If R is partial order on B, then prove R is partial order on $B \times B$

Question

Now let $(x,y) \in B \times B$. So $x \in B$ and $y \in B$. Since $x,y \in B$ and R is a partial order on $B$ so all conditions on $x$ and $y$ are satisfied. I wonder how to do this ?

Answer 1

If $R$ is a partial order on $B$, then $S \subseteq B^2 \times B^2$ defined by $$(a,b) S (c,d) \;\Leftrightarrow\; aRc \,\&\, bRd$$ is a partial order on $B^2$.
This is straightforward to prove.

Of course, as some other user put it in an answer, there is not only one.
Still this one is often seen as the most natural, since given posets $(P_1,\leq_1)$ and $(P_2,\leq_2)$, the direct product is defined as $(P,\leq)$, having $P=P_1\times P_2$ (cartesian product) as its base set, and $\leq$ defined by $$(p_1,p_2) \leq (q_1,q_2) \;\Leftrightarrow\; p_1 \leq_1 q_1 \,\&\, p_2 \leq_2 q_2.$$ Again, it is straightforward to prove that $(P,\leq)$ thus defined is indeed a poset, and I would recommend you to do this.
Then just consider $(P_1,\leq_1)=(P_2,\leq_2)=(B,R)$, and you have the desired result.

Answer 2

$R$ cannot be a partial order on $B \times B$ as a partial order consists of pairs: $R \subseteq B \times B$ but $R \nsubseteq \left(B \times B\right) \times \left(B \times B\right)$. I think you actually mean something different: "Does $R$ induce such a partial order (in a canonical way)"? Considering $\mathbb{R}$ and $\mathbb{C}$ might help.