Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question

2 Answers 2

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.

share|improve this answer
    
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).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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