Order relation on the union of two ordered sets.

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The following exercise comes from the book Introduction to lattice and order, Second edition. David and Priestley.

Let $P$ and $Q$ two ordered sets, such that $P\cap Q\ne \emptyset$. Give formal definition of the ordered sets $P \cup Q$ and $P\oplus Q$ [Hint. Define appropriate orders on $( \{0\} \times P) \cup( \{1\} \times Q)$ ].

My ideas

Without reading the hint, I thought that the relation $\le_{P\cup Q}$ defined as

$a\le_{P\cup Q} b\iff (a\le_Pb, \mbox{if }a,b\in P) \vee(a\le_Qb, \mbox{if }a,b\in Q\setminus P)$

is an order relation on $P\cup Q$. After reading the hint, I convince myself that I made a mistake, but I can't see where. How can I use the hint to solve the problem?

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It's not always possible to define an order on $P \cup Q$ nicely. For consider $P = Q = \{0, 1\}$, but $P$ has the order where $0 < 1$ and $Q$ has the order where $1 < 0$. There is no nice way to "smush" these together in a way which is compatible with both orders.

However, consider the case where the orders of $P$ and $Q$ agree on the set $P \cap Q$. Then, it's possible to combine the orders by simply taking the transitive closure of the union of the relations. If we're working with the $\leq$ presentation of partial orders, then we can write this more simply as $\leq_{P \cup Q} = \leq_P \cup \leq_Q \cup \{(p, q) | p \in P, q \in Q, \exists c \in P \cap Q, p \leq_P c \land c \leq_Q q\} \cup \{(q, p) | p \in P, q \in Q, \exists c \in P \cap Q, c \leq_P p \land q \leq_Q c\}$.