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I've read these words: "How many ultra products $∏_Uℕ$ exist up to isomorphism, where $U$ is a non-principal ultrafilter over $ℕ$? If continuum hypothesis(CH) holds, then obviously just one ..."

i don't know why "obviously" just one? CH just tell the cardinality, not the structral of ultrafilter. even $U_1$ and $U_2$ have same cardinality c, how to make sure their structural can't be different?

where can I find the proof detail or the insight which make this "obvious"?

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3  
For a significant, and non-trivial generalizations, see here. –  Andres Caicedo Jul 15 '13 at 2:35

2 Answers 2

The point is that the ultrapower of any structure $\mathcal{M}$ by a nonprincipal ultrafilter $\mu$ on $\mathbb{N}$ is countably saturated, that is, it realizes any $n$-type with a countable number of parameters, by simple argument: if $p(x)$ is a 1-type consistent with the diagram of the ultrapower of $\mathcal{M}$, a structure in a countable language, by the ultrafilter $\mu$, and $p(x)$ uses only countably many parameters from $\mathcal{M}$, then for each $i$ let $x_i\in M$ witness the first $i$ many instances of the type, so that the function $i\mapsto x_i$ realizes all the assertions of the type for all sufficiently large $i$, and thus the function $\langle x_i\mid i\in\mathbb{N}\rangle$ realizes the whole type. So now by induction we realize all consistent $n$-types.

It follows now from CH that if any two ultrapowers of a countable structure $\mathcal{M}$ by $\mu_0$ and $\mu_1$ are both saturated models of size $\omega_1$, with the same theory, then being saturated, they are isomorphic by a back-and-forth argument. We enumerate both of them in type $\omega_1$, and build up an isomorphism in stages by mapping the next element of one side to any element of the other realizing the same type over the elements that have been determined in the isomorphism so far.

So the general conclusion---the content of the "obvious" claim that you mention---is that under CH, there is up to isomorphism at most one ultrapower of any countable structure in a countable language by a nonprincipal ultrafilter on $\mathbb{N}$, and indeed, at most one ultrapower up to isomorphism of any structure in a countable language of size at most $\omega_1$.

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Under the continuum hypothesis, one only obtains a result about the uniqueness of the ultrapower, but the continuum hypothesis does not imply that there is up-to-isomorphism one ultrafilter on a countable set. The notion of Rudin-Keisler equivalence formalizes the notion of whether two ultrafilters are structurally equivalent. In fact, two ultrafilters are Rudin-Keisler equivalent if and only if they are isomorphic in the category of ultrafilters. It is well known that for every infinite cardinal $\kappa$, there are $2^{2^{\kappa}}$ Rudin-Keisler types of ultrafilters on sets of size $\kappa$.

Furthermore, the uniqueness of the ultrapower only follows if you have countably many function, relation, and constant symbols. However, we do not obtain a uniqueness result if you consider uncountably many function symbols, relation symbols, and constant symbols.

Let $\mathfrak{C}(\mathbb{N})$ be the structure over the natural numbers where function is a fundamental operation and every relation is a fundamental relation. Then there are $2^{\aleph_{0}}$ function and relation symbols. In this case, there are $2^{2^{\aleph_{0}}}$ non-isomorphic ultrapowers $\mathfrak{C}(\mathbb{N})^{\mathcal{U}}$. This is because there are $2^{2^{\aleph_{0}}}$ distinct Rudin-Keisler types of ultrafilters on a countable set, and if $\mathcal{U},\mathcal{V}$ are ultrafilters on a countable set, then $\mathfrak{C}(\mathbb{N})^{\mathcal{U}}$ and $\mathfrak{C}(\mathbb{N})^{\mathcal{V}}$ are isomorphic if and only if the ultrafilters $\mathcal{U}$ and $\mathcal{V}$ are Rudin-Keisler equivalent.

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