0
$\begingroup$

Hi, I would like some help on something that in some ways has already been touched in the past here. I saw these relevant questions being answered, but I am still unable to understand some things. I would be grateful if someone could help me.

Let $M \models$ ZFC that is countable and let $p(x) = {0 \in x, \ldots, n \in x, \ldots} \cup {x \in \omega }$ be a type where $0, \ldots, n, \dots$ are the finite ordinals of $M$, and $\omega$ the first limit ordinal of $M$. Now let $N$ be an $\omega_1$-saturated model s.t. $M \prec N$. Now $p(x)$ obviously is fin. satisfiable in $M$, therefore it is satisfiable in $N$, by let's say some set $a$. Both $M$ and $N$ are ZFC models, therefore natural numbers and $\omega$ are absolute for $M, N$. So how is it possible to have an ordinal $a$ (since $a \in \omega$) which is finite (in $N$ and therefore in $M$) but different from all $0, \dots, n, \ldots$?

In the same way, given any countable set $b$ of finite sets of $M$, I can always find a saturated model $N$ s.t. $M \prec N$, with some $c \in_N b$ and $c$ infinite. Is this correct?

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ If $a$ is a natural number of $M$, it is also a natural number of $N$, but not vice versa, so being a natural number is not absolute, only upwards absolute. What you are adding is a new natural number. Of course, $M$ doesn't see it. Similarly, "finite" is not absolute. There are finite objects in $N$ that $M$ has no way of seeing. If $a$ is in $M$ and $N$ thinks $a$ is finite, then of course it is finite in $M$, but that is weaker than saying that "finite" is absolute. $\endgroup$
    – Andrés E. Caicedo
    Commented Mar 9, 2011 at 20:53
  • 1
    $\begingroup$ @Sumac: Your belief that "natural numbers and $\omega$ are absolute" would be correct if you were dealing with transitive models or, more generally, if $M$ were transitive from the point of view of $N$. But in the situation you consider, there is no transitivity available. $\endgroup$
    – Andreas Blass
    Commented Mar 9, 2011 at 21:14
  • $\begingroup$ @Andreas Blass, Andres Caicedo : Thank you very much for your fast response. Indeed I forgot to notice that since $N$ won't be transitive (even if $M$ is) then most of the concepts like "finite" etc won't be absolute. Just to be sure I understand though... If we had extended a countable transitive model of ZFC $M$ to some $M[G]$ through forcing, since $M[G]$ is also c.t.m. of ZFC, we wouldn't be able to add any natural numbers in $M[G]$. $\endgroup$
    – Sumac
    Commented Mar 9, 2011 at 22:12
  • $\begingroup$ @Sumac: Yes, that's correct; in this case we have absoluteness of "finite" and "natural number". But the extension $M[G]$ is not countably saturated, and (unless $M=M[G]$) it is not an elementary extension of $M$. $\endgroup$
    – Andrés E. Caicedo
    Commented Mar 9, 2011 at 23:00
  • $\begingroup$ @Andreas and Sumac: isn't the problem really well foundedness, not transitivity? $\endgroup$
    – Justin Moore
    Commented Mar 10, 2011 at 1:49

1 Answer 1

Reset to default
4
$\begingroup$

I'm addressing both the question and the comments, but possibly this question should be closed. First let's be clear what we mean by a model of set theory. A model is a set $M$ (or perhaps a class) with a binary relation $E$. $(M,E)$ satisfies Foundation if for every $x$ in $M$, there is an $E$-minimal $y$ in $M$ such that $y E x$. $M$ is well founded if for every $x \subseteq M$, there is an $E$-minimal $y$ in $M$ such that $y$ is in $x$. If $E$ is $\in$, then $M$ is transitive if every element of $M$ is a subset of $M$. Well foundedness is implicit in transitivity.

The problem is that your model need not be well founded. The Axiom of Foundation only implies that the model of set theory you are working with does not have an element witnessing that it is ill founded. But certainly such a witness can exist outside. The Henkin Construction essentially never will result in a well founded model. A simpler example is given by a non standard model of PA. The model satisfies the induction scheme (which is the analog here of Foundation) but any well founded model of PA is isomorphic to the standard model.

Furthermore, any well founded model $(M,E)$ of ZFC is isomorphic to a transitive set (or class) with the membership relation. This is the Mostowski collapse at work. So transitivity is not the issue here. The failure of finiteness to be absolute is tied up in the ill foundedness of the model in question.

Improve this answer
$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .