bio | website | math.ucla.edu/~justinpa |
---|---|---|
location | UCLA | |
age | ||
visits | member for | 4 years, 4 months |
seen | Jun 25 '13 at 0:32 | |
stats | profile views | 2,153 |
UCLA grad student studying set theory.
Mar 20 |
awarded | Nice Answer |
Dec 8 |
awarded | Yearling |
May 24 |
awarded | Nice Answer |
Mar 6 |
awarded | Nice Question |
Dec 8 |
awarded | Yearling |
Dec 3 |
comment |
Why does the Solovay-Tennenbaum theorem work?
With that in mind I'd suggest that a good 'next' iterated forcing argument to look at is Baumgartner's construction of a model where all $\aleph_1$-dense set of reals are isomorphic, since the iteration only uses ccc forcings, but the result doesn't itself doesn't follow from MA (The original paper is freely available at the FM archive matwbn.icm.edu.pl/ksiazki/fm/fm79/fm79111.pdf, and here's an expository note presenting the same result: scholarworks.sjsu.edu/etd_theses/3834) |
Dec 3 |
comment |
Why does the Solovay-Tennenbaum theorem work?
When I'm in situations where I feel like I understand the mechanism of a proof but don't 'grok' them I am reminded of the quote of von Neumann: "In mathematics you don't understand things. You just get used to them." |
Nov 26 |
comment |
Which of these relations on partial orders allows us to identify forcing equivalence?
I see, so it looks like the map $q$ to $[q\in\tau]$ isn't even well-defined in that case, since forcing-wise RO($\mathbb{Q}$) consists only of nonzero elements. If it is well-defined it's a complete embedding, and that will only happen if (and only if) for any $q$ we can find a $p$ forcing $q$ into $\tau$. |
Nov 26 |
comment |
Which of these relations on partial orders allows us to identify forcing equivalence?
The paper "On the Alaoglu-Birkhoff equivalence of posets" by Todorcevic and Zapletal seems relevant, since it discusses the relationship among several natural preorderings on posets, including $\lhd_1$, Tukey reducibility and others. The paper is available open access at projecteuclid.org/… |
Nov 26 |
comment |
Which of these relations on partial orders allows us to identify forcing equivalence?
If I'm interpreting the definitions correctly, I think $\lhd_1$ and $\lhd_4$ are equivalent as long as the forcing notions in question are separative. Certainly $\lhd_4$ implies $\lhd_1$ since a separative forcing $\mathbb{P}$ is forcing equivalent to its Boolean completion RO($\mathbb{P}$). And if $\mathbb{Q}\lhd_1\mathbb{P}$ then in particular there is $\tau$ a RO($\mathbb{P}$)-name for a RO($\mathbb{Q}$)-generic. The map sending q in RO($\mathbb{Q}$) to the Boolean value $[q\in\tau]$ (calculated in RO($\mathbb{P}$) is a complete embedding. |
Nov 24 |
awarded | Nice Question |
Sep 17 |
comment |
Theory of (definable) ideals on a multi-dimensional countable set
Any/all of the above. All the structural results in the literature on 'definable' ideals I know of would follow from determinacy. I'm willing to make large cardinal assumptions here, so I have a fairly large umbrella in mind. But if something about multidimensional ideals can be extracted from stronger definability assumptions I'd be happy to hear about them.. |
Sep 17 |
asked | Theory of (definable) ideals on a multi-dimensional countable set |
Jun 27 |
comment |
Forcing over set theory versus forcing over arithmetic
Can you recommend a good general reference for forcing over models of arithmetic? (for someone who knows the set theory side, but not the arithmetic side very well) |
May 28 |
revised |
Mathias forcing with Ramsey ultrafilters, and Cohen reals
added 86 characters in body; deleted 25 characters in body |
May 28 |
comment |
Mathias forcing with Ramsey ultrafilters, and Cohen reals
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 |
revised |
Mathias forcing with Ramsey ultrafilters, and Cohen reals
added 1461 characters in body |
May 14 |
comment |
Mathias forcing with Ramsey ultrafilters, and Cohen reals
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 |
comment |
Mathias forcing with Ramsey ultrafilters, and Cohen reals
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 |
revised |
Mathias forcing with Ramsey ultrafilters, and Cohen reals
added 25 characters in body |