TITLE: Why does CH imply that there is a unique ultrapower of $\mathbb{N}$?
QUESTION [19 upvotes]: 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"?

REPLY [8 votes]: 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.