# What is the definition of the well-founded part of a model of set theory?

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.

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

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 $M$). – Andrés E. 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). – Andrés E. 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