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Hi,

I have a doubt concerning Kunen's exposition of forcing in his classical book (arguably $the$ book on forcing). When dealing with Countable Transitive Models to set up the forcing machinery, Kunen is always very careful with letting this C.T.M. to model $only$ finite fragments of ZFC. I recently read in one of the answers to this MO question that the point is that CON(ZFC) cannot prove the existence of countable transitive models of ZFC, and I don't understand why not... wouldn't this be just a matter of taking a set model for ZFC (by consistency of ZFC, which we are assuming), which without loss of generality can be countable (by L\"owenheim-Skolem) and then apply the Mostowski collapsing lemma to this in order to get a C.T.M. of the $full$ ZFC?

Also, a professor once told me that Kunen did things this way in order to avoid assuming CON(ZFC), but I didn't understand this explanation either (isn't it pointless avoiding to assume CON(ZFC)... if the negation holds, everything would be provable anyways!!!)

I'm pretty sure there's something about this issue that I'm not taking into account, I would like to know what that is... I'm kindly asking for your help with that!

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  • $\begingroup$ I don't understand your point. A model of ZFC is just a set with some relation on it. That is not enough to apply the Mostowski collapsing lemma. For this you need a special kind of model and Löwenheim-Skolem doesn't guarantee you the existence of a countable such model. $\endgroup$
    – Michael Greinecker
    Commented Mar 3, 2011 at 10:34
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    $\begingroup$ While it is not the most important reason (see oktan's answer), it's perfectly possible that ZFC is consistent, and yet it proves its own inconsistency. $\endgroup$
    – Emil Jeřábek
    Commented Mar 3, 2011 at 11:16
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    $\begingroup$ It is true that if you're going to study ZFC then you're likely to assume (in your head) that ZFC is consistent. But there are technical problems getting the theory itself to believe it's consistent. If you add Con(ZFC) as an axiom, you get a new theory ZFC'; by the incompleteness theorem ZFC' cannot prove Con(ZFC'), so you can't make models of ZFC' in ZFC'. The easiest solution is to just stick with ZFC, and remember it doesn't prove Con(ZFC). A good way to think about forcing while you learn it is to pretend you're actually working in a model of ZFC' but constructing models of just ZFC. $\endgroup$
    – Carl Mummert
    Commented Mar 3, 2011 at 12:46
  • $\begingroup$ Carl, what is the problem with assuming the consistency of ZFC? A typical consistency proof shows a statement like "Con(ZFC) implies $\neg$Con(ZFC+CH)". Now such a statement would have to be proved assuming something, typically ZFC itself, since we believe that a true mathematical statement is one that is provable in ZFC. Goedels proof of the consistency of CH does exactly that: Start with a model of ZF and produce from it a model of ZFC+CH. Unfortunately with forcing there are some technical problems with such an approach, and one way around it is to force over c.t.m.s. $\endgroup$
    – Stefan Geschke
    Commented Mar 3, 2011 at 18:26
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    $\begingroup$ David, the problem is that if ZFC is consistent, then the completeness theorem gives you a model of the form $(M,E)$, where $M$ is a set and $E$ is a binary relation on that set. Typically the binary relation $E$ is not the usual $\in$. While $\in$ is wellfounded by the axiom of regularity in the real world, $(M,E)$ satisfies the axiom of regularity, but still $E$ can be illfounded in the real world and $M$ just doesn't know an infinite $E$-descending sequence. In this case $(M,E)$ is not isomorphic to a structure of the form $(X,\in)$. $\endgroup$
    – Stefan Geschke
    Commented Mar 3, 2011 at 20:44

3 Answers 3

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Here are some examples that might help in understanding. If ZFC is consistent, then it follows we have a set model $M$ of the theory. Consider a nonprincipal ultrafilter $U$ on $\omega$ and let $M^{\omega}/U$ be the induced ultrapower. $M^{\omega}/U$ is a model of ZFC, but it cannot be well-founded because its $\omega$ has nonstandard elements. Specifically, for any strictly increasing function $g: \mathbb{N} \rightarrow \mathbb{N}$ and $n \in \mathbb{N}$, we have $M^{\omega}/U \models \omega > (g)_{U} > n$ where the $\omega$ here is of course the nonstandard one.

