Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It is clear that given a family $(\mathfrak{A}_i)_{i\in I}$ of $L$-structures their ultraproduct may depend on the choice of the ultrafilter (for this question I am only considering non-principal ultrafilters).

Is there an easy example as to why given and $L$-structure $\mathfrak{A}$ the choice of the (nonprincipal) ultrafilter $\mathcal{U}$ on $I$ may affect the isomorphism class of $\mathfrak{A}^I/\mathcal{U}$?

share|cite|improve this question
up vote 9 down vote accepted

Let $L$ have unary predicate symbols $\dot X$ for all subsets $X$ of $\mathbb N$, and let $\mathfrak A$ be the structure with universe $\mathbb N$ in which each $\dot X$ has the obvious interpretation $X$. In any elementary extension $\mathfrak B$ of $\mathfrak A$, define the type of any element $b$ to be the set of those $X\subset\mathbb N$ for which $\dot X(b)$ is true in $\mathfrak B$.

Now consider ultrapowers of $\mathfrak A$ by ultrafilters $\mathcal U$ on $\mathbb N$. Notice that $[i]_{\scr U}$, the equivalence class in $\mathfrak A^{\mathbb N}/\mathcal U$ of the identity function $i$, has type $\mathcal U$. So every ultrafilter on $\mathbb N$ occurs as the type of an element in such an ultrapower. But any one ultrapower realizes only $2^{\aleph_0}$ types, as it has only $2^{\aleph_0}$ elements. And isomorphic ultrapowers realize the same types. So to realize all $2^{2^{\aleph_0}}$ non-principal ultrafilters on $\mathbb N$, there must be $2^{2^{\aleph_0}}$ non-isomorphic ultrapowers.

If you are unhappy with my use of an uncountable language $L$, then it becomes necessary to work harder, and in particular to use uncountable index sets $I$, but it is still possible to get non-isomorphic ultrapowers of any structure, even for a countable language --- indeed even for the language consisting of just equality. The key phrase to look up here is "regular ultrafilter".

share|cite|improve this answer
Ha, great, this was very clear. This then makes me think of two other questions though. 1) In the Countable language and countable index case is there always one ultrapower modulo isomorphism? 2) For a more general case are there necessary and/or sufficient conditions known for a structure $\mathfrak{A}$ such that they imply that the ultrapower is unique, independent of the size of the index? Maybe if there are none, is there like a middle ground by taking some conditions on $\mathfrak{A}$ and conditions on the size of $I$ depending on $\mathfrak{A}$? – Santiago C. Jan 17 '13 at 22:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.