# Difference between revisions of "Funcoid bases/Failed condition"

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<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 | + | 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 | + | 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 | + | 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 $S_1 = \operatorname{up} 1^{\mathsf{FCD}}$).