Importance: Medium ✭✭
Author(s): Porton, Victor
Subject: Topology
Keywords:
Recomm. for undergrads: no
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:
    \item $ 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 $ for every $ 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.


* indicates original appearance(s) of problem.

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