Upgrading a completary multifuncoid

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Posted	by:	porton
	on:	February 4th, 2012

Let $\mho$ be a set, $\mathfrak{F}$ be the set of filters on $\mho$ ordered reverse to set-theoretic inclusion, $\mathfrak{P}$ be the set of principal filters on $\mho$ , let $n$ be an index set. Consider the filtrator $\left( \mathfrak{F}^n ; \mathfrak{P}^n \right)$ .

Conjecture If $f$ is a completary multifuncoid of the form $\mathfrak{P}^n$ , then $E^{\ast} f$ is a completary multifuncoid of the form $\mathfrak{F}^n$ .

See below for definition of all concepts and symbols used to in this conjecture.

Refer to this Web site for the theory which I now attempt to generalize.

Definition A filtrator is a pair $\left( \mathfrak{A}; \mathfrak{Z} \right)$ of a poset $\mathfrak{A}$ and its subset $\mathfrak{Z}$ .

Having fixed a filtrator, we define:

Definition $\ensuremath{\operatorname{up}}x = \left\{ Y \in \mathfrak{Z} \hspace{0.5em} | \hspace{0.5em} Y \geqslant x \right\}$ for every $X \in \mathfrak{A}$ .

Definition $E^{\ast} K = \left\{ L \in \mathfrak{A} \hspace{0.5em} | \hspace{0.5em} \ensuremath{\operatorname{up}}L \subseteq K \right\}$ (upgrading the set $K$ ) for every $K \in \mathscr{P} \mathfrak{Z}$ .

Definition Let $\mathfrak{A}$ is a family of join-semilattice. A completary multifuncoid of the form $\mathfrak{A}$ is an $f \in \mathscr{P} \prod \mathfrak{A}$ such that we have that:

$L_0 \cup L_1 \in f \Leftrightarrow \exists c \in \left\{ 0, 1 \right\}^n : \left( \lambda i \in n : L_{c \left( i_{} \right)} i \right) \in f$

$L_0, L_1 \in \prod \mathfrak{A}$

\item If $L \in \prod \mathfrak{A}$ and $L_i = 0^{\mathfrak{A}_i}$ for some $i$ then $\neg f L$ .

$\mathfrak{A}^n$ is a function space over a poset $\mathfrak{A}$ that is $a\le b\Leftrightarrow \forall i\in n:a_i\le b_i$ for $a,b\in\mathfrak{A}^n$ .

For finite $n$ this problem is equivalent to Upgrading a multifuncoid .

It is not hard to prove this conjecture for the case $\ensuremath{\operatorname{card}}n \leqslant 2$ using the techniques from this my article. But I failed to prove it for $\ensuremath{\operatorname{card}}n = 3$ and above.