$\le_A$ is linear order on $A$ and $\le_B$ is linear order on $B$, relation $\le$ on $A\times B$ denote by: $\lt a_1,b_1\gt \le \lt a_1, b_2 \gt$ $\iff$ $a_1 \le_A a_2 \land b_1 \le_B b_2$
Question is to proof:
a) $\le$ is order relation on $A \times B$,
b) $\le$ is linear order if and only if $A$ has no more than 1 element, or $B$ has no more than 1 element.
I have already solved a) and it was easy and I solved b) $A$ or $B$ has no more than 1 element $\Rightarrow$ $\le$ is linear order,
but i have difficulties in second implication proof. I started by: So if $\le$ is linear then we take $\lt a_1,b_1 \gt,\lt a_2,b_2 \gt \in A\times B$ Then we know that $\lt a_1,b_1 \gt \le\lt a_2,b_2 \gt or \lt a_2,b_2 \gt \le\lt a_1,b_1 \gt$
I think that if we want to prove that $A$ or $B$ has no more than 1 element, whe should prove that (for example in case of $A$), $a_1 = a_2$ and we can get it if we show that $a_2 \le_A a_1$ because $\le_A$ is linear order on $A$ (if aRb and bRa => a=b)
So in case of $\lt a_1,b_1 \gt \le\lt a_2,b_2 \gt$ we know that $a_1 \le_B a_2 \land b_1 \le_B b_2$
And we know that $\le_A$ is linear order so we know that for $a_1,a_2 \in A$ $ a_1 \le_A a_2$ or $a_2 \le_A a_1$
But i have no idea how to show that and i cant figure out any other way to prove it
If my english is not clear Im sorry, and if you don't understand the problem because of my wrong translation english definitions and names, let me know, I will correct my question.
If $a_1,a_2 \in A$ and $b_1,b_2 \in B$, with $a_1<a_2$ and $b_1<b_2$ then what happens with $\langle a_1,b_2\rangle$ and $\langle a_2,b_1\rangle$?
Which is the smaller?