A unique ultrafilter extending a union of filters? - MathOverflow

A unique ultrafilter extending a union of filters?

2011-08-12T21:50:05Z

Original Question:

Let $\mathcal{P}(\omega)/fin$ denote the Boolean algebra formed from $\mathcal{P}(\omega)$ by modding out by the ideal $fin$ of finite subsets of $\omega$. As a first pass at the intended question, consider the following:

Question 0: Are there two filters $F$ and $G$ in $\mathcal{P}(\omega)/fin$ such that there is a unique ultrafilter extending $F \cup G$?

The answer, of course, is yes: consider the case where $F$ is already an ultrafilter, and $G$ is some filter such that $G \subseteq F$. We might therefore ask (what seems to be) a harder question. Given an ideal $I$ in $\mathcal{P}(\omega)/fin$, notice that

$$\{a \in \mathcal{P}(\omega)/fin \: : \: a \geq I\}$$

(where $a \geq I$ iff $(\forall b \in I)[a \geq b]$) is a filter; call such filters regular filters (I made up this terminology, and I would be glad to know if there is already a word for such objects). Now we can ask:

Question 1: Are there two regular filters $F$ and $G$ such that there is a unique ultrafilter extending $F \cup G$?

Via Stone duality, this question can be rephrased (I believe) in topological terms:

Question 1$'$: Are there two regular closed subsets $C,D \subset \omega^{*}$ such that $C \cap D$ is a singleton?

Here, a regular closed set is simply a set which is equal to the closure of its interior, and $\omega^{*} = \beta \omega \setminus \omega$, the space of all non-principal ultrafilters on $\omega$ (i.e. the Stone space of $\mathcal{P}(\omega)/fin$). If $C$ and $D$ are witnesses to a positive answer to question 1$'$, then $int(C)$ and $int(D)$ must be disjoint, in which case the ideals corresponding to these open sets form a gap; this is the basis for my original interest in this question.

Update:

Based on the suggestions given by Andreas Blass, it turns out we have the following consistency result.

Theorem: Under CH, there exist regular filters $F$ and $G$ such that $F \cup G$ extends to a unique ultrafilter.

Proof. (sketch)

Let $\{c_{\alpha} \: : \: \alpha < \omega_{1}\}$ be an enumeration of all elements of $\mathcal{P}(\omega)/fin$. Choose elements $a_{0}, b_{0}$ such that $a_{0} \land b_{0} = 0$, $a_{0} \lor b_{0} < 1$, and furthermore such that either $c_{0} \leq a_{0} \lor b_{0}$, or else $\lnot c_{0} \leq a_{0} \lor b_{0}$.

Now suppose that for $\gamma < \omega_{1}$, we have constructed increasing sequences $\{a_{\alpha} \: : \: \alpha < \gamma\}$ and $\{b_{\alpha} \: : \: \alpha < \gamma\}$ such that, for all $\alpha < \gamma$,

(a) $a_{\alpha} \land b_{\alpha} = 0$;

(b) $a_{\alpha} \lor b_{\alpha} < 1$;

First suppose that $\gamma = \eta + 1$ is a successor ordinal. Let $d \in \{c_{\gamma}, \lnot c_{\gamma}\}$ be such that $$a_{\eta} \lor b_{\eta} \lor d < 1,$$ let ${d_{a}, d_{b}}$ be a (nontrivial, if possible) partition of $d \land \lnot(a_{\eta} \lor b_{\eta})$, and set $a_{\gamma} = a_{\eta} \lor d_{a}$ and $b_{\gamma} = b_{\eta} \lor d_{a}$. Then it is easy to see that $\{a_{\alpha} \: : \: \alpha < \gamma + 1\}$ and $\{b_{\alpha} \: : \: \alpha < \gamma + 1\}$ are increasing sequences satisfying (a) through (c).

