# countably complete filters

Is there any description of the set of countably complete filters on the lattice of dense $G_{\delta}$ subsets of a compact, second countable metric space? [I haven't just dreamt this up: it describes a space of ideals that I am looking at in a C*-algebra.]

-

We shall give a solution in the general lattice theoretic case. In this case, it seems easier to work with ideals rather than filters. More generally, we shall characterize the collection of all $\sigma$-ideals in any $\sigma$-complete join-semilattice with least element $0$. This is a generalization of the duality between join-semilattices with $0$ and algebraic lattices.

We shall call an element $x$ in a complete lattice $L$ Lindelof if whenever $\mathcal{C}\subseteq L$ and $x\leq\bigvee\mathcal{C}$, there is a countable $\mathcal{D}\subseteq\mathcal{C}$ with $x\leq\bigvee\mathcal{D}$. The join of countably many Lindelof elements is Lindelof. If $(x_n)_{n\in\mathcal{N}}$ is a sequence of Lindelof elements and $\bigvee_{n}x_{n}\leq\bigvee\mathcal{C}$, then for each $n$, there is a countable $\mathcal{D}_{n}\subseteq\mathcal{C}$ with $x_{n}\leq\bigvee\mathcal{D}_{n}$. Therefore, we have $\bigvee_{n}x_{n}\leq\bigvee\bigcup_{n}\mathcal{D}_{n}$. Therefore, the Lindelof elements in a complete lattice forms a $\sigma$-complete join-semilattice with $0$. We shall call a complete lattice $\sigma$-algebraic if every element is the join of Lindelof elements.

Theorem: If $A$ is a $\sigma$-complete lattice with $0$, then the lattice of $\sigma$-ideals in $A$ is up to isomorphism the unique $\sigma$-algebraic lattice $L$ where the join-semilattice of Lindelof elements is isomorphic to $A$.

Proof: Let $L$ denote the lattice of $\sigma$ ideals in $A$. If $a\in A$, then let $\downarrow a=\{b\in A|b\leq a\}$. Then clearly each $\downarrow a$ is a $\sigma$-ideal. We claim that the elements of the form $\downarrow a$ are precisely the Lindelof elements in $L$. Assume that $I\in L$ is a Lindelof element. Then $I=\bigvee_{a\in I}\downarrow a$, so there is a sequence $(a_{n})_{n\in\mathbb{N}}$ of elements in $I$ such that $I\leq\bigvee_{n}\downarrow a_{n}$. Then we have $I=\bigvee_{n}\downarrow a_{n}=\downarrow(\bigvee_{n}a_{n})$. Now if $a\in A$, and $\mathcal{C}\subseteq L$ is a collection of ideals with $\downarrow a\subseteq\bigvee\mathcal{C}$, then we have $a\in\bigvee\mathcal{C}$ and $\bigvee\mathcal{C}$ is the ideal generated by $\bigcup\mathcal{C}$. One can clearly see that $\bigvee\mathcal{C}$ is the collection of all elements $x$ where $x\leq\bigvee y_{n}$ when $y_{n}\in\bigcup\mathcal{C}$ for all $n$. Therefore $a\leq\bigvee_{n} y_{n}$ where $y_{n}\in\bigcup\mathcal{C}$ for all $n$, so if $y_{n}\in I_{n}$ and $I_{n}\in\mathcal{C}$, then $a\leq\bigvee_{n}y_{n}\in\bigvee_{n}I_{n}$, so $\downarrow a\leq\bigvee_{n}I_{n}$. Therefore the ideal $\downarrow a$ is Lindelof. We therefore have $\{\downarrow a|a\in A\}$ be precisely the collection of Lindelof elements in $L$. Thus, the join-semilattice of Lindelof elements in $L$ is isomorphic to $A$. Furthermore, if $I\in L$, then $I=\bigvee_{a\in I}\downarrow a$. Therefore the lattice $L$ is $\sigma$-algebraic.

We now claim that $L$ is the unique such lattice. Assume that $M$ is another $\sigma$-algebraic lattice and assume that $A$ is the join-semilattice of Lindelof elements. Then $A$ is an algebraic lattice. Define a map $f:M\rightarrow L$ by letting $f(m)=\{a\in A|a\leq m\}$, and let $g:L\rightarrow M$ be the mapping where $g(I)=\bigvee^{M}I$. Clearly the mappings $f$ and $g$ are order preserving. Furthermore, if $m\in M$, then $g\circ f(m)=g(\{a\in A|a\leq m\})=\bigvee^{M}\{a\in A|a\leq m\}=m$ since $M$ is $\sigma$-algebraic. Similarly, if $I$ is an ideal, then $f\circ g(I)=f(\bigvee^{M}I)=\{a\in A|a\leq\bigvee^{M}I\}$. Clearly $I\subseteq\{a\in A|a\leq\bigvee^{M}I\}$. Similarly, if $a\in A$ and $a\leq\bigvee^{M}I$, then $a\leq\bigvee^{M}\{a_{n}|n\in\mathbb{N}\}$ for some sequence $(a_{n})_{n}$ of elements in $I$. Therefore since $I$ is a $\sigma$-ideal, we have $a\in I$ as well. Therefore we have $\{a\in A|a\leq\bigvee^{M}I\}$ as well, so $I=\{a\in A|a\leq\bigvee^{M}I\}=f(g(I))$. Therefore the mappings $f$ and $g$ are inverse order preserving maps. Therefore $f$ and $g$ are isomorphisms between the lattices $L$ and $M$.

-
Thanks. I will have to think about this. Presumably the definition of a Lindelof element is not quite right? – Douglas Somerset Jun 30 '12 at 19:06
I corrected the definition of Lindelof. Sorry about that. – Joseph Van Name Jul 1 '12 at 23:16

By the Baire Category Theorem the full family of dense $G_\delta$-sets is a countably complete filter. So, unless I'm missing something, the set is a singleton.

-
There can also be some smaller filters. For example, if there are no isolated points, then the co-countable sets form a countably complete filter in the lattice of dense $G_\delta$ sets. And there are lots of minor variants; for example, fix an uncountable subset $A$ of the space and consider the filter of those dense $G_\delta$ sets that contain all but countably many of the points in $A$. Or, if there's a set $A$ of size $\aleph_1$ whose closure is nowhere dense, fix a bijection $f$ from $\omega_1$ to it and form the filter of sets that include $f$ of a club. – Andreas Blass Apr 5 '12 at 20:13
Also if there are no isolated point in the original space $X$ then for any non-empty subset $Y\subseteq X$ the set of dense $G_{\delta}$s containing $Y$ is a countably complete filter whose intersection is $Y$. Thus there are at least $2^X$ distinct countably complete filters. – Douglas Somerset Apr 5 '12 at 20:50
In fact, if we take $X=[0,1]$ then for each subset $Y$ there is a largest countably complete filter of dense $G_{\delta}$s with intersection $Y$ (namely all dense $G_{\delta}$s containing $Y$) and a smallest countably complete filter with intersection $Y$ (namely all co-countable dense $G_{\delta}$s containing $Y$). In the case when $Y=\emptyset$, the question is to describe the countably complete filters of dense $G_{\delta}$s which contain the filter of co-countable sets. – Douglas Somerset Apr 6 '12 at 10:34
I mentally inserted ultra', hence my singleton'. – KP Hart Apr 7 '12 at 20:39