I'm reading some contents on set-theory for my own interest, and I stumbled upon some questions I cannot solve yet.
Let $\mathcal{F}$ be a $\kappa$-complete ultrafilter on $\kappa$, and $\bigcup_{\alpha < \lambda} X_\alpha \in \mathcal{F}$ with $\lambda < \kappa$. Show that $\exists \ \alpha < \lambda$ such that $X_\alpha \in \mathcal{F}$.
I proved that $\mathcal{F}$ is a $\kappa$-complete ultrafilter on $\kappa$ if and only if, for every partition $\{Y_\alpha, \alpha<\lambda\}$ of $\kappa$, there is $\alpha < \lambda$ such that $Y_\alpha \in \mathcal{F}$.
I guess I only have to "transform" my $X_\alpha$'s into a partition of $\kappa$, but I don't see how. Ideally I would create a partition of $\kappa$ from $X_\alpha$'s and only "small elements", i.e. elements not in $\mathcal{F}$. This will directly give me the desired conclusion. Any idea? Thanks
If the ultrafilter is trivial, then the thesis follows easily. So suppose it is not.
Suppose the thesis is false, then it means that $Y_\alpha:=\kappa\setminus X_\alpha$ is in the ultrafilter for every $\alpha<\gamma$. But then, by $\kappa$-completeness, $Y:=\bigcap_{\alpha<\gamma}Y_\alpha$ is an element of the ultrafilter. But then, notice that $Y=\kappa\setminus\bigcup_{\alpha<\gamma}X_\alpha$, which means that the complement of $Y$ is in the ultrafilter as well: this implies that the ultrafilter is trivial, contradicting our assumption.