Skip to content

Example There exist funcoids ${f}$ and ${g}$ such that

$\displaystyle ( \mathsf{RLD})_{\mathrm{out}} (g \circ f) \neq ( \mathsf{RLD})_{\mathrm{out}} g \circ ( \mathsf{RLD})_{\mathrm{out}} f.$

Proof: Take ${f = {( =)} |_{\Omega}}$ and ${g = \mho \times^{\mathsf{FCD}} \left\{ \alpha \right\}}$ for some ${\alpha \in \mho}$. Then ${( \mathsf{RLD})_{\mathrm{out}} f = \emptyset}$ and thus ${( \mathsf{RLD})_{\mathrm{out}} g \circ ( \mathsf{RLD})_{\mathrm{out}} f = \emptyset}$. We have ${g \circ f = \Omega \times^{\mathsf{FCD}} \left\{ \alpha \right\}}$. Let’s prove ${( \mathsf{RLD})_{\mathrm{out}} (\Omega \times^{\mathsf{FCD}} \left\{ \alpha \right\}) = \Omega \times^{\mathsf{RLD}} \left\{ \alpha \right\}}$.

Really:
$( \mathsf{RLD})_{\mathrm{out}} (\Omega \times^{\mathsf{FCD}} \left\{ \alpha \right\}) = \\ \bigcap {\nobreak}^{\mathsf{RLD}} \mathrm{up} (\Omega \times^{\mathsf{FCD}} \left\{ \alpha \right\}) = \\ \bigcap {\nobreak}^{\mathsf{RLD}} \left\{ K \times \left\{ \alpha \right\} \hspace{1em} | \hspace{1em} K \in \mathrm{up} \Omega \right\} = \\ \bigcap {\nobreak}^{\mathfrak{F}} \left\{ K \hspace{1em} | \hspace{1em} K \in \mathrm{up} \Omega \right\} \times^{\mathsf{RLD}} \left\{ \alpha \right\} = \\ \Omega \times^{\mathsf{RLD}} \left\{ \alpha \right\}.$

Thus ${( \mathsf{RLD})_{\mathrm{out}} (g \circ f) = \Omega \times^{\mathsf{RLD}} \left\{ \alpha \right\} \neq \emptyset}$. $\Box$

Advertisements

This site uses Akismet to reduce spam. Learn how your comment data is processed.