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Let $\lambda$ be a cardinal. I would like to prove that for all cardinals $\lambda < \kappa \leq 2^\lambda$, there can't be a $\kappa$-complete non-principal ultrafilter on $\kappa$.

Here is my attempt: $\kappa$ can be seen as a subset of $\{0,1\}^\lambda$. Let $\mathcal U$ be a $\kappa$-complete ultrafilter on $\kappa$. For all ordinals $\alpha < \lambda$, let $E^i_\alpha = \{ f \in \kappa : f(\alpha) = i\}$. As $\mathcal U$ is an ultrafilter, either $E^0_\alpha$ or $E^1_\alpha$ is in $\mathcal U$. Let $F_\alpha$ be the one in $\mathcal U$ (there are $\alpha$ choices to be made here, is AC needed?). By completeness of the filter, $\cap_{\alpha<\lambda} F_\alpha$ is in $\mathcal U$, hence not empty, and $\lvert \cap F_\alpha \rvert \leq 1$, hence $\mathcal U$ is principal.

Is this proof correct?

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2 Answers 2

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It looks fine to me. You didn't use choice when defining $F_{\alpha}$ because your choice isn't arbitrary: exactly one of $E^0_{\alpha}$ and $E^1_{\alpha}$ lies in $\mathcal{U}$, and that's the one you pick.

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First off, in your approach AC is not needed, as the choice is uniquely defined by the ultrafilter $\mathcal{U}$. The rest seems OK to me.

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