Suppose that $\kappa$ is a cardinal, $X$ is a set with $|X|>\kappa$, and $\mathcal{U}\subseteq P(P\_{\kappa}(X))$ is a normal ultrafilter. We say that a collection $C\subseteq P\_{\kappa}(X)$ is a conditional closure system if whenever $D\subseteq C$ and $D\neq\emptyset$, then $\bigcap D\in C$. Let's say that $\mathcal{U}$ is generated by conditional closure systems if for each $A\in\mathcal{U}$ there is a conditional closure system $C\in\mathcal{U}$ with $C\subseteq A$. 


Is a normal ultrafilter $\mathcal{U}\subseteq P(P\_{\kappa}(X))$ necessarily generated by conditional closure systems? Is it even possible that there exists a $\kappa$ and a set $X$ with $|X|>\kappa$ and a $\kappa$-complete normal ultrafilter $\mathcal{U}\subseteq P(P\_{\kappa}(X))$ that is generated by conditional closure systems?



Are normal ultrafilters generated by conditional closure systems?
I claim that if $\lambda,\kappa$ are cardinals with $\lambda>\kappa$ and $\mathcal{U}\subseteq P(P\_{\kappa}(\lambda))$ is a normal ultrafilter, then $\mathcal{U}$ is not generated by conditional closure systems. In fact, I shall now prove that $\mathcal{U}$ is not generated by sets closed under taking finite intersection.


Suppose to the contrary that $\mathcal{U}$ is generated by sets closed under taking finite intersection. Then let $\mathcal{V}\subseteq P(P\_{\kappa}(\kappa^{+}))$ be the ultrafilter where if $R\subseteq P\_{\kappa}(\kappa^{+})$, then $R\in\mathcal{V}$ if and only if
$\{x\in P\_{\kappa}(\lambda)|x\cap\kappa^{+}\in R\}\in\mathcal{U}$. Then $\mathcal{V}$ is a normal ultrafilter on $P(P\_{\kappa}(\kappa^{+})).$


Suppose that $A\in\mathcal{V}$. Then $B:=\{x\in P\_{\kappa}(\lambda)|x\cap\kappa^{+}\in A\}\in\mathcal{U}$. Therefore there is some set $C\in\mathcal{U}$ closed under finite intersections such that $C\subseteq B$. Let $D=\{x\cap\kappa^{+}|x\in C\}$. Then $D$ is closed finite intersections as well. However, if $x\in D$, then $x=x'\cap\kappa^{+}$ for some $x'\in C\subseteq B$, so $x=x'\cap\kappa^{+}\in A$. Therefore $D\subseteq A$. Therefore, since $C\in\mathcal{U}$, we have $D\in\mathcal{V}$. We conclude that $\mathcal{V}$ is generated by sets closed under taking finite intersections.


The idea now is to use the ultrafilter $\mathcal{V}\subseteq P(P\_{\kappa}(\kappa^{+}))$ to obtain a contradiction. To continue the proof, we will need a few definitions. Let $[P\_{\kappa}(X)]\_{\subset}^{n}=\{\{x\_{1},...,x\_{n}\}|x\_{1}\subset x\_{2}\subset...\subset x\_{n};x\_{1},...,x\_{n}\in P\_{\kappa}(X)\}$. Suppose that 
$f:[P\_{\kappa}(X)]\_{\subset}^{n}\rightarrow 2$ is a function. Then we say that a subset $A\subseteq [P\_{\kappa}(X)]\_{\subset}^{n}$ is homogeneous for $f$ if there is some $i\in 2$ such that whenever $x\_{1}\subset x\_{2}\subset...\subset x\_{n}$, we have $f(\{x\_{1},...,x\_{n}\})=i$. An ultrafilter $\mathcal{U}\subseteq P(P\_{\kappa}(X))$ is said to have the partition property if for each $f:[P\_{\kappa}(X)]\_{\subset}^{2}\rightarrow 2$, there is some $A\in\mathcal{U}$ homogeneous for $f$.


Every normal ultrafilter $\mathcal{W}\subseteq P(P\_{\kappa}(\kappa^{+}))$ has the partition property (in fact, if $\mathcal{W}\subseteq P(P\_{\kappa}(\lambda))$ is a normal ultrafilter that does not satisfy the partition property, then $\lambda$ is $\Pi^{2}\_{1}$-indescribable. However, assuming sufficiently strong large cardinal hypotheses, the partition property does not always hold. These facts can be found in Kanamori's book The Higher Infinite).


Now suppose that $g:P\_{\kappa}(\kappa^{+})\rightarrow P\_{\kappa}(\kappa^{+})$ is a function such that $x\subset g(x)$ for each $x\in P\_{\kappa}(\kappa^{+})$. Let
$f:[P\_{\kappa}(\kappa^{+})]^{2}\rightarrow 2$ be the mapping where if $x\subseteq y$, then $f(\{x,y\})=1$ if $g(x)\subseteq y$ and $g(x)=0$ otherwise. Then there is some set $A\in\mathcal{V}$ homogeneous for $f$. Now suppose that $x\in A$. Then since $A\in\mathcal{U}$, there is some set $y\in A$ with $g(x)\subseteq y$. Therefore, $f(\{x,y\})=1$, so $f$ is homogeneous of color $1$. Therefore, if $x\subset y$, then $g(x)\subseteq y$. Now let $C$ be a set closed under taking finite intersections such that $C\in\mathcal{U}$ and $C\subseteq A$.


I now claim that if $x,y\in C$, then either $x\subseteq y$ or $y\subseteq x$. Suppose to the contrary that $x\not\subseteq y$ and $y\not\subseteq x$. Then let $z=x\cap y$. Then $z\in C$ as well. Furthermore, $z\subset x$ and $z\subset y$. Therefore $g(z)\subseteq x$ and $g(z)\subseteq y$, so $g(z)\subseteq x\cap y=z$. This is a contradiction. We conclude that either $x\subseteq y$ or $y\subseteq x$, so $C$ is a linear order. However, one can easily construct a contradiction from the fact that $C$ is a linear order. For instance, suppose that $(x\_{\alpha})\_{\alpha<\kappa}$ is a sequence of sets in $C$ such that if $\alpha<\beta$, then $x\_{\alpha}\subset x\_{\beta}$. Then suppose that $\gamma\in\kappa^{+}\setminus\bigcup\_{\alpha<\kappa}x\_{\alpha}$. Then there is some $x\in C$ with $\gamma\in x$ since $C\in\mathcal{V}$. Then there is some $\alpha$ with $|x|<|x\_{\alpha}|$. However, we have $x,x\_{\alpha}\in C$, but $x\not\subseteq x\_{\alpha}$ and $x\_{\alpha}\not\subseteq x$. This is a contradiction.


