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Let $\mathfrak{A}$ is a poset. For $a, b \in \mathfrak{A}$ we will denote $a \curlyvee b$ if only if there is a non-least element $c$ such that $c \leqslant a \wedge c \leqslant b$.

I call a poset $\mathfrak{A}$ separable if and only if $\forall x \in \mathfrak{A}: \left( x \curlyvee a \Leftrightarrow x \curlyvee b \right) \Rightarrow a = b$.

Let $\mathfrak{A}$ is a family of posets indexed by some set $n$. We introduce a partial order (called product order) on $\prod \mathfrak{A} = \prod_{i \in n} \mathfrak{A}_i$ by the formula (for every $a, b \in \prod \mathfrak{A}$) $$ a \leqslant b \Leftrightarrow \forall i \in n : a_i \leqslant b_i . $$ It is easy to prove that $$ a \curlyvee b \Leftrightarrow \exists i \in \operatorname{dom}\mathfrak{A}: a_i \curlyvee b_i . $$ Let $\mathfrak{A}$ is an indexed family of separable posets. Can we infer that $\prod\mathfrak{A}$ is a separable poset?

I think the answer is ``no'', because I have failed to prove it. But I don't know a counter-example.

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Can you give an example of a separable poset? If you can prove/disprove that the product of two separable posets is separable, then I think this is enough. –  David Roberts Jul 11 '12 at 23:33
    
I wonder if it's worth trying to simplify your notation as a first step. For instance, $a\wedge c$ always satisfies $a\wedge c \le a$ and $a\wedge c\le c$, so you could replace $c\le a\wedge c\le b$ by $c = a\wedge c \le b$ which also then implies $c\le a$. –  Patricia Hersh Jul 12 '12 at 2:38
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@Patricia: I think $c\leq a\land c\leq b$ was intended to mean $(c\leq a)\land (c\leq b)$, not $c\leq (a\land c)\leq b$. In other words, $\land$ is logical conjunction, not lattice meet. After all, the poset wasn't assumed to be a lattice (or semilattice). –  Andreas Blass Jul 12 '12 at 5:05
    
@Porton: Are you assuming that your poset has a least element? Or at least assuming that every two elements have a lower bound? If not, then I don't understand the equivalence that you said is "easy to prove". Specifically, you could have $a_i\curlyvee b_i$ for one index $i$ while for another index $j$ the components $a_j$ and $b_j$ have no lower bound at all. Then the right side of the claimed equivalence would be true but the left would not. Note that, if there is a least element, say 0, then the answer to your question is easily yes, by using an $x$ in which all but one component is 0. –  Andreas Blass Jul 12 '12 at 5:16
    
@Andreas: Thanks! I thought this looked strange. That makes more sense! –  Patricia Hersh Jul 12 '12 at 5:43

1 Answer 1

up vote 4 down vote accepted

Here's a counterexample, taking advantage of the option to have or not have a 0 in a poset. Take the index set to be $\{1,2\}$. Take $\mathfrak A_1$ to be an atomless Boolean algebra minus its 0 element. It is well-known and easy to see that this is separable; if $a\neq b$, then one of $a-b$ and $b-a$ in the Boolean algebra is non-0 and serves as the $x$ required in separability. Let $\mathfrak A_2$ be the poset consisting of three elements, namely a least element 0 and two larger, incomparable elements $p$ and $q$. It is easy to check that this is separable. I claim that the product of these two separable posets is not separable. Let 1 be the top element of $\mathfrak A_1$. Then $(1,p)$ and $(1,q)$ are distinct elements of the product, so separability would require the existence of a pair $x=(x_1,x_2)$ such that $x\curlyvee(1,p)$ but $x\not\curlyvee(1,q)$ or vice versa; by symmetry, I can ignore the "vice versa" possibility. Since $x\curlyvee(1,p)$, there must be $y=(y_1,y_2)$ such that $y_1\leq x_1$, $y_2\leq x_2$, and $y_2\leq p$ (the remaining conditions, that $y_1\leq 1$ and $y$ is not least in the product, are automatic, the first because 1 is the top of $\mathfrak A_1$ and the second because $\mathfrak A_1$ has no least element). Then $y'=(y_1,0)$ satisfies $y'\leq y\leq x$ and $y'\leq(x_1,0)\leq(1,q)$, and $y$ is not least in the product (again because $y_1$ is not least in $\mathfrak A_1$). This contradicts the fact that $x\not\curlyvee(1,q)$.