Also, if you don't want to work with ultrapowers directly, you can simply appeal to the Compactness theorem. Introduce a set of constants $\{c_n| n \in \mathbb{N}\}$ into your language and for each $n \in \mathbb{N}$, define $\varphi_n := c_0 \ni c_1 \ni \ldots \ni c_n$. If ZFC is consistent, then every finite fragment of $ZFC + \{\varphi_n| n \in \mathbb{N}\}$ is consistent so the entire theory is by the Compactness theorem. This then gives us a model $N$ of ZFC that externally can be seen to have an infinite descending chain:

$c_0 \quad \ni_N \quad c_1 \quad \ni_N \quad c_2 \quad \ni_N \quad \ldots \quad \ni_N \quad c_n \ldots$.

Note this is not an actual infinite $\in$-descending chain as Stefan points out, but merely a binary relation on the set $N$. For example, if $N = M^{\omega}/U$ is the ultrapower induced by a nonprincipal ultrafilter $U$ on $\omega$, then $\in_N$ would be defined by $(g)_U \quad\in_N\quad(h)_U$ exactly when $\{n \in \mathbb{N}| g(n) \in h(n)\} \in U$.

You may also be interested in:

Clearing misconceptions: Defining “is a model of ZFC” in ZFC

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  • $\begingroup$ Great explanation!!! I'm finally making some sense of all this issue... thanks a lot!! $\endgroup$
    – David Fernandez-Breton
    Commented Mar 3, 2011 at 23:07
  • $\begingroup$ 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 $i$. $\endgroup$
    – David Fernandez-Breton
    Commented Mar 3, 2011 at 23:22
  • $\begingroup$ 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 $\mathbb{N}$. $\endgroup$
    – Jason
    Commented Mar 4, 2011 at 0:35
  • $\begingroup$ 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 $\in_N$. $\endgroup$
    – Jason
    Commented Mar 4, 2011 at 0:52
  • $\begingroup$ 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. $\endgroup$
    – Jason
    Commented Mar 4, 2011 at 1:12
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It is true that you can use Löwenheim Skolem to get a countable model $M$ of ZFC assuming $Con(ZFC)$. But to use Mostowski you need additionally the well foundedness of that model, which doesn't have to be true, even though that model satisfies the axiom of regularity. The point here is that $M$ 'thinks' that it is wellfounded but from the 'outside' it is not.

Moreover, remember that $CON(ZFC)$ is merely an artihmetical statement, which doesn't tell you anything about the 'real' consistency of $ZFC$. So assuming $\lnot Con (ZFC)$ will not prove you anything you want, it will just prove you that 'there exists a proof for anything you want' (the statement 'there exists a proof...' is again an arithmetical statement).

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  • $\begingroup$ Ok., I'm starting to get the point... but I still can't imagine a model $M$ which thinks it satisfies the Foundation axiom but which is not wellfounded from the outside... I'll keep thinking about it. $\endgroup$
    – David Fernandez-Breton
    Commented Mar 3, 2011 at 19:26
  • $\begingroup$ Hi Stefan, I take issue with your claim that « Con$(ZFC)$ is merely an arithmetical statement ». It is a statement about proofs, which are in textbooks and usually defined as objects in set theory - certain sequences of formulas. It is also customary to "arithmetize" all that, but if the term "arithmetize" is used it is because a priori formulas/proofs are not arithmetical objects and statements about them are not arithmetical statements. Thus i would say that Con$(ZFC)$ is a metatheory statement, either taken as axiom or provable in the base metatheory depending on its consistency strength. $\endgroup$
    – plm
    Commented Oct 17 at 2:19
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If there is any ordinal $\alpha$ such that $L_\alpha$ satisfies ZFC, then consider the least one. This is some countable ordinal $\beta$. It is easy to show that $L_\beta \vDash$ "There is no transitive model of ZFC." However, by absoluteness, it will still think that ZFC is syntactically consistent, and therefore has a (non-well-founded) model. Actually this also shows that any transitive model of ZFC with rank higher than this $\beta$ thinks that there is a transitive model of ZFC, so the existence of transitive models fails only for the very shortest ones.

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