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# Funcoid bases/Failed condition

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