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Given $M$ a wellfounded transitive set model of ZFC, and $x$ a set which is not in $M$, is there always a 'smallest wellfounded transitive model' $M[x]$ of ZFC which extends $M$ and contains $x$?

I believe the answer is NO, because there might not be a 'canonical choice-function' for the set $x$ which we want to add. But I'm not sure how to make that precise.

If the above idea can be made precise and the answer is in fact NO, then the natural follow-up question would be: What if we only demand $M[x]$ to be a model of ZF?

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1 Answer 1

up vote 12 down vote accepted

It is consistent with ZFC that there is no such model at all. For example, when $M$ is countable, let $x$ be a real coding a relation on $\omega$ revealing that the ordinals of $M$ are countable. Thus, any model $N\models{\rm ZFC}$ containing $M$ and the object $x$ will also view the ordinal height of $M$ as a countable ordinal. Thus, the height of $N$ will be strictly taller than the height of $M$. But it is consistent with ZFC that all countable well-founded models of ZFC have the same height, for example, this is true in the second $L_\alpha$ which is a model of ZFC, since all the transitive models inside that model have the same height as the minimal model of ZFC.

And it doesn't matter if you demand ZFC or just ZF.

But on a postive note, if there is at least one model $N$ of ZF extending $M$ and the object $x$, and $x$ is a set of ordinals in $N$, then there is a least model of ZFC containing $M$ and $x$. This can be seen by closing $M$ and $x$ under the Goedel operations inside $N$.

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Regarding your positive node: Is it necessary to demand that $ x $ is a set of ordinals in $ N $? Maybe you can also give a reference to the literature or some research paper for more information on this topic? Thanks in advance! –  Justus87 Jul 17 at 21:11

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