Timeline for A well-behaved $A$ that is almost contained in every element of some filter for a countable arithmetically closed family $\mathfrak X$
Current License: CC BY-SA 3.0
12 events
when toggle format | what | by | license | comment | |
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Sep 8, 2013 at 20:57 | history | edited | François G. Dorais | CC BY-SA 3.0 |
revised and expanded
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Sep 7, 2013 at 23:55 | vote | accept | Victoria Gitman | ||
Sep 7, 2013 at 21:11 | history | edited | François G. Dorais | CC BY-SA 3.0 |
extra details
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Sep 7, 2013 at 20:37 | comment | added | François G. Dorais | (Email me if you want more details on the "little more." These comment boxes are too small...) | |
Sep 7, 2013 at 20:36 | comment | added | François G. Dorais | $F_\sigma$-Mathias forcing is proper. That's not in the paper but I do comment on it. | |
Sep 7, 2013 at 20:35 | comment | added | Victoria Gitman | I see that your forcing satisfies axiom A, so it is proper. | |
Sep 7, 2013 at 20:31 | comment | added | Victoria Gitman | Ideally, I would like a proper forcing notion, where it is sufficient to meet some $\omega_1$-many dense sets to ensure that the arithmetic closure of $\mathfrak X\cup \{A\}$ avoids $C$. But it is great to know about your example because it means that there is no obstacle to what I want to do! | |
Sep 7, 2013 at 20:27 | comment | added | François G. Dorais | Not the stronger cone-avoiding theorem but what you need is just a little more than the relativized version of the cone-avoiding theorem that I do prove in the paper. | |
Sep 7, 2013 at 20:25 | comment | added | Victoria Gitman | Is this all written up in the paper? What are some properties of the forcing? | |
Sep 7, 2013 at 20:22 | history | edited | François G. Dorais | CC BY-SA 3.0 |
conclusion
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Sep 7, 2013 at 20:22 | comment | added | Victoria Gitman | François, I think I can construct such an $A$ also. But what I need is that $C$ is not definable from $A$, so not computable from any jump of $A$. | |
Sep 7, 2013 at 20:09 | history | answered | François G. Dorais | CC BY-SA 3.0 |