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Let $\mathfrak{F}$ is the complete lattice of filters (including the improper filter) on some set, ordered inverse to set-theoretic inclusion.
I will denote $\left\langle f \right\rangle \mathcal{X} = \bigcap^{\mathfrak{F}} \left\{ f \left[ X \right] | X \in \mathcal{X} \right\}$ for every binary relation $f$.f$and filter$\mathcal{X}$. Let$\forall \mathcal{X}\in\mathfrak{F}:\left( \mathcal{X} \cap^{\mathfrak{F}} \mathcal{A} \neq 0^{\mathfrak{F}} \Rightarrow \left( \left\langle f \right\rangle \mathcal{X} \supseteq^{\mathfrak{F}} \mathcal{B} \wedge \left\langle g \right\rangle \mathcal{X} \supseteq^{\mathfrak{F}} \mathcal{B} \right) \right)$for some binary relations$f$and$g$and filters$\mathcal{A}$and$\mathcal{B}$. ($0^{\mathfrak{F}}$is the filter which is the least in our order that is the biggest in set-theoretic order.) Does the implication$\forall \mathcal{X}\in\mathfrak{F}:\left( \mathcal{X} \cap^{\mathfrak{F}} \mathcal{A} \neq 0^{\mathfrak{F}} \Rightarrow \left\langle f \cap g \right\rangle \mathcal{X} \supseteq^{\mathfrak{F}} \mathcal{B} \right)$follow from the above assumption? You can read http://www.mathematics21.org/algebraic-general-topology.html for my related research. 3 Say about the improper filter Let$\mathfrak{F}$is the complete lattice of filters (including the improper filter) on some set, ordered inverse to set-theoretic inclusion. I will denote $\left\langle f \right\rangle \mathcal{X} = \bigcap^{\mathfrak{F}} \left\{ f \left[ X \right] | X \in \mathcal{X} \right\}$ for every binary relation$f$. Let$\forall \mathcal{X}\in\mathfrak{F}:\left( \mathcal{X} \cap^{\mathfrak{F}} \mathcal{A} \neq 0^{\mathfrak{F}} \Rightarrow \left( \left\langle f \right\rangle \mathcal{X} \supseteq^{\mathfrak{F}} \mathcal{B} \wedge \left\langle g \right\rangle \mathcal{X} \supseteq^{\mathfrak{F}} \mathcal{B} \right) \right)$for some binary relations$f$and$g$and filters$\mathcal{A}$and$\mathcal{B}$. ($0^{\mathfrak{F}}$is the filter which is the least in our order that is the biggest in set-theoretic order.) Does the implication$\forall \mathcal{X}\in\mathfrak{F}:\left( \mathcal{X} \cap^{\mathfrak{F}} \mathcal{A} \neq 0^{\mathfrak{F}} \Rightarrow \left\langle f \cap g \right\rangle \mathcal{X} \supseteq^{\mathfrak{F}} \mathcal{B} \right)$follow from the above assumption? You can read http://www.mathematics21.org/algebraic-general-topology.html for my related research. 2 Added missing quantifiers Let$\mathfrak{F}$is the complete lattice of filters on some set, ordered inverse to set-theoretic inclusion. I will denote $\left\langle f \right\rangle \mathcal{X} = \bigcap^{\mathfrak{F}} \left\{ f \left[ X \right] | X \in \mathcal{X} \right\}$ for every binary relation$f$. Let$\mathcal{X} \forall \mathcal{X}\in\mathfrak{F}:\left( \mathcal{X} \cap^{\mathfrak{F}} \mathcal{A} \neq 0^{\mathfrak{F}} \Rightarrow \left( \left\langle f \right\rangle \mathcal{X} \supseteq^{\mathfrak{F}} \mathcal{B} \wedge \left\langle g \right\rangle \mathcal{X} \supseteq^{\mathfrak{F}} \mathcal{B} \right) \right)$for some binary relations$f$and$g$and filters$\mathcal{A}$and$\mathcal{B}$. ($0^{\mathfrak{F}}$is the filter which is the least in our order that is the biggest in set-theoretic order.) Does the implication$\mathcal{X} \forall \mathcal{X}\in\mathfrak{F}:\left( \mathcal{X} \cap^{\mathfrak{F}} \mathcal{A} \neq 0^{\mathfrak{F}} \Rightarrow \left\langle f \cap g \right\rangle \mathcal{X} \supseteq^{\mathfrak{F}} \mathcal{B}$mathcal{B} \right)$ follow from the above assumption?