# Are normal, nonprincipal ultrafilters necessarily closed?

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 $$\kappa$$?

• Wait. If $\kappa$ is regular then normality implies uniformity. No? – Asaf Karagila Mar 22 at 16:54
• @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. – Stefan Mesken Mar 22 at 17:05

## 1 Answer

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 $$\lambda\cap\Delta_{\alpha<\kappa}Y_\alpha=\Delta_{\alpha<\lambda} X_\alpha\in V$$.

• 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! – Andreas Lietz Mar 22 at 21:13