As I am studying set theory, I came to realize that there exists a countable "well-founded" model of ZFC. But I am curious whether countable models can ever be well-founded externally. What would be the way to see whether this is the case or not?

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    $\begingroup$ If it's transitive, then it's automatically well-founded seen from the outside. $\endgroup$ – Henning Makholm Mar 10 '14 at 11:35

When working in set theory, we have a universe of sets, $V$ with an $\in$ relationship. And we work inside of that universe. Since this universe is usually assumed to satisfy the axioms of $\sf ZFC$ this means that $V$ is well-founded, because as a universe of sets, it is the class of all sets.

When we say that $(M,E)$ is a model of $\sf ZFC$ then we say that $M$ and $E$ are sets in $V$ and that $V$ thinks that all the axioms of $\sf ZFC$ are true in $(M,E)$. It should be pointed that $V$ may have "its own version" of $\sf ZFC$ which differs than ours, but let's deal with one obstacle at a time.

When we say that $(M,E)$ is a well-founded model, then we mean that $E$ is a well-founded relation. When we say that $M$ is a transitive model, then we mean that $E=\in$. That is, the membership relation is the same as that of $V$. Since $V$ satisfies the axiom of foundation, this means that $\in$ is well-founded, so $M$ is well-founded.

It should be pointed that every well-founded model of $\sf ZFC$ is isomorphic to a transitive one (and the isomorphism is unique, too). This is by the Mostowski collapse lemma.

  • $\begingroup$ So OK. let's say we work under $(V, \in)$ where $\in$ is well-founded and satisfies ZFC axioms. Then does countable model $(M, \in)$ exist? This is what the answer seems to say.. $\endgroup$ – quanti Mar 10 '14 at 11:42
  • $\begingroup$ No, it doesn't. It just says that when we say that a model is transitive, then we mean bla bla bla. Since $\sf ZFC$ cannot prove its own consistency, it cannot prove that there exists a model, let alone a transitive model (which is a far stronger condition). $\endgroup$ – Asaf Karagila Mar 10 '14 at 12:10
  • $\begingroup$ I was meaning that if we assume that the universe $(V, \in)$ that is a class (or set) model of ZFC exists, then does the existence of $(M,\in)$ with same $\in$ a $V$ follow from existence of $V$? $\endgroup$ – quanti Mar 10 '14 at 12:19
  • $\begingroup$ Again. No. If that would be the case then $\sf ZFC$ would prove far more than its own consistency. Which is not the case, due to Godel's incompleteness theorem. Looking at your other question, it's quite a huge leap in five weeks from basic predicate logic, to countable models of set theory. Perhaps it is better to build a stronger basis before climbing up the pyramid? $\endgroup$ – Asaf Karagila Mar 10 '14 at 12:21
  • $\begingroup$ @AsafKaragila: If we had a set size model, then shouldn't the existence of a countable model follow by Downward Lowenhiem Skolem? However if you have a class sized model, then the answer would be know because of issue related to Tarski's undefinability of truth $\endgroup$ – UserB1234 Mar 12 '14 at 0:08

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