Our question arises from wondering about the systems of natural numbers in models ZFC + Con(ZFC) and ZFC + $\neg$Con(ZFC). In thinking of the systems of natural numbers of these models, we came to the following remark and questions. If we take the ultrapower of a model of Peano arithmetic we get a model of Peano arithmetic with non-standard natural numbers like <1, 2, 3, ...> which have infinitely many natural numbers below them. My question is: if we take an ultrapower, $M$, of a model of ZFC, $V$, does $M$ have non-standard natural numbers?
$\begingroup$
$\endgroup$
3
-
3$\begingroup$ It depends on what ultrafilter you are using. If you mean a non-principal ultrafilter on a countable set, for example, the answer is yes, because you can define a model of PA in ZFC. But you could take a (non-principal) $\sigma$-complete ultrafilter on some large index set, and then the ultrapower of the standard model of PA is standard, and the ultrapower of the ZFC model is also standard. See also mathoverflow.net/questions/56479/… $\endgroup$– Andrés E. CaicedoCommented Jul 25, 2012 at 2:43
-
$\begingroup$ But which models of PA do you define in models of PA defined in the two models from the question: ZFC + Con(ZFC) and ZFC + $\neg$Con(ZFC)? I am interested, yes, in a non-principal ultrafilter. $\endgroup$– user10290Commented Jul 25, 2012 at 3:02
-
$\begingroup$ Note: I think that ulpowers of the univese will have the standard natural numbers, but for a reason different then those given so far. $\endgroup$– user10290Commented Jul 25, 2012 at 3:06
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
An ultrapower of a model $M$ of ZFC will have non-standard natural numbers if and only if either $M$ has non-standard natural numbers or the ultrafilter used in forming the ultrapower is countably incomplete. The natural numbers of the ultrapower of $M$ with respect to an ultrafilter $U$ are isomorphic to the ultrapower with respect to U of the natural numbers of $M$. (More generally, the operations of $U$-ultrapower and taking the subset of a model defined by a particular formula always commute up to canonical isomorphism.) The ultrapower will satisfy Con($PA$) if and only if $M$ does, by Los's theorem.
The preceding was written with "ordinary" ultrapowers in mind. You could also form "internal" ultrapowers of a model $M$ of ZFC. You'd use an element $U$ of $M$ that satisfies in $M$ the formula "$U$ is an ultrafilter on a set $I$", and elements of the ultrapower would be represented by functions on $I$ in the sense of $M$ (i.e., elements $f$ of $M$ that satisfy in $M$ the formula "$f$ is a function on $I$"). Everything I wrote in the first paragraph would still be correct provided you interpret "countably complete" in the sense of $M$.
Finally, one can, in some situations, form ultrapowers where the ultrafilter is external to the model $M$ but the functions representing ultrapower elements are internal to $M$. Similar (but not identical) results hold for these also.
answered Jul 25, 2012 at 3:22
$\begingroup$
$\endgroup$
2
In each copy of $V$ there's some set is a model of $PA$. This sequence of sets corresponds to a set in $M$ which, since being a model of $PA$ is a sentence in the language of $ZFC$, is a model of $PA$. Since there is a unique model of $PA$, this is the model of $PA$.
This set is an ultrapower of a model of $PA$, so it contains nonstandard natural numbers. (Edit: As Andreas points out, only if the ultrapower is countably incomplete.)
answered Jul 25, 2012 at 2:34
-
$\begingroup$ Your answer depends on a unique model of $PA$. But, also you are saying that this model of $PA$ in the ultrapower has non-standard numbers. Does this mean that the unique model of $PA$ has non-stardard natural numbers. $\endgroup$– user10290Commented Jul 25, 2012 at 2:57
-
$\begingroup$ Non-standard from the global point of view in which $M$ is "a model" and not "the entire universe". From the point of view of $M$ it's standard. $\endgroup$ Commented Jul 25, 2012 at 14:10