Let $T$ be a nonempty ordered set considered as an index set. Let $(P_{t})_{t\in T}$ be a family of pairwise disjoint nonempty ordered sets that are all disjoint from $T$. We define lexicographic sum $L\{P_{t}\ |\ t\in T\}$ to be the union $\bigcup _{t\in T}P_t$ ordered by $p_1\leq p_2$ iff
- There are distinct $t_1, t_2\in T$ with $t_1<t_2$, so that $p_i\in P_{t_i}$, or
- There is a $t\in T$ so that $p_1,p_2\in P_t$ and $p_1\leq_{P_t}p_2$.
(Ordered Sets (Bernd Schröder) p. 173)
What does the condition on $T$ achieve here? Why must $P_t\cap T=\emptyset$ for all $t\in T$? Also, given an arbitrary nonempty family of nonempty pairwise disjoint ordered sets $(X_i)_{i\in I}$, how does one ensure that the disjointness condition actually holds for the index set $I$?
I concur with the sentiment in the comment that I see no reason for $T$ to be disjoint from $P_t$. As a definitional point, a family $(P_t)_{t \in T}$ is a shorthand itself for some function $f$ defined on $T$ that maps $t$ to some set $f(t)$ in the universe of sets, denoted by $P_t$. We're using the axiom schema of replacement (in ZF) to justify this "sloppiness".
As to ensuring disjointness, for a set $A$ and sets $t \neq t'$ we have that $A \times \{t\}$ is disjoint from $A \times \{t'\}$ (by the definition of pairs) but both are essentially copies of $A$ (any order structure can trivially be copied over bijectively). So we never lose generality by assuming sets to be pairwise disjoint in such contexts.