Uniform normal filter on $\kappa$ is $\kappa$-complete and contains the club filter.

Question

The question reads:

Edit: The definition for uniform filter I was given is as follows: A filter $\mathcal{F}$ on $\kappa$ is uniform iff $\mathcal{F}$ contains all co-bounded sets in $\kappa$ iff $\mathcal{F}$ contains all intervals $(\alpha, \kappa)$ for $\alpha < \kappa$. This may not be standard, it seems uniformity usually only applies to ultrafilters.

"Work in $\sf ZF$. Assume $\kappa > \omega$ is regular and $\mathcal{F}$ is a uniform normal filter on $\kappa$.

1. Prove that $\mathcal{F}$ is $\kappa$-complete.
2. Prove that $\mathcal{F}^{\text{club}}_\kappa \subseteq \mathcal{F}$

Hint: For 1 for an ordinal $\lambda < \kappa$ and a sequence of sets $\langle A_\xi \mid \xi < \lambda \rangle$ of sets in $\mathcal{F}$, extend this sequence to a sequence of length $\kappa$ whose diagonal intersection differs from $\bigcap_{\xi < \lambda} A_\xi$ by a non-stationary set. For 2 given a club set $C \subseteq \kappa$, try to express $C$ as a diagonal intersection of intervals of the form $(\alpha, \kappa)$. Alternatively, using choice, use pressing down."

So I believe, I found the extended sequence for 1, but I'm not sure why having the diagonal intersection differ by a non-stationary set helps. For 2, I couldn't figure out the intervals. I tried enumerating the club $C$ in increasing order, and tried to create the intervals that way, but I can't seem to get anywhere. I believe, that it is sufficient to just consider club $C$ since any $X \in \mathcal{F}^{\text{club}}_\kappa$ contains a club so then by the upward-closure of $\mathcal{F}$ we will get $X \in \mathcal{F}$. I'm not sure how to do the pressing down argument, but I would love to see that.

My (partial) solution to 1.

Let $\lambda < \kappa$ and $\langle A_\xi \mid \xi < \lambda \rangle \subseteq \mathcal{F}$. Let $A := \bigcap_{\xi <\kappa} A_\xi$. Define a new sequence $\langle A'_\xi \mid \xi < \kappa \rangle$ by

$$A'_\xi = \left\{ \begin{array}{ll} A_\xi & \quad \xi < \lambda \\ (\delta, \kappa) & \quad \lambda \leq \delta < \kappa \end{array} \right.$$

Since $\mathcal{F}$ is uniform, $(\delta, \kappa) \in \mathcal{F}$ for all $\delta < \kappa$ so that $\langle A'_\xi \mid \xi < \kappa \rangle$ is a sequence in $\mathcal{F}$. By normality, $A' := \bigtriangleup_{\xi < \kappa} A'_\xi \in \mathcal{F}$. Then $(A - A') \cup (A' - A) \subseteq [0, \lambda]$ which is bounded, so it is non-stationary.

Answer 1

We have that $A\triangle A'$ is bounded; as you noted, thus by your definition of uniform filter we have that $(A\triangle A')^c\in\mathcal F$, but $(A\triangle A')^c=(A^c\cap A'^c)\cup(A'\cap A),$ hence as $A'\in\mathcal F$ we obtain, intersecting, that $(A'\cap A)\in\mathcal F$, which implies $A\in\mathcal F$.

For $(2)$, let $X_{\alpha}=(\alpha+1,\kappa),$ for each $\alpha<\kappa$, then each $X_{\alpha}\in\mathcal F$, and clearly $C_0=\triangle_{\alpha<\kappa}X_{\alpha}\in\mathcal F$, and $C_0$ is the set of all limit ordinals $<\kappa$.

Now let $C$ be a closed unbounded set. Let $\{a_{\xi}:\xi<\kappa\}$ be the increasing continuous enumeration of $C$. For each $\alpha<\kappa$ let $Y_{\alpha}=(a_{\alpha},\kappa)$. Let $\xi\in C_0\cap\triangle_{\alpha<\kappa}Y_{\alpha},$ then $a_{\alpha}<\xi$ for all $\alpha<\xi$, thus if $\beta=\lim_{\alpha\to\xi}a_{\alpha},$ then $\beta=a_{\xi}$ and $\beta\leq\xi$, thus $\xi\in C$. Hence $C\supseteq C_0\cap\triangle_{\alpha<\kappa}Y_{\alpha}$, but $\triangle_{\alpha<\kappa}Y_{\alpha}\in\mathcal F$, therefore $C\in\mathcal F$.