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I'm trying to do exercise 7.11 of Jech's "Set Theory": If $D$ and $E$ are ultrafilters on $\omega$, then $D\leq E$ and $E\leq D$ implies that $D\equiv E$, where $\leq$ is the Rudin-Keisler ordering, and in this book is defined $D\equiv E$ iff there exists a bijection $f:\omega\rightarrow\omega$ such that $f(D)=E$, there exists a previous result in which one proves that :

  1. if $f:\omega\rightarrow\omega$ is such that $f(D)=D$, then $\{n:f(n)=n\}\in D$.

I also read this definition in wikipedia, under this definition the problem is trivial because of $1$, but I do not know how to proceed under Jech's definition, any help will be appreciated.

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Since $D\le_{\text{RK}}E$, there is a function $f_0:\omega\to\omega$ such that $d\in D$ iff $f_0^{-1}[d]\in E$. Similarly, there is a function $f_1:\omega\to\omega$ such that $e\in E$ iff $f_1^{-1}[e]\in D$. Applying (1) to $f_1\circ f_0$ or $f_0\circ f_1$, you get sets $d\in D$ and $e\in E$ and a bijection $h:d\to e$ such that for $x\subseteq e$, $x\in E$ iff $h^{-1}[x]\in D$. Choose any $d'\in D$ such that $d'\subseteq d$ and $d\setminus d'$ is infinite, let $e'=h[d']$, and let $g_0=h\upharpoonright d'$. Clearly $g_0$ is a bijection from $d'$ to $e'$, and for $x\subseteq e'$ we have $x\in E$ iff $g_0^{-1}[x]\in D$. Let $g_1:\omega\setminus d'\to\omega\setminus e'$ be any bijection; then $g=g_0\cup g_1$ is a bijection from $\omega$ to $\omega$, and $x\in E$ iff $g^{-1}[x]\in D$, as desired.

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well, I had thought instead to take $d$ and $e$ coinfinite and consider your bijection $h$, and let $h'$ be any bijection between the complements of $d$ and $e$, and put $g=h\cup h'$, and again thank you for another useful answer –  rush fan Dec 27 '12 at 19:15
    
@rushfan: It would be nice, but you can’t guarantee that $d$ and $e$ can be chosen to be cofinite: it may be that $h$ is badly behaved on some infinite set not in $D$. (And you’re welcome.) –  Brian M. Scott Dec 27 '12 at 19:19

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