# Cardinality of a set of pairwise non-order-isomorphic ultrafilters on $\omega$

It is well known that there are $2^{2^{\aleph_0}}$ many non-principal ultrafilters on $\omega$. Is there a set ${\frak U}$ of non-principal ultrafilters on $\omega$ with $|{\frak U}| = 2^{2^{\aleph_0}}$ such that for ${\cal U}_1\neq {\cal U}_2\in{\frak U}$ the partially ordered sets $({\cal U}_1,\subseteq)$ and $({\cal U}_2,\subseteq)$ are not order-isomorphic?

• Two ultrafilters $\mathcal{U}_{1},\mathcal{U}_{2}$ on $\omega$ are isomorphic as posets if and only if they are Rudin-Keisler equivalent. There are $2^{2^{\aleph_{0}}}$ different Rudin-Keisler classes of ultrafilters on $\omega$. – Joseph Van Name Jul 19 '17 at 13:57
• Oh - the Rudin-Keisler equivalence is just what I needed, thanks! Can you quickly put this into an answer? – Dominic van der Zypen Jul 19 '17 at 14:10
• Rudin–Keisler equivalence of uniform ultrafilters clearly induces an isomorphism of the two partially ordered sets, but I don’t see why the converse should hold. In fact, I don’t even see why it couldn’t be the case that all nonprincipal ultrafilters on $\omega$ are isomorphic as partial orders. – Emil Jeřábek Jul 19 '17 at 15:46
• I gave a proof of how a poset isomorphism induces a Rudin-Keisler isomorphism. – Joseph Van Name Jul 19 '17 at 18:39

Let me prove that a two ultrafilters $\mathcal{U}_{1},\mathcal{U}_{2}$ on $\omega$ are isomorphic as posets if and only if they are Rudin-Kielser equivalent. Suppose that $\phi:\mathcal{U}_{1}\rightarrow\mathcal{U}_{2}$ is a poset isomorphism. Then whenever $x\in\omega$, the element $\{x\}^{c}$ is a coatom in the poset $\mathcal{U}_{2}$. Therefore, $\phi(\{x\}^{c})$ is a coatom in the poset $\mathcal{U}_{2}$, so $\phi(\{x\}^{c})=\{y\}^{c}$ for some $y\in\omega$. Therefore, there is a unique function $f:\mathcal{U}_{1}\rightarrow\mathcal{U}_{2}$ such that $\phi(\{x\}^{c})=\{f(x)\}^{c}$. Similarly, there is a unique function $g:\mathcal{U}_{2}\rightarrow\mathcal{U}_{1}$ such that $\phi^{-1}(\{x\}^{c})=\{g(x)\}^{c}$. However, it is easy to see that the functions $f,g$ are inverses. Suppose that $R\in\mathcal{U}_{1}$. Then I claim that $\phi(R)=f[R]$.
If $x\not\in R$, then $R\subseteq\{x\}^{c}$, so $\phi(R)\subseteq\phi(\{x\}^{c})=\{f(x)\}^{c}$, hence $f(x)\not\in\phi(R)$. If $x\in R$, then $R\not\subseteq\{x\}^{c}$, hence $\phi(R)\not\subseteq\phi(\{x\}^{c})=\{f(x)\}^{c}$, so $f(x)\in\phi(R)$. Therefore, $\phi(R)=f[R]$. Therefore, since $R\in\mathcal{U}_{1}\rightarrow f[R]\in\mathcal{U}_{2}$ and similarly $S\in\mathcal{U}_{2}\rightarrow g[S]\in\mathcal{U}_{1}$, we conclude that $\mathcal{U}_{1},\mathcal{U}_{2}$ are Rudin-Keisler equivalent. It is well-known that there are $2^{2^{\aleph_{0}}}$ many Rudin-Keisler isomorphism classes on $\omega$.