# $\kappa$-ultrafilter on $\kappa$, creating a partition from $\bigcup_{\alpha < \lambda} X_\alpha \in$ filter

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

## 1 Answer

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.