# ultrafilters' succession

hi

I'n looking for a increasing and bounded ultrafilters' succession in natural numbers with Rudin-Keisler order, actually I need to prove there is that succession the idea is

$U_1,U_2,....$ with $U$ supreme and for all n $U_n < U$ for $h_n$ in Rudin-Keisle order

and por all n $U_n < U_{n+1}$ for $g_n$ in Rudin-Keisle order

and $h_n(m)= g_{n+1}/ocircle h_{n+1} (m)$ with m a natural number

I have no idea how build the $h_n$ funtions

thanks, sorry for my awful english

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## 2 Answers

Most of your question sounds as if you just want an increasing $\omega$-sequence $(U_n)$ in the RK-order, with an upper bound $U$. Although you want these ultrafilters to be on $\omega$, it's a little easier to see the idea for a construction if you use some other countable sets as follows; you can always transfer the result to $\omega$ by suitable bijections. Let $S$ be the set of all those $\omega$-sequences $(a_k)$ of natural numbers in which all but finitely many components $a_n$ are zero. For each natural number $n$, let $h_n:S\to\omega^n$ be the "truncation" function that sends any $(a_k)\in S$ to the tuple of its first $n$ components $(a_k)_{k<n}$. Notice that $h_n=h_{n+1}\circ p_n$ where $p_n:\omega^{n+1}\to\omega^n$ is the projection to the first $n$ components. Let $\mathcal X$ be the collection of those subsets $X\subseteq S$ such that, for some $n$, $p_n$ is one-to-one on $h_{n+1}(X)$. After checking that no finitely many sets in $\mathcal X$ cover all of $S$, we can let $U$ be any ultrafilter on $S$ that is disjoint from $\mathcal X$, and we can let $U_n=h_n(U)$. Then $p_n$ sends $U_{n+1}$ to $U_n$ and is not one-to-one on any set in $U_n$ (because $U$ contains no set from $\mathcal X$). So $p_n$ witnesses that $U_n<_{RK}U_{n+1}$. And of course the $h_n$'s witness that $U$ is an upper bound for all the $U_n$'s.

I worry, though, that the word "supreme" in the question might mean that you want $U$ to be not only an upper bound but a least upper bound for the $U_n$'s. I don't know how to achieve that.

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My (ancient) Ph.D. thesis, a scanned pdf of which is available on my web site, contains the stronger result that, in the Rudin-Keisler ordering of ultrafilters on $\omega$, any continuum many (or fewer) ultrafilters have a (strict) upper bound. The idea is similar to the answer above but using, instead of the functions $h_n$, a family of continuum many independent functions $\omega\to\omega$. – Andreas Blass Dec 23 '11 at 22:17
If $I$ is a set of cardinality continuum, then I was thinking simply to take a countable dense subset $A\subseteq\mathbb{N}^{I}\subseteq(\beta\mathbb{N})^{I}$. Then the inclusion map $A\hookrightarrow(\beta\mathbb{N})^{I}$ extends to a countinuous surjection $\iota:\beta A\rightarrow(\beta\mathbb{N})^{I}$. In particular, if $x_{i}\in\beta\mathbb{N}$ for $i\in I$, then there is some $x\in\beta A$ with $\iota(x)=(x_{i})_{i\in I}$. Therefore $\pi_{i}\iota(x)=x_{i}$ for $i\in I$ where each $\pi_{i}$ is the projection, so $x_{i}\leq_{RK}x$ for $i\in I$. – Joseph Van Name Jan 7 '13 at 2:56
@Joseph Van Name: I think your proof is essentially the same as mine. The existence of a countable dense subset in $\mathbb N^I$ is the same fact as the existence of an $I$-indexed family of independent functions $\mathbb N\to\mathbb N$. – Andreas Blass Jan 7 '13 at 14:22

You don't say what the context is, but let's suppose at first that the question is asked in the large cardinal context of $\kappa$-complete ultrafilters on a measurable cardinal $\kappa$, which is a context where one often considers such questions.

The first thing to say is that there may be no such example, even when there is a measurable cardinal. For example, in the canonical inner model $L[\mu]$ with one measurable cardinal $\kappa$, there is exactly one normal measure $\mu$ on $\kappa$, and every $\kappa$-complete ultrafilter on $\kappa$ is Rudin-Kiesler equivalent to a finite power $\mu^n$ of $\mu$. In particular, although these powers do indeed form an increasing chain $$\mu^1\lt_{RK}\ \mu^2\lt_{RK}\ \mu^3\lt_{RK}\ \cdots$$ in the Rudin-Kiesler order, there can be no measure on top of this chain as you request. In short, it is consistent with a measurable cardinal that the height of the Rudin-Kiesler order is $\omega$.

From a larger large cardinal assumption, however, one can achieve higher Rudin-Kiseler ranks. For example, if $\kappa$ is $\kappa+2$-strong, then there must be ultrafilters $U$ on $\kappa$ with Rudin-Kiesler rank $\omega$, giving rise to the situation of your question. To see this, let $j:V\to M$ with $V_{\kappa+2}\subset M$ and we may assume $M^\kappa\subset M$. Let $X_1=\{j(f)(\kappa)\mid f:\kappa\to V\}$ be the seed hull of $\kappa$, which is elementary in $M$ and collapses to the ultrapower $j_1:V\to M_1$ of the induced normal measure $U_1$ generated by $j$. Consider the first ordinal missing from $X_1$, call it $\delta_1$, and let $X_2$ be the see hull $\{j(f)(\delta_0,\delta_1)\mid f:\kappa^2\to V\}$, using $\delta_0=\kappa$, and similarly define $\delta_n$ for each $n$. The ultrafilter $U_n$ induced by $A\in U_n\iff (\delta_0,\ldots,\delta_{n-1})\in j(A)$ is Rudin-Kiesler below $U$, and they form an increasing chain as desired. The chain is strictly increaing precisely because no $X_n$ can exhaust all of $M$ below $j(\kappa)$.

In the general case, or even in the case of ultrafilters on $\omega$, it remains easy to build increasing chains in the Rudin-Kiesler order. For example, the successive products of a fixed ultrafilter form a strictly increasing chain $$U\lt_{RK} U^2\lt_{RK} U^3\lt_{Rk}\cdots$$ To place an ultrafilter $U$ on top, however, takes some more delicate work (and I see now that Andreas Blass has posted an answer indicated how to undertake that delicate work).

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