$\textbf{Question}$: In expositions to forcing, why do we insist on not adding new ordinals to a countable transitive model $M$ of ZFC?

For example, after ruling out transitive proper class models as possible candidates for satisfying (ZFC plus) $\textbf{V} \neq \textbf{L}$, Kunen writes on pages 185

"We shall describe a general procedure for finding countable transitive models $N$ of ZFC such that $M \subseteq N$ and $o(M) = o(N)$."

Why insist on $o(M) = o(N)$? Later, he goes on to give examples of "non-generic" objects that shouldn't be added to $M$ as they code ordinals $> o(M)$ (Exercise (A3)). Why is this an issue?

  • 1
    $\begingroup$ As JDH pointed out this is a feature, but it is a nice one to have since the usual absoluteness results hold for models with the same ordinals $\endgroup$ Nov 7, 2019 at 21:41
  • $\begingroup$ Because, “God created the ordinals.” $\endgroup$ Dec 31, 2019 at 20:09

3 Answers 3


We don't insist on this as a prior goal. Rather, it just happens to be the case that forcing does not add new ordinals, whether you would want it to or not. Thus, this is a feature of forcing rather than a goal. Meanwhile, there are other model-construction methods that do add new ordinals.

But meanwhile, as far as getting models of $V\neq L$ is concerned, it suffices to have a nontrivial extension with the same ordinals, for then the larger model will have to satisfy $V\neq L$.

  • 5
    $\begingroup$ Even worse, regarding $V\neq L$. Adding ordinals might make $V=L$ true again! $\endgroup$
    – Asaf Karagila
    Nov 3, 2019 at 14:06
  • 3
    $\begingroup$ @AsafKaragila Or one might say, "even better"! This effect is part of my argument that we cannot rule out $V=L$ based on maximization arguments, since it is based on a width-only vision of potentialism. See jdh.hamkins.org/multiverse-perspective-on-constructibility. $\endgroup$ Nov 3, 2019 at 15:05
  • 6
    $\begingroup$ Better is just worse after applying a rotation matrix... :) $\endgroup$
    – Asaf Karagila
    Nov 3, 2019 at 15:07
  • 4
    $\begingroup$ Just like politics, I suppose. $\endgroup$ Nov 3, 2019 at 15:10

You might consider the following quote from Cohen's paper, "The Discovery of Forcing", Rocky Mountain Journal of Mathematics, Volume 32, number 4, 2002, pg. 1091 (found under title on the Web):

So we are starting with a countable standard model $M$, and we wish to to adjoin new elements and still obtain a model. An important decision is that no new ordinals are to be created. Just as Godel did not remove any ordinals in the constructible universe, a kind of converse decision is made not to add any new ordinals.

Since (as you probably know from your readings about forcing) Prof. Cohen was the creator of the forcing technique, forcing was (at its inception), designed not to add new ordinals. The natural question to ask now is, "Why?" (this, of course, is the question you already asked).

A clue as to why is found in Jech's Set theory: Third Millenium Edition, Chapter 13, in Theorem 13.28 and the preceding paragraph:

If $M$ is a transitive model of $ZFC$, then the Axiom of Choice in $M$ enables us to code all sets in $M$ by sets of ordinals and the model is determined by its sets of ordinals. The precise statement of this fact is: if $M$ and $N$ are two transitive models of $ZFC$ with the same sets of ordinals, then $M$ = $N$ [in which case, $M$ and $N$ have identical ordinals--my comment]. In fact, a slightly stronger assertion is true (on the other hand, one cannot prove that $M$ = $N$ if neither model satisfies $AC$.)

Theorem 13.28. Let $M$ and $N$ be transitive models of $ZF$ and assume that the Axiom of Choice holds in $M$. If $M$ and $N$ have the same sets of ordinals, i.e., $P^{M}$($Ord^{M}$) = $P^{N}$($Ord^{N}$) [$P^{M,N}$ is just the power set operation restricted to the models $M$, $N$, respectively and $Ord^{M,N}$ are just the ordinals of $M$ and $N$, respectively (Kunen uses essentially the same definition for $Ord^{M, N}$)--my comment], $M$ = $N$.

