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The assumption that there exists a standard model of ZFC in a given universe is stronger than the assumption that there exists a model.

So, existence of a model of ZFC and existence of a standard model of ZFC are only assumption, not theorem?

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    $\begingroup$ What is a standard model of ZFC? $\endgroup$ – William May 30 '12 at 1:19
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    $\begingroup$ You should improve your question by linking or giving the definitions of terms that aren't wide know. That'll help clarifyng the reading for those who will answer your question and for people who may use your question to get rid of their doubts. $\endgroup$ – Paulo Henrique May 30 '12 at 1:40
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    $\begingroup$ If you quote something, please provide the source. $\endgroup$ – Michael Greinecker May 30 '12 at 7:14
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By model theory, the existence of a model of ZFC is equivalent to ZFC being consistent. By the incompleteness theorem, ZFC cannot prove its own consistency. Therefore, the existence of a model of ZFC is not a theorem of ZFC---if that is what you are asking about.

I am not sure whether "standard model" is a rigorous term. For example, the term standard model of the Peano's arithmetic refers to $\omega$, the usual natural numbers. I guess vaguely the term "standard" would mean the structure from which some theory was inspired by. However, by the above, it is not known whether ZFC has a model, so I can't imagine what a standard model would mean. However, if ZFC does have a model, then there are models with special properties like $V = L$.

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    $\begingroup$ A standard model of ZFC is one in which the model's $\epsilon$ relation is the real $\epsilon$ relation restricted to the domain of the model. This is a use of "standard model" in a different sense than "standard model of PA" - there may be many standard models of ZFC. $\endgroup$ – Alex Kruckman May 30 '12 at 1:31
  • $\begingroup$ @AlexKruckman When you mean real $\in$, you are assuming that $ZFC$ already has a model, right? So this definition of standard model is relative to some model of $ZFC$. So this means that if $\mathcal{M}$ is a model of ZFC, then $\mathcal{M}$ itself is a standard model relative to the "real" $\in$ relation of $\mathcal{M}$. $\endgroup$ – William May 30 '12 at 1:37
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    $\begingroup$ No, I'm not making that assumption. I am assuming that our metatheory is ZFC. Now a model of ZFC is a set $M$ together with an interpretation of a symbol, call it $\varepsilon$ to distinguish it from the symbol $\in$ of the metatheory, as a binary relation on $M$ such that $(M,\varepsilon)\models \text{ZFC}$. Now the statement "there is a standard model of ZFC" is the assertion: There exists $(M,\varepsilon)\models \text{ZFC}$ and $\forall x,y\in M (x\,\varepsilon\,y)\leftrightarrow (x\,\in\, y)$ $\endgroup$ – Alex Kruckman May 30 '12 at 4:21
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We usually assume that there is a grand universe of sets which is not a set, and we often assume that this universe is a model of ZFC. It does not prove ZFC is consistent, though, because it is too large for us to capture.

It may be the case, however, that inside our universe of sets there is a set $M$ and a binary relation $E$ over $M$ such that $\langle M,E\rangle$ is a model of set theory. This is to say that ZFC is consistent, or that it has a model.

From the point of view of $M$ the relation $E$ is $\in$. However, we as all knowing beings, know that $E$ may be something else.

We say that $M$ is well-founded if $E$ is well-founded as a relation in the universe. It is important to remark that $M$ always thinks that $E$ is well-founded, but it might be the case that $M$ does not know about any decreasing sequence.

Well-founded models are good. They are good because we have a theorem known as the Mostowski collapse lemma which tells us that if $\langle A,R\rangle$ is well-founded then there is some $B$ such that $\langle A,R\rangle\cong\langle B,\in\rangle$ as ordered sets, and $B$ is transitive (i.e. $x\in B$ and $y\in x$ imply that $y\in B$).

Now suppose that $\langle M,E\rangle$ is a well-founded model of ZFC, namely it is a model of ZFC and $E$ is really well-founded, in this case we can collapse it and have some $\langle N,\in\rangle$ which is a model of ZFC and $\in$ is the restriction of $\in$ to $N$. Such models are standard models of ZFC.

Now comes the punch: the existence of a well-founded model is stronger than simply the existence of a model. (Since the true $\in$ is well-founded, standard models are always well-founded.)

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    $\begingroup$ It does seem intuitively clear that the existence of a wellfounded model is stronger than a non-wellfounded model, but how does one prove this? Is there some simple consequence of the existence of wellfounded models that does not follow from the existence of non-wellfounded models? $\endgroup$ – Zhen Lin May 30 '12 at 7:02
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    $\begingroup$ @Zhen: See this answer on MathOverflow. $\endgroup$ – Asaf Karagila May 30 '12 at 7:06
  • $\begingroup$ @Zhen: See also this and the links in the introduction (you may also want to read about Worldly cardinals, but that's not related to this question here). $\endgroup$ – Asaf Karagila May 30 '12 at 7:10
  • $\begingroup$ Could you please elaborate on what you mean by that a proper class (like $V$) "is too large for us to capture"? My understanding is that for consistency to follow from a thing we somehow know is a model, that thing needs to be a set within the theory in question, i.e., from its own perspective (?). A class model doesn't work for deducing consistency, but I don't understand why. $\endgroup$ – Ioannis Filippidis Dec 8 '16 at 19:37
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    $\begingroup$ @Ioannis: You are correct, and that is what I meant by the comment. Class models are not sets, so they cannot be used to prove consistency within the theory. If they could, then $V$, being a model of ZFC, would have always satisfied that ZFC is consistent. $\endgroup$ – Asaf Karagila Dec 9 '16 at 5:53

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