Suppose now that $\gamma$ is a limit ordinal. Observe that the sequence $\{\lnot(a_{\alpha} \lor b_{\alpha}) \: : \: \alpha < \gamma\}$ is countable and strictly decreasing. It follows that there exists a nonzero lower bound of this sequence; equivalently, there exists an $e < 1$ such that $e \geq a_{\alpha}$ and $e \geq b_{\alpha}$ for all $\alpha < \gamma$. Moreover, since $\{a_{\alpha} \: : \: \alpha < \gamma\}$ and $\{b_{\alpha} \: : \: \alpha < \gamma\}$ are both countable, they cannot form a gap; hence there exist $a, b \in \mathcal{P}(\omega)/fin$ such that, for all $\alpha < \gamma$, $a \geq a_{\alpha}$ and $b \geq b_{\alpha}$. Replacing $a$ and $b$ by $a \land e$ and $b \land e$, if necessary, we may assume that $a, b \leq e$. Now we can repeat the argument given for the case where $\gamma$ is a successor, replacing $a_{\eta}$ by $a$ and $b_{\eta}$ by $b$.

Thus we obtain $\{a_{\alpha} \: : \: \alpha < \omega_{1}\}$ and $\{b_{\alpha} \: : \: \alpha < \omega_{1}\}$. I claim that these form a gap. If not, then there is some $\beta < \omega_{1}$ such that $c_{\beta} \geq \{a_{\alpha} \: : \: \alpha < \gamma\}$ and $\lnot c_{\beta} \geq \{b_{\alpha} \: : \: \alpha < \gamma\}$; on the other hand, we know that either $c_{\beta} \leq a_{\beta} \lor b_{\beta}$, or $\lnot c_{\beta} \leq a_{\beta} \lor b_{\beta}$, each of which readily yields a contradiction.

I claim also that for every two-element partition $\{p, q\}$ in $\mathcal{P}(\omega)/fin$, one element, say $p$, is such that $\{a_{\alpha} \land p \: : \: \alpha < \omega_{1}\}$ and $\{b_{\alpha} \land p \: : \: \alpha < \omega_{1}\}$ do not form a gap. Indeed, each such partition must be of the form $\{c_{\beta}, \lnot c_{\beta}\}$ for some $\beta < \omega_{1}$. Without loss of generality, suppose we have $c_{\beta} \leq a_{\beta} \lor b_{\beta}$; then it is not difficult to see that $$a_{\beta} \geq \{a_{\alpha} \land c_{\beta} \: : \: \alpha < \omega_{1}\}$$ and likewise $$b_{\beta} \geq \{b_{\alpha} \land c_{\beta} \: : \: \alpha < \omega_{1}\},$$ from which it follows that these sequences do not form a gap. $\blacksquare$

So I suppose the name of the game here is consistency results, such as

Is a negative answer to Question 1 consistent with ZFC? Can a positive answer be proved under any assumptions weaker than CH?

As this seems to be a rather slippery problem, I would welcome any suggested reading on this topic. And thanks again for the helpful replies already given; they are much appreciated.

Answer by Andreas Blass for A unique ultrafilter extending a union of filters?

2011-08-13T21:13:43Z

This isn't an answer but a suggestion of how to get a consistent answer. The idea is to assume the continuum hypothesis (CH) and build a Hausdorff gap with an additional property.

By a Hausdorff gap, I mean two $\omega_1$ -sequences $(A_\alpha)$ and $(B_\alpha)$ of infinite subsets of $\omega$, each almost increasing (here and below, "almost" means "modulo finite sets"), with each $A_\alpha\cap B_\beta$ finite, and such that no set $C$ almost includes every $A_\alpha$ and is almost disjoint from every $B_\alpha$ . [Caution: The same name, "Hausdorff gap" is used with different meanings by lots of people. Some replace the $B_\alpha$ 's by their complements. Some impose additional requirements, making the gap resemble in certain ways the ones constructed by Hausdorff. The latter people don't all agree on what resemblance to require.] Hausdorff's construction works in ZFC and is somewhat tricky. If one assumes CH, there's a brute force construction: List all the subsets $C$ of $\omega$ in an $\omega_1$ -sequence and construct the gap by induction, defining $A_\alpha$ and $B_\alpha$ at step $\alpha$ in such a way as to ensure that the $\alpha$-th $C$ in the list doesn't violate the definition of Hausdorff gap.

