is the product of totally ordered sets is total order

Question

I was working in products of structures and I am trying to find a counterexample to the following: "the product of totally ordered sets is a totally ordered set." Unfortunately I could not find one. Can someone provide me with a counterexample?

Answer 1

If $(A,<_A)$ and $(B,<_B)$ are partially ordered sets (strict orders), their product is $(A\times B,<_{A\times B})$ where $$ (a_1,b_1)<_{A\times B} (a_2,b_2) \qquad\text{if and only if}\qquad a_1<_A a_2 \quad\text{and}\quad b_1<_B b_2 $$

Now, finding counterexamples to “if $(A,<_A)$ and $(B,<_B)$ are totally ordered then $(A\times B,<_C)$ is totally ordered” is quite easy.

Suppose $a_1<_A a_2$ in $A$ and $b_2<_B b_1$. Consider the two pairs $(a_1,b_1)$ and $(a_2,b_2)$ in $A\times B$. Can you find an instance?