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In any partial order $(P,\leq)$ it is easy to see that every chain generates (i.e., by taking the upwards closure) a filter, and any filter built as a result of the Rasiowa-Sikorski lemma in forcing is of this form. When can we reverse this?

More specifically, I am interested in the partial order of all closed subsets, with nonempty interior, of a metric space, ordered by inclusion. (I suppose the nonempty regular open sets would work, too.) Given a filter, can we say that it is generated by a countable chain?

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2 Answers 2

One interesting case, since you mention forcing, although it isn't your main case, is that ultrafilters on an atomless complete Boolean algebra are never generated by a chain (of any cardinality).

To see this, suppose that $U$ is an ultrafilter in a complete atomless Boolean algebra $\mathbb{B}$ generated by a chain. So the elements of $U$ are precisely the elements above some $b_\alpha$, where $$1=b_0\geq b_1\geq\cdots\geq b_\alpha\geq\cdots$$ is a continuous decreasing sequence in $\mathbb{B}$, for $\alpha\lt\kappa$. It is easy to see that $\bigwedge_\alpha b_\alpha=0$. Consider the corresponding difference antichain $d_\alpha= b_\alpha-b_{\alpha+1}$, and let $a$ be the supremum of $d_\alpha$ for $\alpha$ even. Since the chain is continuous, starts at $1$ and has meet $0$, it follows that the difference antichain is a maximal antichain, and $\neg a$ is the join of $d_\alpha$ for $\alpha$ odd. But neither $a$ nor $\neg a$ is above any $b_\alpha$, since each of them has nontrivial meet with cofinally many $b_\beta$, contradicting that $U$ is an ultrafilter. QED

I find this example illuminating in the case of forcing, since many generic ultrafilters are generated by a chain, such as when you force with a tree, which is a very common way to force. What the argument shows is that the ground model Boolean algebra in these cases is no longer a complete Boolean algebra in the forcing extension.

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Perhaps one can use this same difference antichain idea to construct counterexamples with ultrafilters in the spaces you are considering. –  Joel David Hamkins Aug 18 '13 at 17:58
    
Why can you say that the chain is well-ordered with a continuous enumeration? Could it have order type $\mathbb{R}$ or something like that? –  Iian Smythe Aug 18 '13 at 18:11
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Any cofinal refinement of the chain generates the same filter, and so we can find a (reverse) well-ordered subchain. And we can make it continuous by adding in the right limit values, which also doesn't affect the generated filter. –  Joel David Hamkins Aug 18 '13 at 18:22

$\mathbf{Theorem}$ Let $X$ be a $T_{1}$-space. Let $\mathcal{Z}$ denote the lattice of all closed subsets of $X$. Let $M\subseteq\mathcal{Z}$ be a maximal filter. Then $M=\{C\in\mathcal{Z}|x\in C\}$ for some $x\in X$ or $M$ is not countably generated.

$\mathbf{Proof}$ Assume that $M$ is countably generated, and $M\neq\{C\in\mathcal{Z}|x\in C\}$ for all $x\in X$. Then $\bigcap M=\emptyset$. Otherwise, if $x\in\bigcap M$, then $M\subseteq\{C\in\mathcal{Z}|x\in C\}$, so $M=\{C\in\mathcal{Z}|x\in C\}$ by maximality. Assume $M$ is countably generated by some decreasing sequence $(C_{n})_{n}$ of closed sets. Without loss of generality, we may assume that the sequence $(C_{n})_{n}$ is strictly decreasing. Let $x_{n}\in C_{n}\setminus C_{n-1}$ for all $n$. Let $U_{n}=C_{n}^{c}$ for all $n$. Then $(U_{n})_{n}$ is a countable open cover of $X$. However, if $A\subseteq\{x_{n}|n\in\mathbb{N}\}$, then $A\cap U_{n}$ is finite for all n, hence $A\cap U_{n}$ is closed in $U_{n}$ for all $n$. Therefore, since $(U_{n})_{n}$ covers $X$, the set $A$ is closed. Therefore, the set $C=\{x_{2n}|n\in\mathbb{N}\}$ is a closed set in $X$, but $C_{n}\not\subseteq C$ for all $n$ contradicting the maximality of $M$. QED

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+1. Great! It is the same every-other-bit-of-the-difference idea as in my answer. But how does the OP's "nonempty interior" requirement affect this? (Incidentally, I think you want to say $C_n\not\subseteq C$ in the last sentence, rather than $C\not\subseteq C_n$ as you do, and then something about the complementary set.) –  Joel David Hamkins Aug 19 '13 at 18:08
    
Great! It's interesting that in the examples I was thinking of, it is clear that $\bigcap M\neq\emptyset$, since I am taking filters generated by decreasing chains of nonempty closed sets in complete metric spaces. Not sure how the nonempty interior effects this, though. –  Iian Smythe Aug 19 '13 at 23:54

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