# What happens when you iterate Cohen reals?

There are a few classical theorems in set theory:

1. The finite support iteration of ccc forcing is ccc.
2. The countable support iteration of proper forcing is proper.
3. The finite support iteration of length $$\omega$$, Cohen forcing is a Cohen forcing.

But what happens when you iterate Cohen forcing with a countable support iteration of length $$\omega$$?

What can you say about the full/countable support iteration of length $$\omega$$ of Cohen forcings?
• I think it's consistent that it collapses cardinals: unless I'm mistaken, I think it should collapse $\mathfrak{c}^V$ to $\mathfrak{d}^V$. (The reason: if $\mathcal D$ is a dominating family of functions in the ground model, and if $g: \omega \times \omega \rightarrow 2$ is the generic object added by your forcing, then a density argument shows that $\{ g \circ f \,:\, f \in \mathcal D \}$ contains $(2^\omega)^V$.) – Will Brian Nov 15 '18 at 13:54
• In my previous comment, I meant the function $n \mapsto g(n,f(n))$, not $g \circ f$ (which isn't well-defined). – Will Brian Nov 15 '18 at 14:03
• Regarding Mohammad's link, it's important to specify what exactly you are iterating when you have infinite supports. If you take a nice enough full name at each stage then what you wrote is true and the whole iteration is proper. But if you take the check name at each stage, you just get back the product, which will collapse $\omega_1$. – Miha Habič Nov 15 '18 at 15:22