Proving $F . G$ is the greatest lower bound

Question

This is one of the problem I have been solving from Velleman's How to Prove book:

Suppose $A$ is a set. If $F$ and $G$ are partitions of $A$, then we'll say that $F$ refines $G$ if $\forall X \in F \exists Y \in G(X \subseteq Y)$. Let $P$ be the set of all partitions of $A$, and let $R = \{(F,G) \in P \times P \mid F \text{ refines } G \}$

(...) Skip the first two questions

c) Suppose $F$ and $G$ are partitions of $A$. Prove that $F . G$ is the greatest lower bound of the set $\{F,G\}$ in the partial order $R$.

Note that $F. G$ is defined from an earlier question which is like this:

Suppose $F$ and $G$ are partitions of a set $A$. We define a new family of sets $F . G$ as follows:

$F . G = \{Z \in \mathbb{P}(A) \mid Z \neq \emptyset \text{ and } \exists X \in F \exists Y \in G(Z = X \cap Y)\}$

Now in my attempt I have proved that $F . G$ is the lower bound of the set $\{F,G\}$. All that is remaining for me is to prove that it is the greatest lower bound. Here is my attempt for that:

Let $L$ be the set of all lower bound. Suppose $H \in L$ is the greatest lower bound. Now to prove that $F . G$ is the greatest lower bound, we have to prove that $H R F . G$. Let $Z$ be an arbitrary element in $F . G$ and some $Y \in H$. Since $H \in L$, it follows that for $\forall A \in \{F,G\} (HRA)$. So, $HRF$ and $HRG$. From $HRF$, it follows that $Y \subseteq B$ for some $B \in F$. Similarly, $Y \subseteq C$ for some $C \in G$. So, $B \cap C \subseteq Y$.

Now I'm stuck after this. How to proceed from here?

Answer 1

HINT: I’ll write $F\preceq G$ instead of $F\mathrel{R}G$.

You can’t assume that $H$ is the greatest lower bound, because at this point in the argument you don’t know that the set $\{F,G\}$ necessarily even has a greatest lower bound. What you need to do is show that if $H$ is any lower bound for $\{F,G\}$, then $H\preceq F\cdot G$. In order to do this, you must show that if $H\preceq F$ and $H\preceq G$, then for each $X\in H$ there is a $Y\in F\cdot G$ such that $X\subseteq Y$. Every $Y\in F\cdot G$ is of the form $Y=U\cap V$ for some $U\in F$ and $V\in G$, so what you really need to prove is that

for each $X\in H$ there are a $U\in F$ and a $V\in G$ such that $X\subseteq U\cap V$.

This follows almost immediately from the hypothesis that $H\preceq F$ and $H\preceq G$.