Timeline for capturing small sets in small factors
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Feb 16, 2016 at 23:52 | answer | added | Monroe Eskew | timeline score: 7 | |
| Jun 20, 2014 at 14:42 | vote | accept | Monroe Eskew | ||
| Jun 19, 2014 at 17:44 | answer | added | Yair Hayut | timeline score: 5 | |
| Dec 2, 2013 at 19:18 | comment | added | Yair Hayut | I deleted my previous comments since they were completely wrong, as almost disjoint forcing shows. I think that we can follow Mohammad's suggestion to show that at least the minimal $\lambda < \kappa$ in which $2^\lambda \geq \kappa$, if exists, must be singular. The idea is to use $P = Add(\lambda,\kappa) \ast C$ where $C$ is a forcing that codes (using $\kappa$ almost disjoint sets of $2^{<\lambda}$ in $V$) the generic of $Add(\lambda,\kappa)$. The code itself will be a subset of $2^{<\lambda} < \kappa$, but there is not small sub-forcing $Q$ adding it. | |
| Dec 1, 2013 at 9:31 | comment | added | Mohammad Golshani | Dear Monroe, what is the situation at successor cardinals? Maybe by a coding method you can show that no successor cardinal has the above mentioned property? | |
| Nov 30, 2013 at 5:16 | history | edited | Monroe Eskew | CC BY-SA 3.0 |
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| Nov 29, 2013 at 23:34 | history | edited | Monroe Eskew | CC BY-SA 3.0 |
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| Nov 29, 2013 at 23:24 | history | edited | Monroe Eskew | CC BY-SA 3.0 |
partial solution!
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| Nov 29, 2013 at 22:32 | comment | added | Monroe Eskew | You can show this by a "pseudo-generic tower" argument. If $\langle p_\alpha \rangle_{\alpha<\kappa} \subseteq P$ is an antichain in the extension by $\kappa$-closed $Q$, then we can get a descending sequence of conditions $\langle q_\alpha \rangle_{\alpha<\kappa} \subseteq Q$ such that $q_\alpha$ decides the value of $\dot{p}_\alpha$, and the sequence of decided values is an antichain in $V$. | |
| Nov 29, 2013 at 21:41 | comment | added | Trevor Wilson | I wasn't sure whether $\kappa$-closed forcing preserved the $\kappa$-c.c. Does it? (Sorry if this is obvious.) | |
| Nov 29, 2013 at 20:17 | comment | added | Monroe Eskew | It seems right to me. Take a $P$-name $\tau$ and collapse $P$ to have cardinality $\kappa$ with $\kappa$-closed forcing. $P$ retains the $\kappa$-c.c., and if it is still weakly compact then the extension says there are $Q,\sigma$ with the property. Now $Q,\sigma$ are in $V$ by the closure, and the statement $\Vdash_P \sigma = \tau$ is absolute. What's the worry? | |
| Nov 29, 2013 at 17:27 | comment | added | Trevor Wilson | [Sorry, I think my comment may have been mistaken, so I deleted it. I'm not sure whether indestructibly weakly compact cardinals $\kappa$ have the property in question 2.] | |
| Nov 27, 2013 at 8:21 | history | edited | Monroe Eskew | CC BY-SA 3.0 |
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| Nov 27, 2013 at 7:10 | history | edited | Monroe Eskew | CC BY-SA 3.0 |
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| Nov 27, 2013 at 6:37 | history | edited | Monroe Eskew | CC BY-SA 3.0 |
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| Nov 27, 2013 at 6:36 | comment | added | Asaf Karagila♦ | In the part about a supercompact, you might want to mean that $j[P]$ is a suborder of $j(P)$, not of $P$. | |
| Nov 27, 2013 at 6:26 | history | asked | Monroe Eskew | CC BY-SA 3.0 |