times \omega, 2, \aleph_0)$, we add $\kappa$ many reals (and hence can make the Continuuum Hypothesis fail).

What happens if instead we iterate adding one real $\kappa$ many times? Would we still get a model for not-CH?

• How are you iterating? (Meaning what are you doing at limit stages? Or perhaps you are iterating in some completely different, non-linear, fashion?) Commented Nov 29, 2012 at 19:32
• @Andres: Just by the fact that the question was asked naively I would guess that this is a finite support iteration. Commented Nov 29, 2012 at 19:40
• Perhaps it is worth to note that even when forcing with finite functions from $\omega$ to 2 you don't add just a single Cohen real but a whole bunch of them. Commented Nov 29, 2012 at 20:59
• @Miha: Well, that really depends on the context. It's like talking about an interval and not about the entire real line. In some contexts this is the same thing and in other contexts it's another. I agree that one has to be careful. But I also think that it is okay to assume that if the reader is familiar with forcing then they are intelligent enough to discern context and understand when one Cohen real is treated as one and when it is treated as many. Regardless, Cohen reals are like cockroaches, if there is one then there are infinitely many... :-) Commented Nov 29, 2012 at 23:32
• @AsafKaragila I think that when you discover that you have continuum many cockroaches, it's time to move. Commented Nov 30, 2012 at 2:01

Note that the definition of a Cohen forcing as $2^{<\omega}$ does not change between models.
Iterating it $\kappa$ many times, or taking the product of $\kappa$ many Cohen posets, or using $\mathbb P$ as you defined it -- all of these have the same consequence.
So to your question, yes. A finite-support iteration of length $\kappa$ of adding a single Cohen at a time would end up with a model of $\lnot$CH.
• @Kris, the issue is that explaining precisely what one means by "the naive way" may be a bit problematic. For example, you could have $M_0\subseteq M_1\subseteq M_2\subseteq\dots$ models of set theory, each $M_{i+1}$ an extension of $M_i$ by Cohen forcing, and yet $M_\omega=\bigcup_n M_n$ is not a model of set theory. Once one sees how to do iterations "internally" these worries disappear, but there are several options on how to proceed. (It is a very interesting topic.) Commented Nov 29, 2012 at 20:22