Let $A$ an infinite set, $D$ an ultrafilter on $A$ and $\kappa$ an infinite cardinal. I want to show the following :

$D$ is $\kappa$-complete iff $\forall\tau<\kappa$ and $\forall$ partition $\{X_\xi:\xi<\tau\}$ of $A$, there exists $\xi_0<\tau$ so that $X_{\xi_0}\in D$

$(\Rightarrow)$ : it's ok.

$(\Leftarrow)$ : let $M=\{X_\xi:\xi<\tau\}\subseteq D$ with $\tau<\kappa$. We want to show that $\bigcap M\in D$.

Let $Y_\xi=X_\xi\setminus\bigcap_{\eta\neq\xi}X_\eta$ for all $\xi<\tau$. It's easy to see that $Y_\xi\cap Y_{\xi'}=\emptyset$ for $\xi\neq\xi'$. Put $Y:=\bigcup Y_\xi$. We have $A=\bigcup Y_\xi\cup(A\setminus Y)$ and $\bigcup X_\xi=Y$. Also $A\setminus Y$ is disjoint from all the $Y_\xi$ so we have a partition of $A$, of cardinality less than $\kappa$. By the hypothesis and because $A\setminus Y\notin D$ ($X_\xi\subseteq Y$ so $Y\in D$), there exists $\xi_0<\tau$ such that $Y_{\xi_0}\in D$. So $A\setminus Y_{\xi_0}=(A\setminus X_{\xi_0})\cup(\bigcap_{\eta\neq\xi_0}X_\eta)\notin D$ so $\bigcap_{\eta\neq\xi_0}X_\eta\notin D$ and by $\bigcap M\subseteq \bigcap_{\eta\neq\xi_0}X_\eta$ we have $\bigcap M\notin D$.

Where is my mistake ? Thanks.


It’s not necessarily true that $Y_\xi\cap Y_{\xi'}=\varnothing$ for $\xi\ne\xi'$. Take $A=\omega$, $\tau=\omega$, and $X_n=\omega\setminus\{n\}$ for $n\in\omega$. Then $Y_n=X_n$ for every $n\in\omega$, and $Y_m\cap Y_n=\omega\setminus\{m,n\}$ for all $m,n\in\omega$.

Try this instead. Suppose that $D$ is not $\kappa$-complete, and let $\tau$ be minimal such that there is a family $\{X_\xi:\xi<\tau\}$ such that $\bigcap_{\xi<\tau}X_\xi\notin D$. Without loss of generality you may assume that $X_\xi\supsetneqq X_{\xi+1}$ for each $\xi<\tau$ and let $Y_\xi=X_\xi\setminus X_{\xi+1}$. Now apply your hypothesis on partitions of $A$ to get a contradiction.

  • $\begingroup$ thanks for your counter-example. I will try to finish the proof... good night. $\endgroup$ – Marc Moretti Mar 20 '13 at 21:39
  • $\begingroup$ @Marc: You’re welcome. If you don’t manage to finish it, just ask. $\endgroup$ – Brian M. Scott Mar 20 '13 at 22:07
  • $\begingroup$ I can construct $X_{\xi+1}\subseteq X_\xi$ (with intersections) but not with $\subsetneq$. If $A=Y\cup(A\setminus Y)$ then I can see that there is no $Y_\xi\in D$ but I have some pb with $A\setminus Y$... $\endgroup$ – Marc Moretti Mar 22 '13 at 7:47
  • 1
    $\begingroup$ @Marc: Start with arbitrary $X_\xi$ such that $\bigcap_{\xi<\tau}X_\xi\notin D$. Let $Y_\eta=\bigcap_{\xi\le\eta}X_\xi\in D$. Let $Z_0=Y_0$. Given $Z_\xi$ for some $\xi<\tau$, there is a least $\eta<\tau$ such that $Z_\xi\supsetneqq Y_\eta$; let $Z_\xi=Y_\eta$. $\langle Z_\xi:\xi<\tau\rangle$ is a strictly decreasing $\tau$-sequence of members of $D$ whose intersection is not in $D$. Your partition then consists of the sets $Z_\xi\setminus Z_{\xi+1}$ for $\xi<\tau$, the set $A\setminus A_0$, and the set $\bigcap_{\xi<\tau}Z_\xi$; none of these is in $D$. $\endgroup$ – Brian M. Scott Mar 22 '13 at 7:59
  • $\begingroup$ Thanks a lot ! constructive answer... $\endgroup$ – Marc Moretti Mar 22 '13 at 8:46

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