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This short article is the first my public writing where I introduce the concept of multidimensional funcoid which I am investigating now.

But the main purpose of this article is to formulate a conjecture (see below). This is the shortest possible writing enough to explain my conjecture to every mathematician.

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

If you solve this my open problem, please send me the solution.

Definition 1 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 2 $\mathrm{up}\,x = \left\{ Y \in \mathfrak{Z} \hspace{0.5em} | \hspace{0.5em} Y \geqslant x \right\}$ for every ${X \in \mathfrak{A}}$.

Definition 3 $E^{\ast} K = \left\{ L \in \mathfrak{A} \hspace{0.5em} | \hspace{0.5em} \mathrm{up}\,L \subseteq K \right\}$ (upgrading the set ${K}$) for every ${K \in \mathscr{P} \mathfrak{Z}}$.

Definition 4 A free star on a join-semilattice ${\mathfrak{A}}$ with least element 0 is a set ${S}$ such that ${0 \not\in S}$ and $\displaystyle \forall A, B \in \mathfrak{A}: \left( A \cup B \in S \Leftrightarrow A \in S \vee B \in S \right) .$

Definition 5 Let ${\mathfrak{A}}$ be a family of posets, $f \in \mathscr{P} \prod \mathfrak{A}$ ( $\prod \mathfrak{A}$ has the order of function space of posets), ${i \in \mathrm{dom}\,\mathfrak{A}}$, ${L \in \prod \mathfrak{A}|_{\left( \mathrm{dom}\,\mathfrak{A} \right) \setminus \left\{ i \right\}}}$. Then $\displaystyle \left( \mathrm{val}\,f \right)_i L = \left\{ X \in \mathfrak{A}_i \hspace{0.5em} | \hspace{0.5em} L \cup \left\{ (i ; X) \right\} \in f \right\} .$

Definition 6 Let ${\mathfrak{A}}$ is a family of posets. A multidimensional funcoid (or multifuncoid for short) of the form ${\mathfrak{A}}$ is an ${f \in \mathscr{P} \prod \mathfrak{A}}$ such that we have that:

1. ${\left( \mathrm{val} f \right)_i L}$ is a free star for every ${i \in \mathrm{dom} \mathfrak{A}}$, ${L \in \prod \mathfrak{A}|_{\left( \mathrm{dom} \mathfrak{A} \right) \setminus \left\{ i \right\}}}$.
2. ${f}$ is an upper set. $\mathfrak{A}^n$ 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 $a,b\in\mathfrak{A}^n$.

Conjecture 7 Let $\mho$ be a set, ${\mathfrak{F}}$ be the set of f.o. on $\mho$, ${\mathfrak{P}}$ be the set of principal f.o. on $\mho$, let ${n}$ be an index set. Consider the filtrator ${\left( \mathfrak{F}^n ; \mathfrak{P}^n \right)}$. If ${f}$ is a multifuncoid of the form ${\mathfrak{P}^n}$, then ${E^{\ast} f}$ is a multifuncoid of the form ${\mathfrak{F}^n}$.

It is not hard to prove this conjecture for the case $\mathrm{card}\,n\le 2$ using the techniques from this my article. But it’s not easy to prove it for $\mathrm{card}\,n= 3$ and above. I failed to find a general solution.

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