4
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ 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 $\omega$. $\endgroup$ – Joseph Van Name Jul 19 '17 at 13:57
  • $\begingroup$ Oh - the Rudin-Keisler equivalence is just what I needed, thanks! Can you quickly put this into an answer? $\endgroup$ – Dominic van der Zypen Jul 19 '17 at 14:10
  • 2
    $\begingroup$ 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. $\endgroup$ – Emil Jeřábek Jul 19 '17 at 15:46
  • $\begingroup$ I gave a proof of how a poset isomorphism induces a Rudin-Keisler isomorphism. $\endgroup$ – Joseph Van Name Jul 19 '17 at 18:39
10
$\begingroup$

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 $\omega$.

$\endgroup$
  • $\begingroup$ This theorem apparently first time appears in this paper of Walter Rudin (theorem 1.5) and was pointed out to Rudin by Hewitt Kenyon $\endgroup$ – ar.grig 14 hours ago

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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