Upgrading a completary multifuncoid

OpenYear of origin: 2018

Posted online: 2018-12-01 21:14:41Z by Victor Lvovich Porton26

Cite as: P-181201.15

  • General Topology
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Problem's Description

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 $.

The definitions follow:

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 $ \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} \; | \; \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 $ for every $ L_0, L_1 \in \prod \mathfrak{A} $.

- 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 $.

  1. BookIs an originAlgebraic General Topology. Volume 1

    year of publication: 2018fulltext


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  • Created at: 2018-12-01 21:14:41Z