if and only if
1) $a\le c$ and $a \ne c$
2) $a = c$ and $b\le d$
Prove that $(A\times A,\preceq)$ is also a partially ordered set.
So to prove this I would start with trying to find 1) Reflexivity, 2) anti-symmetricy and 3) Transitivity right?
How would I do that, does the $A\times A$ mean anything special in this case?
For Reflexivity would i say 1) since $\le$ is reflexive then that means $A\le A$ ... I just dont know how to begin this?
The order $\preceq$ on $A\times A$ is known as the lexicographic order induced by $\le$. $A\times A$ is the Cartesian product of $A$ with itself: $A\times A=\{\langle a,b\rangle:a,b\in A\}$. To show that $\preceq$ is a partial order on $A\times A$, you do indeed have to show that it’s reflexive, transitive, and antisymmetric.
Reflexive: You must show that for every $\langle a,b\rangle\in A\times A$, $\langle a,b\rangle\preceq\langle a,b\rangle$. This is completely straightforward: just check the definition of $\preceq$.
Transitive: You must show that if $\langle a_0,b_0\rangle,\langle a_1,b_1\rangle,\langle a_2,b_2\rangle\in A\times A$, $\langle a_0,b_0\rangle\preceq\langle a_1,b_1\rangle$, and $\langle a_1,b_1\rangle\preceq\langle a_2,b_2\rangle$, then $\langle a_0,b_0\rangle\preceq\langle a_2,b_2\rangle$. Since the definition of $\preceq$ has two clauses, you’ll have to consider several cases. I’ll do one of them to get you started. Suppose that $\langle a_0,b_0\rangle\preceq\langle a_1,b_1\rangle$ because $a_0\le a_1$ and $a_0\ne a_1$. We know that $a_1\le a_2$ (why?), so $a_0\le a_2$ and $a_0\ne a_2$, and therefore $\langle a_0,b_0\rangle\preceq\langle a_2,b_2\rangle$, as desired.
Antisymmetry: You must show that if $\langle a_0,b_0\rangle,\langle a_1,b_1\rangle\in A\times A$, $\langle a_0,b_0\rangle\preceq\langle a_1,b_1\rangle$, and $\langle a_1,b_1\rangle\preceq\langle a_0,b_0\rangle$, then $\langle a_0,b_0\rangle=\langle a_1,b_1\rangle$. Here again I’ll get you started. Suppose that $\langle a_0,b_0\rangle\preceq\langle a_1,b_1\rangle$ because $a_0\le a_1$ and $a_0\ne a_1$; then $a_1\not\le a_0$ (why?), and it is therefore impossible to have $\langle a_1,b_1\rangle\preceq\langle a_0,b_0\rangle$. Thus, we must have $a_0=a_1$ and $b_0\le b_1$. Now finish the argument by showing that $b_0=b_1$.