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 $latex {\left( \mathfrak{A}; \mathfrak{Z} \right)}&fg=000000$ of a poset $latex {\mathfrak{A}}&fg=000000$ and its subset $latex {\mathfrak{Z}}&fg=000000$.

Having fixed a filtrator, we define:

Definition 2 $latex \mathrm{up}\,x = \left\{ Y \in \mathfrak{Z} \hspace{0.5em} | \hspace{0.5em} Y \geqslant x \right\}&fg=000000$ for every $latex {X \in \mathfrak{A}}&fg=000000$.

Definition 3 $latex E^{\ast} K = \left\{ L \in \mathfrak{A} \hspace{0.5em} | \hspace{0.5em} \mathrm{up}\,L \subseteq K \right\}&fg=000000$ (upgrading the set $latex {K}&fg=000000$) for every $latex {K \in \mathscr{P} \mathfrak{Z}}&fg=000000$.

Definition 4 A free star on a join-semilattice $latex {\mathfrak{A}}&fg=000000$ with least element 0 is a set $latex {S}&fg=000000$ such that $latex {0 \not\in S}&fg=000000$ and

$latex \displaystyle \forall A, B \in \mathfrak{A}: \left( A \cup B \in S \Leftrightarrow A \in S \vee B \in S \right) . &fg=000000$

Definition 5 Let $latex {\mathfrak{A}}&fg=000000$ be a family of posets, $latex f \in \mathscr{P} \prod \mathfrak{A}&fg=000000$ ($latex \prod \mathfrak{A}&fg=000000$ has the order of function space of posets), $latex {i \in \mathrm{dom}\,\mathfrak{A}}&fg=000000$, $latex {L \in \prod \mathfrak{A}|_{\left( \mathrm{dom}\,\mathfrak{A} \right) \setminus \left\{ i \right\}}}&fg=000000$. Then

$latex \displaystyle \left( \mathrm{val}\,f \right)_i L = \left\{ X \in \mathfrak{A}_i \hspace{0.5em} | \hspace{0.5em} L \cup \left\{ (i ; X) \right\} \in f \right\} . &fg=000000$

Definition 6 Let $latex {\mathfrak{A}}&fg=000000$ is a family of posets. A multidimensional funcoid (or multifuncoid for short) of the form $latex {\mathfrak{A}}&fg=000000$ is an $latex {f \in \mathscr{P} \prod \mathfrak{A}}&fg=000000$ such that we have that:

  1. $latex {\left( \mathrm{val} f \right)_i L}&fg=000000$ is a free star for every $latex {i \in \mathrm{dom} \mathfrak{A}}&fg=000000$, $latex {L \in \prod \mathfrak{A}|_{\left( \mathrm{dom} \mathfrak{A} \right) \setminus \left\{ i \right\}}}&fg=000000$.
  2. $latex {f}&fg=000000$ is an upper set.

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

Conjecture 7 Let $latex \mho&fg=000000$ be a set, $latex {\mathfrak{F}}&fg=000000$ be the set of f.o. on $latex \mho&fg=000000$, $latex {\mathfrak{P}}&fg=000000$ be the set of principal f.o. on $latex \mho&fg=000000$, let $latex {n}&fg=000000$ be an index set. Consider the filtrator $latex {\left( \mathfrak{F}^n ; \mathfrak{P}^n \right)}&fg=000000$. If $latex {f}&fg=000000$ is a multifuncoid of the form $latex {\mathfrak{P}^n}&fg=000000$, then $latex {E^{\ast} f}&fg=000000$ is a multifuncoid of the form $latex {\mathfrak{F}^n}&fg=000000$.

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

2 thoughts on “Conjecture: Upgrading a multifuncoid

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