Existence of an $\omega$-nonstandard model of ZFC from compactness - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T19:22:53Z http://mathoverflow.net/feeds/question/40821 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/40821/existence-of-an-omega-nonstandard-model-of-zfc-from-compactness Existence of an $\omega$-nonstandard model of ZFC from compactness Marc Alcobé García 2010-10-02T09:21:21Z 2010-10-02T12:33:17Z <p>I have read several times that assuming Con(ZFC), and using compactness it can be proved the existence of a model of ZFC with an ill-founded $\omega$. How is that? Any reference will be welcome.</p> http://mathoverflow.net/questions/40821/existence-of-an-omega-nonstandard-model-of-zfc-from-compactness/40826#40826 Answer by Joel David Hamkins for Existence of an $\omega$-nonstandard model of ZFC from compactness Joel David Hamkins 2010-10-02T11:10:11Z 2010-10-02T11:10:11Z <p>This is a standard application of the Compactness Theorem, and works basically the same in producing nonstandard models of ZFC as it does for producing nonstandard models of PA or real-closed fields.</p> <p>Consider the theory $T$, in the language of set theory augmented with an additional constant symbol $c$, consisting of all the ZFC axioms, plus the assertions that $c$ is a natural number, but not equal to $0$, not equal to $1$, and so on, including for each natural number $n$ the statement $\varphi_n$ that $c$ is not equal to $n$. (Note that all such $n$ are definable in set theory, and so we use the definition of $n$ in $\varphi_n$ when asserting that $c$ is not $n$.)</p> <p>If there is a model $M$ of ZFC, then every finite subtheory of $T$ is consistent, since any finite subtheory of $T$ makes only finitely many assertions about $c$, and we may therefore interpret $c$ as any natural number of $M$ not mentioned in the subtheory.</p> <p>Thus, by compactness, $T$ has a model. Any such model of $T$ will be $\omega$-nonstandard, since the interpretation of $c$ in the model will be a nonstandard natural number.</p> <p>There are numerous other ways to produce $\omega$-nonstandard models of ZFC from existing models, the most common being ultrapowers.</p> <p>The conclusion is that if there is any model of ZFC, then there are nonstandard models of ZFC. The converse of this is not true, for if there are any transitive models of ZFC, then there is an $\in$-minimal such model $M$, and being standard, $M$ will have the same arithmetic as the ambient universe, thus thinking Con(ZFC), but being minimal, will have no transitive models of ZFC inside it.</p> http://mathoverflow.net/questions/40821/existence-of-an-omega-nonstandard-model-of-zfc-from-compactness/40830#40830 Answer by Carl Mummert for Existence of an $\omega$-nonstandard model of ZFC from compactness Carl Mummert 2010-10-02T12:30:43Z 2010-10-02T12:30:43Z <p>I'm not sure, but maybe the question is why a non-&omega;-model of ZFC must have infinite decreasing $\in$ chains. This follows from two facts:</p> <ul> <li><p>Viewed externally, any nonstandard model of Peano arithmetic is non-well-founded. Given a nonstandard number $n$, consider the sequence $n-1$, $n-2$, ...</p></li> <li><p>The standard construction of a model of Peano arithmetic within ZFC uses $\in$ for the order relation on the natural numbers. So if the natural numbers within a model of set theory are not well founded, then the model's set inclusion relation is also not well founded. These are, formally, different meanings of "well founded". </p></li> </ul> <p>It's interesting to ask how the model can think that its natural numbers are well founded, given that the descending sequence above was completely concrete. The answer is that if that sequence is defined within the model, as $a_k = n-k$, then $k$ can be any number in the model, even a nonstandard one, and the model verifies that this sequence reaches $0$ after $n$ steps. Only from an external viewpoint can we limit $k$ to an external set of "standard" natural numbers. </p>