Timeline for Kunen's use of Countable Transitive Models
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Mar 4, 2011 at 1:12 | comment | added | Jason | The ultrapower will actually be able to construct sequences for which the $\langle f_i| i \in \mathbb{N}\rangle$ form proper initial segments because from the point of view of $N$, we have $(f_{i+1}) = (f_i) - 1$ for all $i \in \mathbb{N}$, again by Łoś's theorem. For example, it can construct what it believes to be an $(\text{identity})_U$-sequence this way, but that looks like a finite sequence from its point of view so it doesn't contradict internal regularity. | |
| Mar 4, 2011 at 0:52 | comment | added | Jason | For example, in the ultrapower $N$, we can let each $c_i$ be the equivalence class of the function $f_i$ defined by $f_i(n) := n \stackrel{\cdot}{-} i$ (cutoff subtraction: when $n \geq i$, $f_i(n) = n - i$; otherwise $f_i(n) = 0$). In this way, $f_0$ is the equivalence class of the identity function on the true $\omega$ and the $\omega$ of the ultrapower contains this descending sequence of $f_i$ under $\in_N$. Alternatively, the equivalence class of the identity function contains the infinite sequence of $(f_i)$ for $i \in \mathbb{N}^+$ under $\in_N$. | |
| Mar 4, 2011 at 0:35 | comment | added | Jason | Basically: In general, it will be the equivalence class of the sequence with constant value $\omega^M$ by Łoś's theorem so if $M$ has a standard $\omega$, then yes. Actually I assumed $M$ was transitive when I defined the relation on $N$ (it should be $\{n \in \mathbb{N}| g(n) \in_M h(n)\} \in U$ in general). To your second question, it depends what you mean by for all $i$. Whatever value $c_0$ assumes will satisfy that property for all true $n \in \mathbb{N}$, but your model cannot construct this chain because it does not have the standard $\mathbb{N}$. | |
| Mar 3, 2011 at 23:22 | comment | added | David Fernandez-Breton | Just a quick question... when you construct the ultrapower $M^\omega/U$ in the first paragraph, is the $\omega$ of $M^\omega/U$ just the equivalence class of the sequence with constant value $\omega$?... and regarding the constructed sequence $c_o\ni_N c_1\ni_N\cdots$, I'm guessing there's no element $A$ of the model such that $c_1\in_N A$ for every $i$. | |
| Mar 3, 2011 at 23:07 | vote | accept | David Fernandez-Breton | ||
| Mar 3, 2011 at 23:07 | comment | added | David Fernandez-Breton | Great explanation!!! I'm finally making some sense of all this issue... thanks a lot!! | |
| Mar 3, 2011 at 21:55 | history | edited | Jason | CC BY-SA 2.5 |
clarified; got rid of example, which is only relevant to relativization
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| Mar 3, 2011 at 21:05 | history | edited | Jason | CC BY-SA 2.5 |
added clarification to infinite descending sequence
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| Mar 3, 2011 at 20:49 | history | edited | Jason | CC BY-SA 2.5 |
got rid of unnecessary existential quantifiers
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| Mar 3, 2011 at 20:36 | history | answered | Jason | CC BY-SA 2.5 |