Consider (for the moment) any Hausdorff gap; let $I$ and $J$ be the ideals generated by the set of $A_\alpha$ 's and the set of $B_\alpha$ 's, respectively. Let $F$ and $G$ be the filters obtained from $I$ and $J$ as in the question. Then $F\cup G$ generates a proper filter. [Otherwise, there would be some set $C\in F$ with $\omega-C\in G$, and that would violate the definition of Hausdorff gap.] For this filter to be an ultrafilter, you need that, whenever $\omega$ is partitioned into two pieces, then one of the pieces, say $P$, has the property that $(A_\alpha\cap P)$ and $(B_\alpha\cap P)$ do not form a Hausdorff gap, i.e., there should be a set $C$ that almost includes every $A_\alpha\cap P$ and is almost disjoint from every $B_\alpha\cap P$ . An arbitrary Hausdorff gap is unlikely to satisfy this. Nevertheless, it seems to me that this requirement can be met by listing all the relevant partitions in an $\omega_1$ -sequence and, at each stage $\alpha$ in the construction of the gap, ensuring that one of the pieces of the $\alpha$-th partition has the property we want.

If this works, is shows that, under CH, the answer to Question 1 is positive. I would expect that, if it works, then a modification of the argument would work under Martin's axiom (or just $\mathfrak p=\mathfrak c$). [EDIT: I retract the preceding sentence. I envisioned a $\mathfrak c$ long construction of a $(\mathfrak c,\mathfrak c)$ gap, but such a construction runs the risk of ending prematurely with, for example, a Hausdorff gap.] But it doesn't look likely to me that an argument like this can be made to work in ZFC without any special hypotheses. Even though one can construct Hausdorff gaps in ZFC, I don't see any way to "mix" that $\omega_1$ long construction with handling all $\mathfrak c$ partitions if $\mathfrak c>\aleph_1$ . Notice, though, that the argument doesn't need the full strength of a Hausdorff gap; it doesn't matter that the sequences $(A_\alpha)$ and $(B_\alpha)$ are almost increasing, so we could work with something like Lusin gaps instead. Unfortunately, I don't see how that extra freedom helps.

Answer by Joel David Hamkins for A unique ultrafilter extending a union of filters?

2011-08-14T00:34:01Z

Your question is both interesting and tricky, and although I don't have an answer to the specific question you asked, I do have some observations about the situation in other Boolean algebras.

First is the trivial observation that in a Boolean algebra with atoms, the union of two incomparable regular filters certainly can generate an ultrafilter as follows: let $a$ and $b$ be two non-atomic elements whose meet $a\wedge b$ is an atom. Let $F$ be the principal filter consisting of all elements $\geq a$ and let $G$ be the principal filter of all elements $\geq b$. These are both regular in your terminology, witnessed by the corresponding principal ideals below $a$ and $b$, respectively. But the filter generated by the union $F\cup G$ will contain both $a$ and $b$ and hence $a\wedge b$, and this will be the principal ultrafilter generated by this atom.

More generally, weakening your regularity property, I observe that in any Boolean algebra there can be distinct incomparable nontrivial filters whose union generates an ultrafilter, as follows. Let $\mathbb{B}$ be any Boolean algebra with an ultrafilter $\mu$ on $\mathbb{B}$, and suppose $a\in\mu$ with $a\neq 0,1$. Let $F$ be the filter $\{x\vee \neg a\mid x\in \mu\}$ and let $G$ be any regular subfilter of $\mu$ containing $a$, such as the principal filter on $a$, consisting of all $y\geq a$. Note that $F\not\subset G\not\subset F$. Note also that $F\cup G$ has $a$ and has $x\vee \neg a$ for any $x\in\mu$, and so it has $x\wedge a=(x\vee \neg a)\wedge a$. Thus, the union filter generates $\mu$. But also, $F\cup G\subset\mu$ and so the filter it generates is contained in $\mu$, and so $F\cup G$ is contained in a unique ultrafilter, $\mu$.

