MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.]

share|cite|improve this question
up vote 1 down vote accepted

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 $M$.

share|cite|improve this answer
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.