Suppose $R$ is a partial order on $A$ and $S$ is a partial order on $B$. Define a relation $L$ on $A×B$ as follows: $L=\{((a,b),(a',b'))\in (A×B)×(A× B) \mid aRa', \text{and if } a=a' \text{ then } bSb'\}$. To show that $L$ is a partial order on $A×B$.
I am okay with proving that $L$ is reflexive and antisymmetric on $A×B$. How do you prove that $L$ is transitive on $A×B$? Also if $R$ and $S$ are total orders, will $L$ be also a total order? If it is, how do you prove it?
This is an old inactive question. But I will answer just to remove it from unanswered category.
In general, say, $A$ and $B$ are two sets equip with two binary operations $R$ and $S.$ If both $R, S$ are reflexive, irreflexive, symmetric, antisymmetric, transitive, or antitransitivity then so does the product binary operation (typically denoted by $R\times S$ ) on $A\times B$ defined as $(a,b)(R\times S)(a',b')$ iff $aRa'$ and $bSb'.$ In other words, Cartesian product preserve both equivalence relations and partial orderings (but not total relations, in general). However here you are given the not the product ordering, but the lexicographical order, which is a total ordering.
For your first question, suppose $(a,b)L(a',b')$ and $(a',b')L(a'',b''),$ then we have following four cases:
Now carefully apply the transitivity of both $R$ and $S$ to obtain the conclusion in the remaining cases.
For your second question, lets take any $(a,b), (a',b')\in A\times B.$ Assuming the totality of $R$ and $S$ we have either $aRa'$ or $a'Ra$ or $a=a'.$ The result is true in first two cases, and in the last (equality) case use the totality of $S.$ There we have either $bSb'$ or $b'Sb$ or $b=b',$ hence finally we have $(a,b)L(a',b')$ or $(a',b')L(a,b)$ or $(a,b)=(a',b').$