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I have a naive and open-ended question:

How can one remove a set from a model of set theory in such a way that the result is again a model of set theory? Directly related: what kinds of sets can be removed?

For example: If $U$ is an ultrafilter on a measurable cardinal $\kappa$ then the ultrapower $M:=$Ult$(V,U)$ is a model of set theory that does not contain $U$. In an abuse of notation, this model is occasionally written as $V^{\kappa}/U$, mimicking the notation for modding out by some relation.

Another example: If $M[G]$ is a forcing extension, then $M$ would be an inner model with the set $G$ removed. However, for the life of me I don't see a way to find $M$ from within $M[G]$ even if the original forcing poset has been identified.

Are there other examples of this?

Murphy
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    "this model is occasionally written as $V/U$, mimicking the notation for modding out by some relation." It should be written instead as $V^\kappa/U$, since the ultrapower is literally the result of modding out $V^\kappa$ by the equivalence relation of equality modulo $U$. – Andrés E. Caicedo Mar 24 '14 at 07:36
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    $M$ is definable within $M[G]$. This is a result of Laver. See here. – Andrés E. Caicedo Mar 24 '14 at 07:38
  • Thanks for the correction and the link. I'm looking at Laver's paper now. It appears though that one must already know that M[G] is a forcing extension. So given a model N containing a poset $\mathbb{P}$ and a candidate for a set that may have been added by $\mathbb{P}$, one may not necessarily be able to "undo" the forcing. I assume this relates to the Ground Axiom. – Murphy Mar 24 '14 at 08:20
  • @Murphy: You can, since ground models are definable from parameter, ask if there exists a set that as a parameter defines a class which satisfies $\sf ZFC$, and in that class there is a partial order and it has a generic in the [full] universe, and that generic extension is everything. This is an immensely long statement, but it is a statement in the language of set theory. – Asaf Karagila Mar 24 '14 at 09:29

2 Answers2

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What kind of sets can be removed is a slightly strange question.

Two models of $\sf ZFC$ with the same sets of ordinals (and in particular, the same ordinals) are equal. Therefore if the inner model is strictly different than the universe, we have removed sets of ordinals, sets of sets of ordinals, and so on.

Think of a rank function, where sets of ordinals are of rank $0$, and sets of rank $\alpha+1$ are sets whose element are all of rank $\leq\alpha$; and at limit steps we consider sets whose elements are of ranks unbounded below $\alpha$. It is clear that if we remove sets of rank $\alpha$ then we remove sets of every rank above it. And if we add sets of rank $\alpha$, then we add sets of every rank above it as well.

So as it turns out, in models of $\sf ZFC$ rank $0$ determines everything (it is not necessarily the case in models of $\sf ZF$, though).

If so, what sort of sets of ordinals can we remove? Well, we can sometimes remove real numbers, and other times we can remove subsets of $\omega_1$, and other times we can remove subsets of larger cardinals, and so on. In fact, sometimes (e.g. when the universe is a class forcing extension) we can remove a subset from each and every infinite cardinal.

Of course, sometimes we can do better. In ultrapowers of the universe, as you mentioned, we keep all the subsets of $\kappa$, but we can rather identify a particular set of subsets of $\kappa$ that is removed. Namely, the ultrafilter that was used to construct the ultrapower.

Similarly, as I wrote in my comment, we can identify that the universe is a generic extension over some inner model. And then we can identify sets which are not in that inner model due to their genericity.

However, we can't remove all the sets using the process of taking inner models (we can, sometimes, remove sets by chopping off the universe, or going to a set transitive model inside the universe, but I assume that's not quite the answer you're looking for). For example, if $x\in L$ then we cannot possibly remove $x$ by taking inner models; in particular we cannot change the ordinals themselves. Beyond that, it's hard to give a concrete answer, because pretty much everything is possible (that is, if $x\notin L$ then there is an inner model which omits $x$, namely $L$ itself).

Asaf Karagila
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Here are some other ways to get models of $ZFC$ that exclude certain sets.

If your set occurs at sufficiently high $V_\kappa$ and there are sufficiently many inaccessibles below this $\kappa$, sometimes $V_\lambda$ for some $\lambda < \kappa$ inaccessible is a model of $ZFC$ which does not have a particular set.

For example if $U$ is a ultrafilter on a measureable cardinal $\kappa$. There are inaccessible $\lambda < \kappa$. $V_\lambda \models \text{ZFC}$ and $U \notin V_\lambda$. By some absoluteness, if you do this at the least measureable cardinal, you can even model with no measureable cardinals.

Also if $0^\sharp$ exists and the set you want to remove is uncountable, then $L_{\omega_1} \models ZFC$ and this set is not in $L_{\omega_1}$.

William
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