Proving that R is a partial Order. 
Define the relation $\Bbb R \times \Bbb R$ by $(a,b) \; R$ $ (x,y)$ iff $a \le x$ and $b \le y$ , prove that R is a partial ordering for $\Bbb R\times\Bbb R $ .

A partial order is if R is reflexive on A, antisymmetric and transitive. One must prove these properties true. My question for this problem is trying to comprehend why this problem is antisymmetric and why it is transitive.
$(i)$  $R $ is reflexive as we say $x=a$ and $y = b$. Thus we can conclude that that $x\le x , y\le y$. $(x,y)R(x,y)$. 
$(ii)$ if $a\le x$ and $x\le a$ then $x= a $
If $ b \le y $ and $ y \le b$ then $y =b$. Since you interchange these would it not be symmetric?
$(iii)$ Suppose $(a,b) R(x,y)$ and $(x,y) R (c,d)$ 
This is as far as I got for transitive. Any advice on how prove this partial ordering true would be appreciated.
 A: *

*Of course $R$ is reflexive. For all $(a,b)\in\mathbb{R}\times\mathbb{R}$ it holds $(a,b)R(a,b)$ because $a\leq a$ and $b\leq b$;

*If $(a,b)R(x,y)$ and $(x,y)R(a,b)$, then both $a\leq x,b\leq y$ and $x\leq a,y\leq b$. It follows $a=x$ and $b=y$, so $(a,b)=(x,y)$

*Let $(a,b)R(x,y)$ and $(x,y)R(c,d)$. Hence, $a\leq x$ and $x\leq c$, so $a\leq c$. The same for $b\leq d$. Both imply $(a,b)R(c,d)$.

A: If $P$ and $Q$ are ordered sets, then the product $P\times Q$ is an ordered set, with ordering: $(p,q) \leq (u,v)$ iff $p \leq u$ and $q \leq v$. 
If $P$ and $Q$ are both partially ordered sets, then $P\times Q$ is also a partially ordered set (Poset).
A: If P and Q are two posets . Then by the definition of poset P and Q both are reflexive,anti-symmetric and transitive . For PxQ to be a poset : 
Let a, a', a"∈P and b,b',b"∈Q
1. Reflexivity:- (a,b)≤(a,b) for all (a,b)∈P×Q 
2.Symmetry:- (a,b), (a',b')∈P×Q then a=a',b=b' hence 
antisymmetric.
3.Transitivity:-(a,b),(a',b'),(a'',b'')∈P×Q as a≤a',a'≤b''thena≤a'' as well as b≤b',b'≤b'' then b≤b'' 
P×Q is transitive.
Hence all the Poset properties are satisfied therefore P×Q is also a Poset.
