chap-quasiatoms.tex

\chapter{Quasi-atoms}

\begin{defn}
\emph{Quasi-atoms} funcoid~$\mathscr{A}$ is the funcoid $A\rightarrow \atoms^{\mathfrak{A}} A$ defined by the formula
$\rsupfun{\mathscr{A}} X = \atoms^{\mathfrak{A}} X$.
\end{defn}

This really defines a funcoid because $\atoms^{\mathfrak{A}} \bot = \emptyset$ and
$\atoms^{\mathfrak{A}}(X\cup Y) = \atoms^{\mathfrak{A}}X\cup\atoms^{\mathfrak{A}}Y$.

\begin{obvious}
$\mathscr{A}$ is a co-complete funcoid.
\end{obvious}

\begin{prop}
$\rsupfun{\mathscr{A}^{-1}} Y = \bigsqcup Y$.
\end{prop}

\begin{proof}
$Y \nasymp \left\langle \mathscr{A} \right\rangle^{\ast} X \Leftrightarrow Y
\nasymp \atoms^{\mathfrak{A}} X \Leftrightarrow \exists x \in
\atoms^{\mathfrak{A}} X, y \in Y : x \nasymp y \Leftrightarrow \exists y
\in Y : X \nasymp y \Leftrightarrow \text{(because $X$ is a principal filter)}
\Leftrightarrow X \nasymp \bigsqcup Y$.
\end{proof}

Note
$\rsupfun{\mathscr{A}} \mathcal{X} =
\bigsqcap^{\mathscr{F}}_{X \in \up \mathcal{X}}
\atoms^{\mathfrak{A}} X$;

$\rsupfun{\mathscr{A}^{-1}} \mathcal{Y} =
\bigsqcap^{\mathscr{F}}_{Y \in \up \mathcal{Y}} \bigsqcup Y$
($\mathcal{Y}$ is filter on the set of ultrafilters).

Can $\atoms^{\mathfrak{A}} \mathcal{X}$ be restored knowing $\supfun{\mathscr{A}} \mathcal{X}$?
Can $\bigsqcup \mathcal{Y}$ be restored knowing $\supfun{\mathscr{A}^{-1}} \mathcal{X}$?

\begin{prop}
(Provided that~$A$ is infinite) $\mathscr{A}$ is not complete.
\end{prop}

\begin{proof}
Take a nonprincipal ultrafilter~$x$. Then $\rsupfun{\mathscr{A}^{-1}}\{x\} = \bigsqcup\{x\} = x$
is a nonprincipal filter.
\end{proof}

\begin{conjecture}
There is such filter~$\mathcal{X}$ that $\rsupfun{\mathscr{A}} \mathcal{X}$ is non-principal.
\end{conjecture}

Does quasi-atoms funcoid define a more elegant replacement of $\atoms^{\mathfrak{A}}$? Does this concept have any use?