Timeline for What can we learn about an elementary embedding from the image of the ordinals?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Feb 27, 2012 at 1:30 | comment | added | Joel David Hamkins | For example, one usually denotes forcing extensions by $V[G]$, and the meaning is the same as here. | |
| Feb 26, 2012 at 21:52 | comment | added | Mike Shulman | Ah, I see. By $M[A]$ you mean a thing that I would have been inclined to write write $M[\{A\}]$. To me, the notation $M[A]$ naturally suggests to adjoin to $M$ all the elements of $A$ (cf. for example a group ring $\mathbb{Z}[G]$). Thanks! | |
| Feb 26, 2012 at 19:58 | comment | added | Joel David Hamkins | Although $j''\text{Ord}$ is contained in $M$, it is generally not a class in $M$, and the point is that adding it as a class allows one to construct new sets not in $M$. For example, imagine that $j''\theta$ is unbounded in $j(\theta)$ and $\theta\lt j(\theta)$, but $M$ thinks $j(\theta)$ is regular. So $j''\theta$ would reveal that $M$ was wrong about the regularity of $j(\theta)$. In general, for the reasons Jonas mentioned in the question, one cannot have $j''V$ a class in $M$, since then $M$ would be able to construct every set in $V$ and conequently $M=V$, which is impossible. | |
| Feb 26, 2012 at 17:02 | comment | added | Mike Shulman | Okay. Those were my guesses, but then I am confused. Since $j$ maps $V$ into $M$, and $ORD\subseteq V$, it seems to me that $j''ORD \subseteq M$. So why does "adjoining" $j''ORD$ to $M$ do anything at all, when $j''ORD$ is already contained in $M$? | |
| Feb 26, 2012 at 4:44 | comment | added | jonasreitz | Oops - Joel beat me to the reply -- thanks, Joel! | |
| Feb 26, 2012 at 4:43 | comment | added | jonasreitz | Hi Mike -- by $j''A$ I mean "the image of $A$ under the map $j$", or $\{j(a) | a \in A\}$. The notation $M[A]$ indicates the extension of the model $M$ by the element $A$ -- that is, the structure we get by starting with $M$, adding $A$, and closing under definability (analogous to to building a field extension by adding an element and closing under plus and times). In the case where $A$ is a proper class and thus can't be "added to the structure" (as with $ORD$), we proceed by adding a predicate for $A$ and closing under definability in this extended language. | |
| Feb 26, 2012 at 4:35 | comment | added | Joel David Hamkins | Mike, I believe that Jonas intends that $j$ is a large cardinal elementary embedding of the universe $V$ into a transitive class $M$. The double quote notation $j''X$ means the pointwise image of $X$ under $j$, which might elsewhere be denoted $j[X]$. And $M[j''\text{Ord}]$ means the model obtained from $M$ by adjoining the class $j''\text{Ord}$---for example, close under the Goedel operations, but allowing also the characteristic function of that class. In my solution, I argue that every set in the universe $V$ is constructible from an object in $M$ and a small piece of $j''\text{Ord}$. | |
| Feb 26, 2012 at 4:34 | vote | accept | jonasreitz | ||
| Feb 26, 2012 at 4:06 | comment | added | Mike Shulman | I would like to understand this question and its answer, but as a non-set-theorist I'm not familiar with the notation. Can you say what $j''$ and $M[-]$ mean? | |
| Feb 26, 2012 at 2:59 | answer | added | Joel David Hamkins | timeline score: 8 | |
| Feb 26, 2012 at 2:35 | history | edited | jonasreitz | CC BY-SA 3.0 |
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| Feb 26, 2012 at 1:51 | history | asked | jonasreitz | CC BY-SA 3.0 |