For transitive model $M$, is $L^M$ really a model of ZFC?

There are two concepts of "models" in set theory: one says that for each axiom $\phi$ of ZFC, $\phi^M$ holds. Remaining one just says that $M\models \mathsf{ZFC}$. First one is a schema rather a single statement. Second one is a single statement and $M$ possibly satisfies the non-standard axioms of ZFC.

We know that $L$ is an inner model of ZFC, that is, every relativized axiom over $L$ holds. My question is, if $M$ is a transitive model of ZFC, then $L^M\models \mathsf{ZFC}$? $L^M$ should be $L_\gamma$ for some $\gamma$, and $L^M$ satisfies all "standard" axioms of ZFC. Can it possible that some non-standard axiom fails on $L^M$? Thanks for any help.

• What axiom (schema) are you using? There are multiple common axiomatizations of ZFC in use, AFAIK. And relatedly, how are you defining 'non-standard' axioms; as axioms whose Godel numbers correspond to non-standard integers? Those may not be well-formed, and it's not immediately clear how one can specifically speak of non-standard axioms... – Steven Stadnicki Jan 2 '16 at 11:48
• @StevenStadnicki I wonder what axioms I used are important in my question, but for convenience, I assume the axioms in Wikipedia. – Hanul Jeon Jan 2 '16 at 11:53

2 Answers

Given an axiom $\varphi$ of $\sf ZFC$, it is provable from $\sf ZF$ that $\varphi^L$ holds. This means that as a meta-theorem, if $M$ is a model of $\sf ZF$, then $L^M$ is a model of $\sf ZFC$.

You are correct to assume that if $M$ is a transitive model, then $L^M$ is $L_\gamma$ for some $\gamma$. Or if $M$ is a class, then $L^M$ is just $L$ again.

But then your question takes a turn into unclear waters. Non-standard axioms occur when you have a non-standard model, with non-standard integers, and internally it has new axioms of $\sf ZFC$ which need not hold externally. If $M$ is transitive, then this is certainly not the case.

Moreover, the truth is a meta-theoretic concept, namely it is not part of $M$ but rather a part of $V$. So asking whether or not a statement is true or false is done in the meta-theory, and non-standard axioms live - by definition - internally to non-standard models, which are models which disagree with the meta-theory by definition.

So in other words, the meta-theory is not aware of any "non-standard axioms", so it certainly cannot tell you whether or not they are true or false in $L^M$, even if $M$ was a non-standard model.

• I concern when $V$ is non-standard universe, that is, when $V$ has a non-standard natural numbers. As far as I imagine, when $V$ is non-standard, our formalized ZFC could have non-standard axiom, but meta-theory does not recognize such non-standard ones. However, even if that case your answer implies that $L^M$ could not satisfy whole $\mathsf{ZFC}$ formalized in $V$, right? – Hanul Jeon Jan 2 '16 at 12:02
• But the point is that it doesn't matter. The proof that $L^M$ is a model of $\sf ZFC$ happens in $V$, and therefore it satisfies all the axioms that $V$ knows about. You want to take this proof to the meta-meta-theory, and then argue that because you only proved this in the meta-meta-theory, there are axioms from the meta-theory which might not hold in $L^M$. But the proof goes through in the meta-theory, so even if you do jump to the meta-meta-theory, it will not invalidate the one from the meta-theory. – Asaf Karagila Jan 2 '16 at 12:05
• I am bit confused. I have not considered or thought the meta-meta-theory. In my belief, all natural numbers in meta-theory could be standard unless $V$ has non-standard natural numbers, and every (standard?) formulas for $V$ are coded by naturals in the meta-theory. I conceive and ask my question over that belief. Have I gotten the wrong belief? – Hanul Jeon Jan 2 '16 at 12:21
• We say that a model has non-standard integers if it disagrees with its meta-theory. So by definition, the meta-theory only has standard integers. If you want to involve meta-meta-theory, then you can have a meta-theory which disagrees with its meta-theory (the meta-meta-theory), but then everything becomes messy, and there's very little point in doing so. So mostly we work inside $V$ with a model $M$. So $V$ cannot have non-standard integers. – Asaf Karagila Jan 2 '16 at 12:50
• I noticed you unaccepted. That's fine, but I'd be happy to know what I can do to give you a better answer. – Asaf Karagila Jan 16 '16 at 15:13

The problem I imagine is that formulas are not an object in $V$. As far as I know formulas live outside $V$. As I described it makes some technical problem.

For example, consider the theory $$ZFC + \exists c:\text{c is countable transitive and \phi^c for each axiom \phi of ZFC}.$$ (It is described in Ch. 9, section 9 (1b) in Kunen's old set theory. I have seen its name but I forgot it.) Such theory is not $\omega$-consistent and if $V$ is a model of such theory then $V$ should have non-standard naturals. (Is it right?) Now we gaze the inside of $V$. By Gödel, $c$ does not think that itself satisfies (formalized) ZFC, though it satisfies all "standard" axioms of ZFC.

I think my question is related such situation. When proving $L$ is an inner model of ZFC, we provide $\phi^L$ holds for each axiom $\phi$. But I realized that if we apply same argument to proving $L^M$ is an inner model of $M$ then the argument says that for each (coded) axioms of ZFC $\ulcorner\phi\urcorner$, $M\models \ulcorner\phi^L\urcorner$. By induction for $\ulcorner\phi\urcorner$ we can check that $M\models \ulcorner\phi^L\urcorner$ if and only if $L^M\models \ulcorner\phi\urcorner$. Thus $L^M$ is really a model of ZFC (and in fact a model of $ZFC+ V=L$.)

Is my reasoning right?

• The idea is generally correct. Some remarks: (1) The theory is due to Feferman, not sure if it has a common name, though; (2) if $V$ satisfies the theory it's not necessarily the case that $V$ should have non-standard integers, but it might be the case; (3) $M^L$ should be $L^M$, I'm guessing. – Asaf Karagila Jan 17 '16 at 21:32
• @Asaf Yep, $M^L$ is a typo of $L^M$ so I will edit it. – Hanul Jeon Jan 18 '16 at 5:52