Funcoid bases/Disproof

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Proposition 2 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 $S_1 = \bigcup_{F \in S_0} \{ \operatorname{up} G \mid G\in\operatorname{up}^{\Gamma} F \}$ (that is $S_1 = \bigcup_{F\in\operatorname{up}^{\Gamma} 1^{\mathsf{FCD}}} \operatorname{up} F$). 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 $1^{\mathsf{FCD}}$.

2 holds for $S_0$ because 3 implies 2.

But 2 holds also for $S_1$ because:

Suppose $X_0, \ldots, X_n \in S_1$. Then $X_i \sqsupseteq F_i$ where $F_i \in S_0$. Consequently (take into account that $\Gamma$ is a sublattice of $\mathsf{FCD}$), $X_0, \ldots, X_n \sqsupseteq F_0 \sqcap^{\mathsf{FCD}} \ldots \sqcap^{\mathsf{FCD}} F_n$ and so $X_0 \sqcap^{\mathsf{FCD}} \ldots \sqcap^{\mathsf{FCD}} X_n = X_0 \sqcap \ldots \sqcap X_n \sqsupseteq F_0 \sqcap^{\mathsf{FCD}} \ldots \sqcap^{\mathsf{FCD}} F_n \sqsupseteq 1^{\mathsf{FCD}}$. Thus $X_0 \sqcap^{\mathsf{FCD}} \ldots \sqcap^{\mathsf{FCD}} X_n \in \operatorname{up}^{\Gamma} 1^{\mathsf{FCD}} \subseteq S_1$. $\operatorname{up}(X_0 \sqcap^{\mathsf{FCD}} \ldots \sqcap^{\mathsf{FCD}} X_n) \subseteq S_1$ as $S_1$ is an upper set.

3 cannot hold for both $S_0$ and $S_1$ because $\operatorname{up} f_0 = S_0$ and $\operatorname{up} f_1 = S_1$ imply $f_0 = \bigwedge^{\mathsf{FCD}} S_0 = 1^{\mathsf{FCD}} = \bigwedge^{\mathsf{FCD}}\operatorname{up}^{\Gamma} 1^{\mathsf{FCD}} = \bigwedge^{\mathsf{FCD}} S_1 = f_1$ and thus $S_0 = S_1$.