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Funcoid bases/Another reduce to ultrafilters

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Statement If $\forall X_0, \ldots, X_n \in S : \operatorname{up} (X_0 \sqcap^{\mathsf{FCD}} \ldots \sqcap^{\mathsf{FCD}} X_n) \subseteq S$ (for every natural $n$), then there exists a funcoid $f$ such that $S = \operatorname{up} f$.

Proof attempt Let's denote $f = \bigwedge^{\mathsf{FCD}} S$.

$\operatorname{up} f \supseteq S$ is obvious. It remains to prove $Y \in \operatorname{up} f \Rightarrow Y \in S$.

Let $Y \in \operatorname{up} f$. Then $Y \in \operatorname{up} \bigwedge^{\mathsf{FCD}} S$. Then for every ultrafilter $x$ we have $\langle Y \rangle x \sqsupseteq \left\langle \bigwedge^{\mathsf{FCD}} S \right\rangle x = \bigwedge_{X \in S} \langle X \rangle x$.

$\operatorname{up} \langle Y \rangle x \subseteq \operatorname{up} \left\langle \bigwedge^{\mathsf{FCD}} S \right\rangle x = \\ \operatorname{up} \bigwedge_{X \in S} \langle X \rangle x = \\ \left\{ K_0 \sqcap^{\mathfrak{Z}} \ldots \sqcap^{\mathfrak{Z}} K_n \mid i = 0, \ldots, n, X_i \in S, K_i \in \operatorname{up} \langle X_i \rangle x \right\} = \\ \bigcup \left\{ \operatorname{up} (\langle X_0 \rangle x \sqcap \ldots \sqcap \langle X_n \rangle x) \mid n = 0, 1, \ldots \right\} = \\ \bigcup \left\{ \operatorname{up} \left\langle X_0 \sqcap^{\mathsf{FCD}} \ldots \sqcap^{\mathsf{FCD}} X_n \right\rangle x \mid n = 0, 1, \ldots \right\} = \\ \bigcup \left\{ \bigcup_{F \in \operatorname{up} (X_0 \sqcap^{\mathsf{FCD}} \ldots \sqcap^{\mathsf{FCD}} X_n)} \operatorname{up} \langle F \rangle x \mid n = 0, 1, \ldots \right\} = \\ \bigcup \left\{ \operatorname{up} \langle F \rangle x \mid n = 0, 1, \ldots, F \in \operatorname{up} (X_0 \sqcap^{\mathsf{FCD}} \ldots \sqcap^{\mathsf{FCD}} X_n) \right\} \subseteq \bigcup \left\{ \operatorname{up} \langle F \rangle x \mid F \in S \right\}$.

Here the proof is stuck.

Above used the lemma from Funcoid bases/Reduce to meets of applying funcoids to ultrafilters.