Let $U$ be a normal (closed under diagonal intersections), nonprincipal (doesn't contain singletons) ultrafilter on some uncountable cardinal $\kappa$.

Q. Is $U$ $<\kappa$-complete (closed under intersections of length $< \kappa$)?

If $U$ is uniform (every member of $U$ has size $\kappa$), the answer is positive:

Proof. Let $(X_i | i < \eta)$ be a sequence of sets in the ultrafilter and $\eta < \kappa$.

Let $$ Y_i = \begin{cases} X_i & \text{, if } i < \eta \\ \kappa & \text{, o/w} \end{cases} $$ Then $\Delta_{i < \kappa} Y_i \cap (\kappa \setminus \eta) \subseteq \bigcap_{i < \eta} X_i$ and the left-hand side is in the filter. Hence the right-hand side is as well. Q.E.D.

Conversely, if $U$ is not uniform, it cannot be $<\kappa$-complete.

Proof. Let $X \in U$ be such that $\mathrm{card}(X) < \kappa$. Then $$ \kappa \setminus X = \bigcap_{\xi \in X} \kappa \setminus \{\xi\} \not \in U $$ witnesses that $U$ is not $< \kappa$-complete. Q.E.D.

So the question really is

Q. Is it possible to have a normal, nonprincipal, non-uniform ultrafilter on some uncountable cardinal $\kappa$?

  • $\begingroup$ Wait. If $\kappa$ is regular then normality implies uniformity. No? $\endgroup$ – Asaf Karagila Mar 22 at 16:54
  • $\begingroup$ @AsafKaragila I thought I had a proof of that but discovered a flaw in it. If you have a proof, I'd like to see it as a partial answer to my question. But I'm also interested in the case that $\kappa$ is singular. $\endgroup$ – Stefan Mesken Mar 22 at 17:05

Yes you can have normal, non-principal, non-uniform ultrafilters. Suppose $\lambda$ is measurable and $U$ is a normal ultrafilter witnessing this. Let $\kappa>\lambda$ be any ordinal and just take $F$ the trivial extension of $U$ from $\lambda$ to $\kappa$, i.e. $$F=\{X\subseteq\kappa\mid X\cap\lambda\in U\}$$ This is certainly a non-principal and non-uniform ultrafilter on $\kappa$. To see that it is normal, let $(X_\alpha)_{\alpha<\kappa}$ be a sequence of sets in $F$. Then $\lambda\cap\Delta_{\alpha<\kappa} X_\alpha=\Delta_{\alpha<\lambda} (X_\alpha\cap\lambda)\in U$ and so $F$ is closed under diagonal intersections.

On the other hand, all such filters arise as $F$ does from $U$ above. Suppose $G$ is a normal, non-principal, non-uniform ultrafilter on $\kappa$. Let $\lambda$ be least so that $\lambda\in G$. Let $V$ be $G$ restricted to $\lambda$, i.e. $$V=\{X\subseteq\lambda\mid X\in G\}$$ Certainly, $V$ is a non-principal ultrafilter on $\lambda$. It is uniform as $\lambda$ is chosen minimal. To see that its normal, let $(X_\alpha)_{\alpha<\lambda}$ be a sequence of sets in $V$. Define $Y_\alpha$ to be $X_\alpha$ if $\alpha<\lambda$ and $Y_\alpha=\kappa$ for $\lambda\leq\alpha<\kappa$. Then $\Delta_{\alpha<\kappa}Y_\alpha\in V$ and so $\lambda\cap\Delta_{\alpha<\kappa}Y_\alpha=\Delta_{\alpha<\lambda} X_\alpha\in V$.

  • $\begingroup$ Maybe I can convince you some more. Note that in your proof of uniformity+normality implies completeness, you do not need full uniformity, only that the filter contains all tail segments. The minimality of $\lambda$ yields that this is true for $V$, hence it must be $<\lambda$-complete and so it is uniform. You're welcome! $\endgroup$ – Andreas Lietz Mar 22 at 21:13

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