Let $(A,R_1)$, $(B,R_2)$ be two posets. Then, $R$ is defined by-$\{(a,b)R(x,y)$ if $aR_1x$ and $bR_2y \}$ is a partial order.
Using this statement, can we say that if $R_1$ and $R_2$ are total orders then, R is a total order?
My book has the following answer however, I think it's wrong.
This is why I think its wrong- it says we cannot compare $(1,2)$ and $(2,1)$. This means that $(1,2)$ and $(2,1)$ must be in the relationship $R$. If such is the case then, $1R_12$(Does exist in $R$) and $2R_21$(Does not exist $R$). (basically, what I am trying to say is that we could have made this argument if $(1,2)$ and $(2,1)$ belonged to the relation $R$ but since they don't($2 !< 1$) we can't make this argument, (it is failing $2R_21)$)
The original question is as follows in my book(question 4)
Starting with relations $R_1$ and $R_2$ on $A$ and $B$ respectively, we define the relation $R$ on $A\times B$ by: $$\forall x_1,x_2\in A:\text{ }\forall y_1,y_2\in B:\bigg[(x_1,y_1)\color{red}{R}(x_2,y_2)\leftrightarrow \big((x_1\color{red}{R_1}x_2)\wedge (y_1\color{red}{R_2}y_2)\big)\bigg].$$
It is straight forward to check that the Reflexive, Antisymmetric, and Transitive properties of a Partial Order follow from both $R_i$'s having that property. This is the content of question (3).
For question (4), consider the following diagram for arbitrary $x_1,x_2\in A\text{ and }y_1,y_2\in B$
$\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }$
The Reflexive Property gives us the dual black arrows and the $\color{darkred}{\text{red}}$ and $\color{blue}{\text{blue}}$ arrows are one sided possibilities for the component relations (Anti-Symmetry causes the one-sidedness, unless the elements are equal). So there are cases to consider here.
$\underline{If\text{ } x_1 = x_2\text{ and }y_1=y_2:}$
Then the red and blue arrows become double sided and Transitivity allows us to connect $(x_1,y_1)$ and $(x_2,y_2)$ in both directions. Hence in this case, they are comparable.
$\underline{\text{Else If } x_1 \neq x_2\text{ and }y_1=y_2:}$
The equality collapses the blue arrows and Transitivity gives us one-directional comparability of $(x_1,y_1)$ and $(x_2,y_2)$ (depending on the option for red).
$\underline{\text{Else If } x_1 = x_2\text{ and }y_1\neq y_2:}$
Similarly, we get one directional comparability of $(x_1,y_1)$ and $(x_2,y_2)$ (dependent on blue's option).
$\underline{\text{Else If }\text{ } x_1 \neq x_2\text{ and }y_1\neq y_2:}$
Both red and blue arrows have options for 1-sidedness. In the event that the arrows line up, Transitivity gives us 1-d comparability. In the other event, when arrows conflict, we are not guaranteed an arrow in $R$ by Transitivity, though this diagonal arrow from $(x_1,y_1)$ to $(x_2,y_2)$ may or may not exist in the relation independently without contradiction. $\color{orange}{\text{In this case, comparability is not guaranteed. Begins hunt for counter-example...}}$
...Finds the books counter example is also labeled 4. With $R_i := "\leq"$, and $A,B:=\{1,2\}$, notice $(x_1,y_1) := (1,2)$ and $(x_2,y_2) := (2,1)$ fall in our latter case. We investigate the existence of the diagonal arrows, given our particular knowledge of how $"\leq"$ works. We come up with the diagram:
$\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }$
and deduce that since the $\color{gold}{\text{gold}}$ component arrows go in opposite directions, the desired arrow in $R$ does not exist.
Finally, this gives us a c.x. where $R_i$'s may be total, but $R$ is not. $[\times]$