Publish here. Donate

Difference between revisions of "Funcoid bases/Failed condition"

From Virtual scientific conference
Jump to: navigation, search
 
Line 9: Line 9:
 
<b>Proof</b>
 
<b>Proof</b>
 
Take $S_0 = \operatorname{up} 1^{\mathsf{FCD}}$ (where $1^{\mathsf{FCD}}$ signifies the diagonal of some infinite
 
Take $S_0 = \operatorname{up} 1^{\mathsf{FCD}}$ (where $1^{\mathsf{FCD}}$ signifies the diagonal of some infinite
set) and $S_1^{\mathsf{FCD}} = \bigcup_{F \in S_0} \operatorname{up}^{\Gamma} F$. Both $S_0$ and $S_1^{\mathsf{FCD}}$
+
set) and $S_1 = \bigcup_{F \in S_0} \operatorname{up}^{\Gamma} F$. Both $S_0$ and $S_1$
are upper sets. $S_0 \neq S_1^{\mathsf{FCD}}$ because $1^{\mathsf{FCD}} \in S_0$ and $1^{\mathsf{FCD}} \notin S_1^{\mathsf{FCD}}$. The $S
+
are upper sets. $S_0 \neq S_1$ because $1^{\mathsf{FCD}} \in S_0$ and $1^{\mathsf{FCD}} \notin S_1$. The $S
 
\cap \Gamma$ for both is the filter on $\Gamma$ defining the funcoid $1^{\mathsf{FCD}}$.
 
\cap \Gamma$ for both is the filter on $\Gamma$ defining the funcoid $1^{\mathsf{FCD}}$.
 
    
 
    
 
Thus in assumption of 4 we have not 3 (as otherwise both $S_0 = \operatorname{up} 1^{\mathsf{FCD}}$
 
Thus in assumption of 4 we have not 3 (as otherwise both $S_0 = \operatorname{up} 1^{\mathsf{FCD}}$
and $S_1^{\mathsf{FCD}} = \operatorname{up} 1^{\mathsf{FCD}}$).
+
and $S_1 = \operatorname{up} 1^{\mathsf{FCD}}$).

Latest revision as of 23:24, 18 April 2017

The following condition:

4. $S$ is an upper set and $S\cap\Gamma$ is a filter on the boolean lattice $\Gamma$ (defined in the chapter "Funcoids are filters" of the book).

was once thought by me to be equivalent to other three conditions. It is not so:

Proposition 4 does not imply 3.

Proof Take $S_0 = \operatorname{up} 1^{\mathsf{FCD}}$ (where $1^{\mathsf{FCD}}$ signifies the diagonal of some infinite set) and $S_1 = \bigcup_{F \in S_0} \operatorname{up}^{\Gamma} F$. Both $S_0$ and $S_1$ are upper sets. $S_0 \neq S_1$ because $1^{\mathsf{FCD}} \in S_0$ and $1^{\mathsf{FCD}} \notin S_1$. The $S \cap \Gamma$ for both is the filter on $\Gamma$ defining the funcoid $1^{\mathsf{FCD}}$.

Thus in assumption of 4 we have not 3 (as otherwise both $S_0 = \operatorname{up} 1^{\mathsf{FCD}}$ and $S_1 = \operatorname{up} 1^{\mathsf{FCD}}$).