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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?

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How are you iterating? (Meaning what are you doing at limit stages? Or perhaps you are iterating in some completely different, non-linear, fashion?) –  Andres Caicedo Nov 29 '12 at 19:32
    
@Andres: Just by the fact that the question was asked naively I would guess that this is a finite support iteration. –  Asaf Karagila Nov 29 '12 at 19:40
    
Kris, also relevant is Stefan's answer: math.stackexchange.com/questions/173559/… –  Asaf Karagila Nov 29 '12 at 19:42
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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. –  Miha Habič Nov 29 '12 at 20:59
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@AsafKaragila I think that when you discover that you have continuum many cockroaches, it's time to move. –  Miha Habič Nov 30 '12 at 2:01
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up vote 3 down vote accepted

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.

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thanks! we've only done basic one-step forcing in lectures, so i didn't even know how you would iterate forcing (apart from in the naive way). –  Kris Nov 29 '12 at 20:01
    
Sorry to ruin your palindrome. ;) –  Arthur Fischer Nov 29 '12 at 20:16
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@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.) –  Andres Caicedo Nov 29 '12 at 20:22
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