In your case of $P(\omega)/\text{Fin}$, you could take $\mu$ to be any ultrafilter containing the set $a$ of even numbers, and let $F$ be all sets in $\mu$, but throwing in the odd numbers to accompany every one, and let $G$ be any subfilter of $\mu$ containing the evens, such as the principal filter on the evens. The union $F\cup G$ will contain the evens and the union of any set in $\mu$ with the odds, and so the filter it generates will contain any set in $\mu$, which is an ultrafilter.

But the filter $F$ in these examples is not regular in your sense, and so this does not answer the question.

Meanwhile, the situation is impossible in a complete atomless Boolean algebra:

Theorem. If $F$ and $G$ are regular filters in a complete atomless Boolean algebra $\mathbb{B}$, then $F\cup G$ does not generate an ultrafilter.

Proof. The main point is that if $F$ is a regular filter, as witnessed by ideal $I$, and $\mathbb{B}$ is complete, then actually $F$ and $I$ are principal. To see this, note that $F$ consists of the upper bounds of $I$, but by completeness, there is a least upper bound, and so $F$ has a minimal element $a=\text{inf}(F)=\text{sup}(I)$. Similarly, the regular filter $G$, witnessed by ideal $J$, is also principal, with $b=\text{inf}(G)=\text{sup}(J)$. It follows that $a\wedge b$ is the least element of the filter generated by $F\cup G$, and so $\langle F\cup G\rangle$ is the principal filter on $a\wedge b$. In an atomless Boolean algebra, however, principal filters are never ultrafilters. QED

The point was that regular ultrafilters in a complete Boolean algebra are the same as principal filters.

A related observation is that if $\mathbb{B}$ is not complete, then we may still complete it, forming the completion $\bar{\mathbb{B}}$, in which $\mathbb{B}$ is dense, and the idea of the previous proof shows:

Lemma. In any Boolean algebra $\mathbb{B}$, the regular filters $F$ are exactly the traces on $\mathbb{B}$ of the principal filters of the completion $\bar{\mathbb{B}}$.

Proof. If $F$ is regular in $\mathbb{B}$, as witnessed by ideal $I$, then $F$ consists precisely of the upper bounds of $I$ in $\mathbb{B}$. Let $a$ be the least upper bound of $I$ in the completion $\bar{\mathbb{B}}$, and let $\bar F$ be the principal filter on $a$ in $\bar{\mathbb{B}}$. It follows that $F\subset\bar F$, and furthermore $\bar F\cap\mathbb{B}=F$, because any element of $\mathbb{B}$ above $a$ in $\bar{\mathbb{B}}$ is an upper bound of $I$ and hence already in $F$. Thus, $F$ is the trace of the principal filter $\bar{F}$ in $\bar{\mathbb{B}}$ on $\mathbb{B}$.QED

And another observation concerns the situation arising in any nonatomic positive instance of the phenomenon:

Lemma. If $F$ and $G$ are regular filters in an atomless Boolean algebra $\mathbb{B}$, witnessed by ideals $I$ and $J$, respectively, and $F\cup G$ generates an ultrafilter, then $I$ and $J$ are orthogonal, in the sense that $i\wedge j=0$ for any $i\in I, j\in J$. Thus, the dual filter to $I$ is contained in $G$ and the dual filter to $J$ is contained in $F$.

Proof. If $i\wedge j\gt 0$ for some $i\in I$ and $j\in J$, then since $F$ consists of the upper bounds of $I$ and $G$ consists of the upper bounds of $J$, it follows that every element of $F\cup G$ is contained in the principal filter above $i\wedge j$, and since $\mathbb{B}$ is atomless this cannot be an ultrafilter. Another way to say $i\wedge j=0$ is to say that $\neg i$ is above $j$ and $\neg j$ is above $i$, and so the statement about dual filters follows. QED

I have called these both "lemmas," but I'm not yet quite sure what the conclusion is that they might be building toward.

