Timeline for Diagonalizing against a non stationary set of functions
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Jul 20, 2015 at 15:22 | vote | accept | Ashutosh | ||
| S Jul 19, 2015 at 20:37 | history | suggested | Ed Dean |
added [lo.logic] tag
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| Jul 19, 2015 at 20:13 | review | Suggested edits | |||
| S Jul 19, 2015 at 20:37 | |||||
| Jul 19, 2015 at 17:48 | answer | added | Avshalom | timeline score: 1 | |
| Jul 8, 2015 at 6:41 | comment | added | Mohammad Golshani | See the paper ``Weakly compact cardinals: a combinatorial proof''. | |
| Jul 8, 2015 at 6:40 | comment | added | Mohammad Golshani | The following theorem of Shelah might be related: Let $\mu$ be strongly inaccessible. Then $\mu$ is weakly compact iff for every family $(f_\alpha: \alpha \to \alpha: \alpha < \mu),$ there exists $f:\mu \to \mu$ such that for every $\alpha<\mu,$ we can find $\beta \in [\alpha, \mu)$ with $f\restriction \alpha = f_\beta \restriction \alpha.$ | |
| Jul 7, 2015 at 20:08 | comment | added | Ashutosh | Assaf pointed me to his blog post where he mentioned a result of Matet that explains this. Here's the link: blog.assafrinot.com/?p=1867 | |
| Jul 7, 2015 at 19:41 | comment | added | Ashutosh | In nutshell, step c in Fremlin's argument seems to be saying that if you can guess every set at a limit stage then you can modify your guessing sequence to guess everything at stationary many stages. I don't think this is always true and i dont see how to do this modification in this specific case. | |
| Jul 7, 2015 at 19:30 | comment | added | Will Brian | I have to admit that I won't really understand the link you give without putting a lot more effort in. That said, I'll share one more observation. If $A$ is such that there is an injective regressive function $\varphi: A \rightarrow \kappa$, then the answer to your original question is yes. (On the image of $\varphi$, define $f$ so that $f(\varphi(i)) \neq f_i(\varphi(i))$ for every $i \in A$. Off of the image of $\varphi$ define $f$ arbitrarily.) Since Fremlin mentions injective regressive functions in his argument, maybe this is relevant? | |
| Jul 7, 2015 at 18:53 | history | edited | Ashutosh | CC BY-SA 3.0 |
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| Jul 7, 2015 at 18:45 | comment | added | Ashutosh | Thanks Will, that pretty much nails it. I was trying to finish a proof of Fremlin's. I will edit to explain where i am stuck. | |
| Jul 7, 2015 at 17:59 | comment | added | Will Brian | There are two trivial ways this can fail. If $0 \in A$ then $f_0$ and $f \upharpoonright 0$ are both the empty function (hence equal). A little less trivially, if $0 \notin A$ but $n \in A$ for every $0 < n < \omega$, then suppose $f_{n+1}$ is the function mapping $m$ to $0$ for $m < n$ and $n$ to $1$. If your condition is satisfied, this forces $f$ to map every $n < \omega$ to $0$. But if $f_\omega$ is constantly $0$, your condition fails at $\omega$. I'm writing this as a comment instead of an answer because I'm sure it's not really what you're wanting. Maybe put some restrictions on $A$? | |
| Jul 7, 2015 at 17:06 | history | edited | Ashutosh | CC BY-SA 3.0 |
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| Jul 7, 2015 at 16:57 | history | asked | Ashutosh | CC BY-SA 3.0 |