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.