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 $R = \{(F,G) \in P \times P \mid F \text{ refines } G \}$
a) Prove that $R$ is a partial order on $P$
Now, I have proved that $R$ is both reflexive and transitive. I'm stuck at the anti-symmetric part. This is my attempt:
Let $F,G$ be arbitrary element in $P$ such that $F R G$ and $G R F$. From $F R G$, it follows that $\forall X \in F \exists Y \in G (X \subseteq Y)$. Using universal instantiation, we get $\exists Y \in G (X_1 \subseteq Y)$. Similarly, from $G R F$, it follows that $\forall A \in G \exists B \in F(A \subseteq B)$. Again using universal instantiation, we get $\exists B \in F(A_1 \subseteq B)$.
Here after I'm stuck at the problem. How to proceed from here ?
HINT: Start with an $X\in F$. Since $F\mathrel{R}G$, there is a $Y\in G$ such that $X\subseteq Y$. Then since $G\mathrel{R}F$, there is an $X'\in F$ such that $Y\subseteq X'$, and it follows that $X\subseteq X'$. But $X$ and $X'$ are both members of the partition $F$, so what does that tell you about them? What can you then infer about $X$ and $Y$?