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I've been trying to understand John Steel's various notes on inner model theory, but the one thing that trips me up is what he calls the well-founded part of a model of set theory. What exactly is the well-founded part of a model? If someone could give me a precise definition (maybe it can be defined using transitive closures, but I don't really know) of the well-founded part of a model, it'd be greatly appreciated.

Addendum

The well-foundedness that I'm referring to is not the internal well-foundedness that comes from assuming the Axiom of Regularity within the model. It's an external property, as viewed from outside the model.

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Brian deleted his answer before I could comment, but AC has nothing to do with the von Neumann construction. –  Asaf Karagila Oct 26 '12 at 16:15
    
I must have mistaken it for something else. Which axiom is it that shows that the von Neumann hierarchy equals the whole universe? –  Haskell Curry Oct 26 '12 at 16:17
    
The fact that every set has a rank is sufficient. This follows from regularity and some replacement (for the transfinite induction defining rank). –  Asaf Karagila Oct 26 '12 at 16:18
    
I mistook it for AC. I now remember that it was Regularity instead. I'm assuming Replacement, of course. Thanks! –  Haskell Curry Oct 26 '12 at 16:20
    
That has nothing to do with your question, really. Just a side note to what you wrote under Brian's answer. –  Asaf Karagila Oct 26 '12 at 16:22

1 Answer 1

up vote 3 down vote accepted

Suppose that $(M,E)$ is a model of ZFC, this is a set in the universe (which is also a model of ZFC, for our purposes).

It is possible that $(M,E)$ is not a well-founded relation. Internally, of course, this is impossible. $M$ does not have any element which is a decreasing sequence in $E$, since $M$ satisfies the axiom of regularity.

However we, as educated men staring at $M$ externally, know that it is possible that $M$ has more than it knows about. One can now ask about the ordinals of $M$. Namely $(Ord^M,E)$ as a linear order. This order has a maximal initial segment which is well-founded.

The well-founded part is the initial part [internally] of $(M,E)$ which is truly well-founded. It is exactly the sets whose [internal] von Neumann rank is an ordinal in the well-founded part of $(Ord^M,E)$.

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Asaf: It is not just the ordinal part. It is the subclass of $M$ consisting of those sets $x\in M$ such that $E$ restricted to $x$ is well-founded. This is a model of KP (though in general, not an inner model of $M$, or even a definable subclass of $M$). –  Andres Caicedo Oct 26 '12 at 16:22
    
@Andres: But the ranks of the well-founded sets are exactly the well-founded ordinals, right? –  Asaf Karagila Oct 26 '12 at 16:23
    
I remember John Steel himself mentioning that the ordinal part is a subset of the well-founded part. Is it possible that we are looking at the external transitive closure of the ordinal part? Just to inform you, I may not be making sense here. –  Haskell Curry Oct 26 '12 at 16:25
    
To be somewhat more precise, given $x\in M$, we can identify it with $x^*=\{y\in M\mid M\models y E x\}$. When I talk of the transitive closure of $x$, I really mean its $*$ version, so we look at $x$, and $x^*$, and $\{z\in M\mid M\models z E \bigcup x\}$ and $\dots$. If $E$ restricted to this collection is well-founded, then we say that $x$ is in the well-founded part of $M$. (From this it easily follows that yes, the ranks of well-founded sets are the well-founded ordinals of the model). –  Andres Caicedo Oct 26 '12 at 16:25
    
@Andres: I think that you're much more capable of writing an answer. I'll add the important correction, but I'd still love reading an answer by you. –  Asaf Karagila Oct 26 '12 at 16:27

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