Conjecture: Upgrading a multifuncoid

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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