Timeline for Mathias forcing with Ramsey ultrafilters, and Cohen reals [closed]
Current License: CC BY-SA 3.0
15 events
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| Jun 11, 2015 at 7:45 | comment | added | Goldstern | A possibly simpler way of getting the Cohen real: Let $g\in \omega^\omega$ be the generic real, a strictly increasing sequence. Let $c(i)=0$ if $g(i)$ and $g(i+1)$ are in the same class of the partition $(A_k:k\in \omega)$, and $c(i)=1$ otherwise. Then $c\in 2^\omega$ is a Cohen real. (The proof uses the fact that the intersections of $ran(g)$ with the $A_k$ are not bounded.) | |
| May 5, 2014 at 19:31 | comment | added | David Fernandez-Breton | Even though this wasn't a real question, I still found it extremely useful (I was trying to prove that certain ultraLaver forcing doesn't add Cohen reals, even though the ultrafilter in question is not a P-point. Thanks to this post, now I know that this is, most likely, impossible). | |
| May 28, 2012 at 17:07 | history | edited | Justin Palumbo | CC BY-SA 3.0 |
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| May 28, 2012 at 17:05 | comment | added | Justin Palumbo | Ah right, yes I was mentally conflating bounded and finite. (The Cohen real I described only needs the $A$ in $\mathcal{U}$ to have unbounded intersection with the $A_k$). Thanks | |
| May 28, 2012 at 14:28 | comment | added | Andreas Blass | In the second paragraph of the edit, the sentence "And if $\mathcal U$ isn't selective, that means ..." seems to describe non-P-points rather than non-selective ultrafilters. In the case of a non-selective P-point, there would be an $A$ that meets each $A_k$ in only finitely many points (but the "finitely many" would not be bounded independently of $k$). | |
| May 28, 2012 at 2:45 | comment | added | François G. Dorais | (Closed per author's request.) | |
| May 28, 2012 at 2:44 | history | closed |
Justin Palumbo François G. Dorais |
not a real question | |
| May 28, 2012 at 2:27 | history | edited | Justin Palumbo | CC BY-SA 3.0 |
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| May 14, 2012 at 20:04 | comment | added | Justin Palumbo | in the meantime I've 'strengthened' the question by also asking about p-points, where it isn't clear that the forcing has the Laver property (and I would guess, perhaps, it does not) | |
| May 14, 2012 at 20:04 | comment | added | Justin Palumbo | Ramiro, your suggestion seems to be right; I think the same diagonal arguments that show vanilla Mathias forcing has the Laver property shows that Mathias forcing relative to a Ramsey ultrafilter does.. if you wanted to add your suggestion as an answer I would certainly upvote it... | |
| May 14, 2012 at 19:35 | history | edited | Justin Palumbo | CC BY-SA 3.0 |
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| May 14, 2012 at 19:24 | comment | added | Justin Palumbo | i didn't think about the Laver property, that's a very good suggestion | |
| May 14, 2012 at 19:23 | comment | added | Justin Palumbo | vanilla Mathias forcing (without an ultrafilter) doesn't add Cohen reals, but Mathias forcing with an ultrafilter very well might; in Shelah's model with no nowhere dense ultrafilter every Mathias forcing with an ultrafilter and indeed every sigma-centered forcing will add a Cohen real | |
| May 14, 2012 at 19:00 | history | edited | Justin Palumbo | CC BY-SA 3.0 |
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| May 14, 2012 at 18:00 | history | asked | Justin Palumbo | CC BY-SA 3.0 |