I want to refer to Exercise 26 part (c) of Section 4.6 of Velleman's 2nd Edition book. The exercise is as follows:
Supose $A$ is a set. If $\mathcal{F}$ and $\mathcal{G}$ are partitions of $A$, then we say that $\mathcal{F}$ refines $\mathcal{G}$ iff $\forall X\in\mathcal{F} \exists Y\in\mathcal{G} (X\subseteq Y)$. Let $P$ be the set of all partitions of $A$, and let $R=\{ (\mathcal{F},\mathcal{G}) \in P\times P \; | \; \mathcal{F} \; \text{refines} \; \mathcal{G} \}$.
Suppose $\mathcal{F}$ and $\mathcal{G}$ are partitions of $A$. Prove that $\mathcal{F}\cdot\mathcal{G} = \{ Z\in\mathscr{P}(A) | Z\neq\varnothing \; \text{and} \; \exists X\in\mathcal{F}\exists Y\in\mathcal{G} (Z=X\cap Y) \}$ is the greates lower bound of the set $\{ \mathcal{F},\mathcal{G} \}$ in the partial order $R$.
My approach was the following: 1) Let $L$ be the set of all lower bounds for $P_1=\{ \mathcal{F},\mathcal{G} \}$. Prove that $\mathcal{F}\cdot\mathcal{G}$ is a lower bound; 2) Prove that $\mathcal{F}\cdot\mathcal{G}$ is the only lower bound.
So eventually I proved that $\mathcal{F}\cdot\mathcal{G}$ is a lower bound. In proving my second step I tried following this strategy: $\forall \mathcal{H}(\mathcal{H}$ is l.b. for $P_1 \rightarrow \mathcal{H}=\mathcal{F}\cdot\mathcal{G}$.
- Let $\mathcal{H}$ arbitrary and suppose that $\mathcal{H}$ is l.b. for $P_1$. Thus, the goal is very straightforward.
- Basically I used universal instantiation for the definition of lower bound, and I got that $(\mathcal{H},\mathcal{F})\in R$ and $(\mathcal{H},\mathcal{G})\in R$. I did the same but using the $\mathcal{F}\cdot\mathcal{G}$.
This is as far as I got. It ocurred to me that I could try proving that $\mathcal{F}\cdot\mathcal{G}$ is symmetric so that I could use transitivity and then antisymmetry to prove that $\mathcal{H}=\mathcal{F}\cdot\mathcal{G}$, but didn't get much progress with that. I also tried proving that $\mathcal{F}\cap\mathcal{G}=\mathcal{F}\cdot\mathcal{G}$ to use that theorem that sayst that the intersection of a family of sets is the glb, but didn't go any better.
I would appreciate any hints as to how I can proceed with my strategy, or any different approach I can take for this proof.
Suppose that $\mathcal{H}$ is any lower bound of $\{\mathcal{F},\mathcal{G}\}$. I'll denote the refinement partial order by $\le$.
Let $H \in \mathcal{H}$. It is the case that $\mathcal{H} \le \mathcal{F}$ so there exists some $F_1 \in \mathcal{F}$ such that $H \subseteq F_1$.
Also $\mathcal{H} \le \mathcal{G}$ so there is a $G_1 \in \mathcal{G}$ such that $H \subseteq G_1$.
It follows that $H \subseteq F_1 \cap G_1$ and by definition, $F_1 \cap G_1 \in \mathcal{F}\cdot \mathcal{G}$. As we can do this for any $H \in \mathcal{H}$ it follows that $\mathcal{H} \le \mathcal{F} \cdot \mathcal{G}$ which shows that $\mathcal{F} \cdot \mathcal{G}$ is maximal among the lower bounds of $\{\mathcal{F}, \mathcal{G}\}$.
So it is the glb in the refinement partial order $\le$, as claimed.