Timeline for What happens when you iterate Cohen reals?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Dec 9 at 10:10 | comment | added | Asaf Karagila♦ | @Lorenzo: I don't know, is the simple answer! I imagine that the answer is positive, since the forcing in not ccc, and we can do some crazy diagonalisation thing where we evade any Cohen dense set from the ground model. But I haven't got the slightest idea as to how to actually prove that. | |
| Dec 9 at 8:58 | comment | added | Lorenzo | @AsafKaragila So, do we know that a full-support iteration of $\omega$-many Cohen reals adds a non-Cohen real (i.e., a real that does not belong to any Cohen extension of the ground model)? | |
| Dec 1, 2018 at 12:21 | comment | added | Mohammad Golshani | @AsafKaragila I don't know if this is useful for you, but you can show that the forcing does not add random reals, and even more. see No random reals in countable support iterations. | |
| Nov 15, 2018 at 15:58 | comment | added | Asaf Karagila♦ | @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 | |
| Nov 15, 2018 at 15:22 | comment | added | Miha Habič | 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 $\omega_1$. | |
| Nov 15, 2018 at 14:03 | comment | added | Will Brian | In my previous comment, I meant the function $n \mapsto g(n,f(n))$, not $g \circ f$ (which isn't well-defined). | |
| Nov 15, 2018 at 13:54 | comment | added | Will Brian | 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 $(2^\omega)^V$.) | |
| Nov 15, 2018 at 13:42 | comment | added | Asaf Karagila♦ | Well, not really related. Just vaguely related, I'd say. :) | |
| Nov 15, 2018 at 13:15 | comment | added | Mohammad Golshani | Related: https://mathoverflow.net/questions/193936/products-of-cohen-forcings | |
| Nov 15, 2018 at 12:32 | history | asked | Asaf Karagila♦ | CC BY-SA 4.0 |