Timeline for Is it necessary that model of theory is a set?
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| Mar 6, 2011 at 8:00 | comment | added | David Fernandez-Breton | So there's no problem here since even if the set model satisfies the ZFC axioms, it need not be an exact copy of the "real" universe of sets... sorry about the really really long comment, but I felt it inappropriate to post a new answer so many time after the question was done. | |
| Mar 6, 2011 at 7:59 | comment | added | David Fernandez-Breton | ...looks less paradoxical, since this set model won't actually be the universe of all sets, it's only going to be certain set with certain binary relation (which the model itself will "think" is the membership relation) that satisfies the axioms. Of course, we expect the "actual" universe of sets to satisfy the axioms as well, but this and the set models for set theory can differ in some other statements that don't follow from the axioms (so for instance, we can have a set model satisfying $CH$ while the "real" universe of sets maybe don't satisfy $CH$). | |
| Mar 6, 2011 at 7:55 | comment | added | David Fernandez-Breton |
This only means that given $x,y\in M$, there must be an element $z\in M$, such that the only elements which bear the relation $E$ with $z$ are $x$ and $y$ (this is, we have $(x,z)\in E$, $(y,z)\in E$ and there is no other $u$ such that $(u,z)\in E$). So $M$ doesn't really needs to contain the unordered pair $\{x,y\}$, it just needs to contain something that $M$ will believe" that is the unordered pair. Similarly, $M$ must contain something that it thinks" is the powerset of any of its elements, etc. From this point of view, the fact that there is a model of set theory which is itself a set
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| Mar 6, 2011 at 7:49 | comment | added | David Fernandez-Breton | It is maybe worth pointing out that this model must not only be a set, but in fact an ordered pair $(M,E)$ such that $M$ is a set and $E$ is a binary relation in $M$ which is going to be the interpretation of the symbol $\in$. So if we take, for example, the axiom of pairing, namely $(\forall x)(\forall y)(\exists z)(\forall u)(u\in z\iff u=x\vee u=y)$ (this is the assertion that given sets $x$ and $y$, there exists the set $\{x,y\}$), we have that this set model must satisfy it: but with the intended interpretation, this doesn't mean that given $x,y\in M$, we must have that $\{x,y\}\in M$. | |
| Feb 18, 2010 at 17:12 | comment | added | kakaz | The aim of asking this question is not to discuss about ZFC at all, but about model theory and models. Unfortunately ZFC is incorporated in it, so probably my question is a idea of looking for "new model theory" which do not use set theory at all - clearly without direct answer. But from answer of Joel D.Hamkins we see, that at least for first order theories consistency is equivalent to having a model as a set, which for me is very interesting and definite statement which is far from being obvious, even if You heard about theorems mentiuoned in comments/answers. | |
| Feb 18, 2010 at 16:51 | history | answered | abcdxyz | CC BY-SA 2.5 |