EDIT: In response to a comment from Porton, suggesting that I misinterpreted a quantifier, I admit that this is a possibility, but I think the result I claimed, that a product of separable posets with least elements is separable, holds under either interpretation. First, the interpretation I had in mind (because it seems the more reasonable concept) is that separability means that $a=b$ follows from $(\forall x)(x\curlyvee a\iff x\curlyvee b)$. As in the question, I'll write $n$ for the index set, $\mathfrak A_i$ for the factor with index $i$, and $\prod\mathfrak A$ for the product. I'll also write 0 for the least element in any poset. To prove separability, I'll assume $a,b\in\prod\mathfrak A$ with $a\neq b$, and I'll find an $x\in\mathfrak A$ such that either $x\curlyvee a$ and $x\not\curlyvee b$ or vice versa. Since $a\neq b$, there is an index $i$ with $a_i\neq b_i$; fix such an $i$. Since $\mathfrak A_i$ is separable, choose some $y\in\mathfrak A_i$ such that either $y\curlyvee a_i$ and $y\not\curlyvee b_i$ or vice versa; without loss of generality, assume the former, and fix a non-0 $z\leq a_i$ with $z\leq y$. Define $x\in\prod\mathfrak A$ by setting $x_i=z$ and $x_j=0$ for all $j\neq i$. Then $x\neq0$ (because $x_i=z\neq0$) and $x\leq a$, so $x\curlyvee a$. Yet $x\not\curlyvee b$, because any $w\in\prod\mathfrak A$ that is below both $x$ and $b$ would have $w_i$ below both $x_i=z$ and $b_i$, which is absurd as $z\leq y$ and $y\not\curlyvee b_i$.

The alternative reading of the definition of separability is that the colon after the quantifier makes the scope of the quantifier the whole implication rather than the antecedent. Under this interpretation, the implication is true for whatever value of $x$ one chooses. I choose the value 0 (since we're assuming the posets have least elements 0) and infer the implication $(0\curlyvee a\iff0\curlyvee b)\implies a=b$. But by definition of $\curlyvee$, we never have $0\curlyvee$ anything, so it is true that $(0\curlyvee a\iff0\curlyvee b)$ because both sides of the biconditional are false. So we infer from our implication that $a=b$. Thus, under this reading of the definition, the only way for a poset with a least element to be separable is to have only one element. And that's preserved by products.

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How is it proved existence of atomless Boolean algebras? –  porton Jul 12 '12 at 18:33
    
Thanks for a solution. The question, whether there are non-separable products of posets each having the least element, yet remains. I will yet read your answer and check it for possible errors. By the way, I solved the REAL problem (I can't explain it in customary terminology) due which I have raised this question, and now this question does not poses for me to be very important, but anyway thanks for a solution. –  porton Jul 12 '12 at 18:39
    
@porton: One example of an atomless Boolean algebra is obtained by starting with the Boolean algebra of subsets of an infinite set $X$, and then forming a quotient algebra by identifying any two subsets of $X$ whose symmetric difference is finite. If you consider only posets that have least elements, then I still believe that, as I said in my second comment on the question, that the answer is easily yes. (In fact, you said, in comment 7, that the answer is trivially yes if every two elements have a lower bound, which is surely true if there is a least element.) –  Andreas Blass Jul 12 '12 at 18:49
    
@Andreas Blass: I don't see a proof that the product of separable posets with least elements is a separable poset. Dear Andreas, it seems likely for me that you confuse $\exists$ and $\forall$ quantifiers. Try to write the proof down and see that there are wrong quantifiers. Or am I wrong? –  porton Jul 12 '12 at 19:04
    
@Andreas Blass: You are right, this proves separability of the product of separable posets with least elements. I was not enough careful. –  porton Jul 15 '12 at 19:24

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