As regards Profs. Hamkins' and Karagila's comments to each other regarding Prof. Hamkins' answer to you, you might ask either (or both) of them for a proof of their assertion, i.e., that adding ordinals to a model of $ZFC$ + $V$ $\ne$ $L$ might (or would) make $V$ = $L$ true again. If they would design their proofs so as to educate, I'm sure you would find them very interesting (and then you would truly know and understand why one "doesn't insist on adding ordinals to a countable transitive model $M$ of $ZFC$ in expositions of forcing").

As regards Prof. Hamkins' assertion that there are "other model-theoretic methods that do add new new ordinals", you might take a look at his answer to S A's mathoverflow question, "Can 'syntactic forcing' add ordinals?". As regards the fact that forcing does not add new ordinals, you might take a look at tomasz's mathstackexchange question, "A question about the proof that forcing extensions don't add ordinals", paying particular attention to Halbeisen's text mentioned, Profs. Caicedo's and Blass's comments, and the theorem in Halbeisen's text that tomasz mentions in his question. You might find this information at least somewhat helpful in your search for an answer.

  • 4
    $\begingroup$ Your added comment, "which in turn is determined by which ordinals are in $M$," is wrong. Two transitive models of ZFC with the same ordinals need not have the same sets of ordinals. $\endgroup$ Nov 12, 2019 at 1:40
  • $\begingroup$ @AndreasBlass: Although you are certainly correct in stating that "Two transitive models of $ZFC$ with the same ordinals need not have the same sets of ordinals", I am unclear why my added statement, "which in turn is determined by which ordinals are in $M$" ('is should be replaced by 'are' in that statement) should imply that models of $ZFC$ having the same ordinals should have the same sets of ordinals. Nevertheless, I have removed the offending "added comment" and placed it where it hopefully won't cause me any problems. $\endgroup$ Nov 12, 2019 at 22:39
  • $\begingroup$ @ThomasBenjamin You wrote that the sets of ordinals in $M$ are determined by the ordinals in $M$. But that's not true: we can have models with the same ordinals but with different sets of ordinals. (E.g. if $M$ is any model of ZFC + $\neg$V=L, then $M$ and $L^M$ have the same ordinals but have different sets of ordinals.) $\endgroup$ Nov 13, 2019 at 3:28
  • $\begingroup$ @NoahSchweber: You are correct in your statement, "we can have models with the same ordinals but with different sets of ordinals (and the example you gave is great as well!), but I think that you and Prof. Blass, and I, are talking at cross purposes here. The phrase I used, "which in turn are determined by which ordinals are in $M$", has different senses, and you and Prof. Blass chose one sense, and I another. The sense I used , namely, that the ordinals in $M$ form the basis from which $P^{M}$ forms the other sets of ordinals (let us assume, for the sake of argument, that we are using set $\endgroup$ Nov 14, 2019 at 0:37
  • $\begingroup$ (cont.) models here), certainly is a sense of the phrase, "which in turn are determined by which ordinals are in $M$. What would be more interesting is providing a proof that adding new (imaginary?) ordinals to the model you speak of in your example will recover $V$ = $L$. Can you and/or Prof. Blass do this? It would certainly if either of you could. Thanks in advance for hour help. $\endgroup$ Nov 14, 2019 at 0:44

I think Paul Cohen has a decent (retrospective) justification in Section 2, Chapter IV of his book "Set Theory and the Continuum Hypothesis". Check out the Corollary on page 110 and the discussion following it.

I am not sure why this issue is ignored by other authors.

  • 4
    $\begingroup$ Perhaps you might supply a quotation? I think most readers will not necessarily look this up. $\endgroup$ Nov 3, 2019 at 15:06
  • 3
    $\begingroup$ I suppose this is what hello is referring to: It is consistent that there is a transitive model and no transitive model with more ordinals. $\endgroup$ Nov 3, 2019 at 18:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.