If $j : V \rightarrow M$ is an elementary embedding, what can we learn in $M$ from $j''ORD$? That is, what is $M[j''ORD]$?

In particular,

Is it $M[j''ORD]$ equal to all of $V$?

If not, do we get a model intermediate between $M$ and $V$? If $\kappa$ is the critical point of $j$, is $\kappa$ still a large cardinal in $M[j''ORD]$? I am thinking of $j$ arising from a measure on $\kappa$, but I'm also interested in the more general situation.

My thinking goes like this: If we have the image of all of $V$, we can reconstruct $V$ itself by taking the Mostowski collapse of $j''V$ (and $j$ is the inverse of the Mostowski collapse). In $M[j''ORD]$, let's consider the class $W$ of sets with rank in $j''ORD$. Does the Mostowski collapse of $W$ yield all of $V$, and if not, what's missing?

EDIT: formatting, clarification of question.

  • $\begingroup$ 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? $\endgroup$ Commented Feb 26, 2012 at 4:06
  • $\begingroup$ 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 $j''\text{Ord}$. $\endgroup$ Commented Feb 26, 2012 at 4:35
  • $\begingroup$ 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. $\endgroup$
    – jonasreitz
    Commented Feb 26, 2012 at 4:43
  • $\begingroup$ Oops - Joel beat me to the reply -- thanks, Joel! $\endgroup$
    – jonasreitz
    Commented Feb 26, 2012 at 4:44
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Feb 26, 2012 at 19:58

1 Answer 1


Nice question, Jonas!

Yes, in the case that $V$ satisfies ZFC and $M\subset V$, then indeed $M[j''\text{Ord}]=V$. To see this, consider first the case of a set of ordinals $A\subset\theta$ in $V$. Notice that from $j''\theta$ we may reconstruct $j\upharpoonright\theta$. Further, $j(A)$ is in $M$, and from $j(A)$ and $j\upharpoonright\theta$ we may easily reconstruct $A$ itself. So every set of ordinals in $V$ is in $M[j''\text{Ord}]$. If $V$ satisfies ZFC, then this suffices, since every set is coded by a set of ordinals. Namely, (as you know) if $a$ is any set, then $\langle \text{TC}(\{a\}),{\in}\rangle\cong\langle\theta,E\rangle$ for some cardinal $\theta$ and some binary relation $E$ on $\theta$, and then using a Gödel pairing function to code $E$ as a single set $A\subset\theta$. So for any set $a$, we find $E$ and then $A$, and by the previous argument $A$ is in $M[j''\text{Ord}]$, and so also $E$ and hence $a$ itself is there. Thus, $M[j''\text{Ord}]=V$, as desired.

The argument relies on the axiom of choice, and in the most general case, I believe this is required.

  • $\begingroup$ Thanks, Joel -- very nice argument! If we drop your condition that $M \subset V$, I think your argument shows we can still recover $V$ but we get $V \subset M[j''ORD]$ instead of equality. Is $V$ a definable class in this case? For this I guess we'd need a more canonical representation of each element -- something like the set $W$ that I described in the question, I'll think about how to make this work. Your argument also shows that we can recover the map $j$ from the image of the ordinals, provided $j$ is definable in $V$. What if $j$ is not 'in $V$', for example an external ultrapower? $\endgroup$
    – jonasreitz
    Commented Feb 26, 2012 at 4:58
  • $\begingroup$ If we can always recover $j$ from $j''ORD$, then it seems $M[j''ORD]$ could properly contain both $M$ and $V$, and more. $\endgroup$
    – jonasreitz
    Commented Feb 26, 2012 at 5:00
  • $\begingroup$ I agree with all that. $\endgroup$ Commented Feb 26, 2012 at 5:21

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