# Meet of two elements of a poset, which is a superset of another poset

Let $\mathfrak{A}$ and $\mathfrak{Z}\subseteq\mathfrak{A}$ be two complete lattices (with $\bigcup$ and $\bigcap$ supremum and infimum), order on which agrees.

I will denote $\operatorname{up} a = \{ x\in\mathfrak{Z} \mid x\geq a \}$ for every $a\in\mathfrak{A}$.

Let the filtrator $(\mathfrak{A},\mathfrak{Z})$ be filtered, that is the following two equivalent (see my free book) conditions hold:

1. $\forall a,b\in\mathfrak{A}: (\operatorname{up}a \supseteq \operatorname{up}b \Rightarrow a\leq b)$;
2. $a = \bigcap^{\mathfrak{A}}\operatorname{up}a$ for every $a\in\mathfrak{A}$.

Let also $\mathfrak{Z}$ be a closed (as a subposet of $\mathfrak{A}$) regarding any supremums.

Conjecture $\operatorname{up} (f \cap^{\mathfrak{A}} g) \subseteq \left\{ F \cap^{\mathfrak{Z}} G \mid F \in \operatorname{up} f, G \in \operatorname{up} g \right\}$ for every $f,g\in\mathfrak{A}$.

This conjecture is known to be true (see my free book) for the special case when $\mathfrak{A}$ is the poset of all filters on a set (ordered reverse set-theoretic inclusion) and $\mathfrak{Z}$ is its subset of principal filters. (Moreover for filters $\subseteq$ becomes equality.)

It (possibly) should also hold:

$$\operatorname{up} (f_0 \cap^{\mathfrak{A}}\dots\cap^{\mathfrak{A}} f_n) \subseteq \left\{ F_0 \cap^{\mathfrak{Z}}\dots\cap^{\mathfrak{Z}} F_n \mid F_0 \in \operatorname{up} f_0, \dots, F_n \in \operatorname{up} f_n \right\}.$$

The above conjecture was formulated in attempt to solve these questions.

It seems that after prayer in tongues and going down anointment of Holy Spirit I proved the first of above conjectures. Please check my proof for errors.

I additionally assume that $\mathfrak{A}$ is a distributive lattice.

$\operatorname{up} (f \sqcap^{\mathfrak{A}} g) \subseteq \bigcup \left\{ \operatorname{up} (F \sqcap^{\mathfrak{A}} G) \mid F \in \operatorname{up} f, G \in \operatorname{up} g \right\}$

because $X \in \operatorname{up} (f \sqcap^{\mathfrak{A}} g) \Rightarrow X \in \left\{ X \sqcup Y \mid Y \in \operatorname{up} (f \sqcap^{\mathfrak{A}} g) \right\} = X \in \left\{ Z \mid Z \in \operatorname{up} (X \sqcup (f \sqcap^{\mathfrak{A}} g)) \right\} \Leftrightarrow X \in \operatorname{up} (X \sqcup (f \sqcap^{\mathfrak{A}} g)) \Rightarrow X \in \operatorname{up} (X \sqcap^{\mathfrak{A}} g) \Rightarrow X \in \bigcup \left\{ \operatorname{up} (F \sqcap^{\mathfrak{A}} g) \mid F \in \operatorname{up} f \right\}$ that is $\operatorname{up} (f \sqcap^{\mathfrak{A}} g) \subseteq \bigcup \left\{ \operatorname{up} (F \sqcap^{\mathfrak{A}} g) \mid F \in \operatorname{up} f \right\}$. Apply this formula twice.

But $\operatorname{up} (F \sqcap^{\mathfrak{A}} G) \subseteq \operatorname{up} (F \sqcap^{\mathfrak{Z}} G)$. Thus follows the thesis.

• We may also need distributivity of $\mathfrak{Z}$ to finish the proof. I leave to finish it as an easy exercise to the reader. – porton Jan 7 '17 at 23:03
• The full proof is currently available in this PDF file: mathematics21.org/binaries/addons.pdf – porton Jan 8 '17 at 14:17
• The second formula (which I wrote "It (possibly) should also hold") is quite easy to prove by induction – porton Jan 8 '17 at 14:23