2
$\begingroup$

Let $\mathbb{B} = \langle B, \wedge, \vee, \leq, \neg, 0, 1 \rangle$ be an atomless complete Boolean algebra.

We call $f$ a splitting function on $\mathbb{B}$ iff

  1. $f : B-\{1\} \longrightarrow B \times B \ \ \ (b \mapsto (b_0, b_1))$,
  2. $b_i \leq b$ for all $b \in B-\{1\}$ and $i \in \{0, 1\}$,
  3. $b_0 \wedge b_1 = 0$ for all $b \in B-\{1\}$.

A splitting function on $\mathbb{B}$ is monotone iff

  1. $b \leq b'$ implies $b_i \leq b_{i}'$, for all $\{b, b'\} \subset B-\{1\}$ and $i \in \{0, 1\}$.

Monotone splitting functions trivially exist: consider eg. the map $b \mapsto (0, b)$. However, it is not so clear to me how non-trivial these can be.

In particular, is there always a monotone splitting function on $\mathbb{B}$ satisfying

  1. $\sup\{b_{i}' : b \leq b'\} = \neg b_{1-i}$ for all $b \in B-\{1\}$ and $i \in \{0, 1\}$?

If not, what additional conditions on $\mathbb{B}$ can guarantee its existence, and in what cases does it fail to exist?

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ Given 4, your condition 5 is a bit hopeless: the supremum $\sup\{c_i:c\ge b\}$, when $b$ is fixed, does not depend on $b$. More precisely, it implies that $b_{1-i}$ does not depend on $b$, so has to be constant. $\endgroup$
    – YCor
    Commented Sep 27, 2018 at 16:54
  • $\begingroup$ Sorry, I don't quite get why the supremum does not depend on $b$. $b$ varies over the domain of $f$. $\endgroup$
    – Zoorado
    Commented Sep 27, 2018 at 17:07
  • 1
    $\begingroup$ For any fixed $b \neq 0$ in the domain, $b_i$ and $(\neg b)_i$ are both defined but $(b \vee \neg b)_i$ is not. Same holds if you replace $\neg b$ with any $c \geq \neg b$ in the previous sentence. $\endgroup$
    – Zoorado
    Commented Sep 28, 2018 at 1:51
  • $\begingroup$ Thanks, you're right. The sup of $c_i$ for $c\ge b$ is the same as the sup for $c$ such that $b\vee c\neq 1$, but that's it. $\endgroup$
    – YCor
    Commented Sep 28, 2018 at 5:41

2 Answers 2

Reset to default
1
$\begingroup$

Claim: Let $a \mapsto (a_0, a_1)$ be any map (for $a < 1_B$) satisfying (1)-(4). Call a condition $b < 1_B$ left-trivial (resp. right-trivial) if $(\forall a \leq b)(a_0 = 0_B)$ (resp. $(\forall a \leq b)(a_1 = 0_B)$). Then we can partition $1_B$ into two trivial conditions.

Proof: Let $D$ be the family of all trivial members of $B \setminus \{0_B, 1_B\}$. Note that if $0_B < x < 1_B$ and $x$ is nontrivial then, then for some $y \leq x$, both $y_0, y_1 > 0_B$. It follows that $y_0$ is right-trivial and $y_1$ is left-trivial. So $D$ is dense in $B$. Let $A$ be a maximal antichain of members of $D$. Let $x$ (resp. y) be the union of all left-trivial (resp. right-trivial) members of $A$. Then $\{x, y\}$ is such a partition.

It follows that for any $z < 1_B$, $z_0 = z \cap y$ and $z_1 = z \cap x$. This determines all such maps.

Improve this answer
$\endgroup$
4
  • $\begingroup$ This is great. So condition 5. is untenable for any non-trivial map. $\endgroup$
    – Zoorado
    Commented Sep 28, 2018 at 17:03
  • 1
    $\begingroup$ Sorry I don't understand almost anything. What do you mean by condition? Just any element $<1$? What are elements of $D$ - those which are left and right trivial or those which are either left or right trivial? Why does $y_0,y_1>0$ imply that $y_0$ is right-trivial and $y_1$ is left-trivial? What do you mean by dense? $\sup=1$ or what? Why is a join of left (right) trivials again left (right) trivial? Why $x\lor y=1$? Why $x\land y=0$? Why does it follow that $z_0=z\land y$ and $z_1=z\land x$? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Sep 28, 2018 at 19:49
  • $\begingroup$ I retract my previous statement. What is shown is that either of $\{a : a_0 = 0\}$ or $\{a : a_1 = 0\}$ (or both) is dense open, but that doesn't mean either all the $a_0$'s or all the $a_1$'s are 0. So I think it may still be possible for condition 5. to hold. $\endgroup$
    – Zoorado
    Commented Sep 29, 2018 at 4:02
  • 1
    $\begingroup$ @Adam I approved the edit, but it seems you created a new account to make it. Please try to stay logged in as yourself. Then you can make comments in response to other comments on your answer. $\endgroup$
    – David White
    Commented Sep 29, 2018 at 13:07
1
$\begingroup$

I think you are asking too much. Assume we have such a function and let $a$ be nonzero such that both $a_0$ and $a_1$ are nonzero. Then $b\le a_0$ implies $b_1=0$ and $b\le a_1$ implies $b_0=0$. If $a$ is nonzero such that $a_0=0$ and $0<a_1<a$ then $b\le a\setminus a_1$ implies both $b_0=0$ and $b_1=0$, so the function is trivial below $a\setminus a_1$. This shows that the union of the three sets $\{a:a_0=a_1=0\}$, $\{a:a_0=0\land a_1=a\}$ and $\{a:a_0=a\land a_1=0\}$ is open and dense in the algebra.

Improve this answer
$\endgroup$
8
  • $\begingroup$ Hmm but $1$ is not in the domain of any splitting function, so $1_0$ and $1_1$ are not defined. $\endgroup$
    – Zoorado
    Commented Sep 28, 2018 at 12:24
  • $\begingroup$ OK, I missed that, but that is immaterial: the function, if it exists, works in $\{b:b\le a\}$ for any nonzero $a$. So you may as well have $1$ in the domain anyway. I amended my answer. $\endgroup$
    – KP Hart
    Commented Sep 28, 2018 at 12:38
  • $\begingroup$ Hmm I don't need all values to be non-zero. But you have a point: if $a_1$ is non-zero then $c_0$ needs to be zero. That gives another property of such a function. I need some time to think if this leads to a contradiction with the 5 conditions. $\endgroup$
    – Zoorado
    Commented Sep 28, 2018 at 14:40
  • $\begingroup$ I thought a bit more and improved my answer a bit. $\endgroup$
    – KP Hart
    Commented Sep 28, 2018 at 19:21
  • 1
    $\begingroup$ In the context of Boolean algebras (even partial orders in general): $D$ is dense means for every non-zero $p$ there is a non-zero $q\in D$ with $q\le p$, and $D$ is open means: if $x\in D$ and $y\le x$ then $y\in D$. $\endgroup$
    – KP Hart
    Commented Sep 30, 2018 at 13:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .