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Let $\mathfrak{A}$ and $\Gamma$ be boolean lattices with $\Gamma$ being a boolean sublattice of $\mathfrak{A}$.

Let us denote $\mathfrak{F}(X)$ the set of filters on a poset $X$ ordered by set theoretic inclusion. Denote $\mathfrak{P}(X)$ the set of principal filters on a poset $X$ ordered by set theoretic inclusion.

Can we warrant that there exists an order embedding from $\mathfrak{F}(\Gamma)$ to $\mathfrak{F}(\mathfrak{A})$ which maps $\mathfrak{P}(\Gamma)$ into $\mathfrak{P}(\mathfrak{A})$ (with principal filters on $\Gamma$ mapped into corresponding (generated by the same element $x$ of $\Gamma$) principal filters on $\mathfrak{A}$)?


In the real example which I try to solve, $\mathfrak{A}$ is the set of binary relations between some (infinite) sets $A$ and $B$ and $\Gamma$ is the set of finite unions of binary cartesian products $X\times Y$ of subsets $X\in\mathscr{P}A$ and $Y\in\mathscr{P}B$ of these sets.

If this does not hold in general, does it hold for this special case?

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  • $\begingroup$ It seems that the answer is no. I will be positively surprised and enjoy if the answer is yes $\endgroup$ – porton Sep 16 '17 at 16:37
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The answer to your question is yes in general.

Let $S$ be a po (particially ordered) set and $C$ a po subset of $S$.
A filter for a po set is any down directed upper subset.

Define
$\operatorname{up}_S A = \{ x \in S \mid \exists a \in A: a \leq x \}$, $A$ subset $S$;

$\operatorname{up}_C A = \{ x \in C \mid \exists a \in C: a \leq x \}$, $A$ subset $C$;

$Fs = \{ F \mid F \text{ filter for }S \}$; $Fc = \{ F \mid F \text{ filter for }C \}$.

Theorem $\operatorname{up}S:Fc \rightarrow Fs$, $F \rightarrow \operatorname{up}_S F$ is subset-order embedding;
if $F$ principle filter in $Fc$, then $\operatorname{up}_S F$ is a principle filter in $Fs$.

Lemmas: for all $F$ in $Fc$, $\operatorname{up}_S F$ in $Fs$;
$F = C \cap \operatorname{up}_S F$.

Proof of theorem. Assume $F$, $G$ in $Fc$. Clearly if $F$ subset $G$, then $\operatorname{up}_S F$ subset $\operatorname{up}_S G$.
If $\operatorname{up}_S F$ subset $\operatorname{up}_S G$, then
$F = C \cap \operatorname{up}_S F$ subset $C \cap \operatorname{up}_S G$. Hence $\operatorname{up}_S$ is order embedding.

Let $F$ be a principle filter in $Fc$. Then
some $a$ in $A$ with $F = \operatorname{up}_C a$; $\operatorname{up}_S \operatorname{up}_C a = \operatorname{up}_S a$. Whence $\operatorname{up}_S F$ is a principle filter.

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  • $\begingroup$ You seems right. But now I've just realized that my question was severely wrong. It is just not what I wanted to ask. $\endgroup$ – porton Sep 21 '17 at 11:49

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