12
$\begingroup$

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 do we already know:

  1. This is a proper forcing. Because Cohen forcing is proper.
  2. This is not a ccc forcing, because we iterate with full support.

What can you say about the full/countable support iteration of length $\omega$ of Cohen forcings?

Specifically, does it add non-Cohen reals? Does it collapse any cardinals?

Improve this question
$\endgroup$
9
  • 1
    $\begingroup$ Related: https://mathoverflow.net/questions/193936/products-of-cohen-forcings $\endgroup$
    – Mohammad Golshani
    Commented Nov 15, 2018 at 13:15
  • 3
    $\begingroup$ 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 $(2^\omega)^V$.) $\endgroup$
    – Will Brian
    Commented Nov 15, 2018 at 13:54
  • 2
    $\begingroup$ In my previous comment, I meant the function $n \mapsto g(n,f(n))$, not $g \circ f$ (which isn't well-defined). $\endgroup$
    – Will Brian
    Commented Nov 15, 2018 at 14:03
  • 6
    $\begingroup$ 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 $\omega_1$. $\endgroup$
    – Miha Habič
    Commented Nov 15, 2018 at 15:22
  • 1
    $\begingroup$ @Miha: I expected that this is clear enough that we iterate in the proper sense of iteration. Also, I figured that people would give me some leeway that I know that iterating with ground model posets is the same as a product. I guess I was wrong... :P $\endgroup$
    – Asaf Karagila ♦
    Commented Nov 15, 2018 at 15:58

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .