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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

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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.

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