Answer by KP Hart for A unique ultrafilter extending a union of filters?

2011-11-25T14:27:19Z

The following is due to Alan Dow:

In any model obtained by adding $\aleph_2$ many Cohen reals to a model of $\mathsf{CH}$ the statement is false. We force with $\mathbb{P}=\operatorname{Fn}(\omega_2,2)$ and we let $\dot{\mathcal{I}}$, $\dot{\mathcal{J}}$ and $\dot u$ be $\mathbb{P}$-names such that $\dot{\mathcal{I}}$ and $\dot{\mathcal{J}}$ are forced to be ideals and $\dot u$ is forced to be the unique ultrafilter that extends the two associated regular filters.

Now let $M$ be an elementary substructure of a suitable large $H(\theta)$ that has cardinality $\aleph_1$ and that is closed under $\omega$-sequences. Let $\delta=M\cap\omega_2$ and $\mathbb{P}_M=\operatorname{Fn}(\delta)$.

By elementarity the $\mathbb{P}_M$-names $\dot{\mathcal{I}}\cap M$ and $\dot{\mathcal{J}}\cap M$ are forced to be ideals and $\dot u\cap M$ is forced to be the unique ultrafilter that extends the two associated regular filters.

Work in $V[G\cap M]$. We write $\mathcal{I}_M$, $\mathcal{I}_M$ and $u_M$ for the interpretations of $\dot{\mathcal{I}}\cap M$, $\dot{\mathcal{J}}\cap M$ and $\dot u\cap M$ in this model.

Note that every element of $u_M$ meets elements of $\mathcal{I}_M$ and $\mathcal{J}_M$. On the other hand if $\mathcal{I}'$ and $\mathcal{J}'$ are countable subfamilies of $\mathcal{I}_M$ and $\mathcal{J}_M$ respectively then some infinite subset, $x$, of $\omega$ separates the elements of $\mathcal{I}'$ from those of $\mathcal{J}'$. This implies that, without loss of generality, for every countable subfamily $\mathcal{J}'$ of $\mathcal{J}_M$ there is an element $x\in u_M$ such that $x\cap y$ is finite for all $y\in\mathcal{J}'$.

Using this one can construct a sequence $\langle b_\alpha:\alpha<\omega_1\rangle$ in $\mathcal{J}_M$ such that every element of $u_M$ contains all but countably many of the $b_\alpha$.

In $V[G]$ consider the next Cohen real $c$, added by $\operatorname{Fn}([\delta,\delta+\omega,2)$, say and assume, without loss of generality, that $c\notin u$.

There must be a set $Y\subseteq c$ that separates $\lbrace x\cap c:x\in\mathcal{I}\rbrace $ from $\lbrace y\cap c:y\in\mathcal{J}\rbrace $. In $V[G\cap M]$ we take names, $\dot c$ and $\dot Y$, for $c$ and $Y$, note that these are $\operatorname{Fn}(C,2)$-names for some countable set $C$. For every $\alpha<\omega_1$ it is forced that $\dot Y\cap b_\alpha$ is finite; by pigeon-holing there will be one $p\in\operatorname{Fn}(C,2)$ and one $n\in\omega$ such that $p$ forces $\dot Y\cap b_\alpha\subseteq n$ for uncountably many $\alpha$. Still in $V[G\cap M]$ let $Y_p=\lbrace k:(\exists q\le p)(q \Vdash k\in\dot Y)\rbrace $. Then $b_\alpha\cap Y_p\subseteq n$ for uncountably many $\alpha$ so that $Y_p\notin u_M$. On the other hand for every $x\in\mathcal{I}_M$ it is forced that $x\cap\dot c\subseteq \dot Y$ (mod finite); this implies that $x\setminus Y_p$ is finite for all $x\in\mathcal{I}_M$, so that $Y_p\in u_M$.