Order preserving map $\mathbb R^2 \mapsto \mathbb R$ What would be an order preserving map $\mathbb R^2 \mapsto \mathbb R$ if $\mathbb R^2$ is arbitrarily ordered by x then y? 
I saw the interleaving answers here: Examples of bijective map from $\mathbb{R}^3\rightarrow \mathbb{R}$ but they don't mention order preserving. 
Simply doing a linear transofrmation $w=5x+7y$ for example isn't order preserving because before $(1,1) < (2,0)$ but after $12 > 10$. 
So what would be an order preserving map? Is interleaving order preserving? 
Edit: By arbitrarily ordering $\mathbb R^2$ I mean using a comparer like this one: 
(a,b)=>{  //a,b are Vector2
    let x = a.x - b.x;
    if (x != 0){
        return x;
    }
    return a.y - b.y;  // a tie with x is broken with y
}

So for example $(1,3) < (1,5) < (2,2)$
Another comparer is to set a known minimum point and then to compare distances from that point. My set of points is known to be "larger" than that minimum. 
For example: 
I set my minimum to $(0,0)$ and then order all the other points relative to their distance from that point. So we'll get: blue(2,3) < red(6,2) < green(9,5) etc.
 A: The ordering you seem to be defining on $\mathbb{R}^2$ is
$$
(a,x)<(b,y)\qquad\text{for}\qquad
(a<b) \text{ or } (a=b \text{ and }x<y)
$$
Suppose $f\colon\mathbb{R}^2\to\mathbb{R}$ is an injective order preserving map. Set
$$
T_a=\{(a,x):x\in\mathbb{R}\},
\qquad
S_a=f(T_a)
$$
Thus the image of $f$ is the disjoint union of the sets $S_a$, for $a\in\mathbb{R}$; moreover, $a<b$ implies that $S_a<S_b$, in the sense that $r\in S_a$ and $s\in S_b$ implies $r<s$.
Note that each $S_a$ is bounded; for instance, from $(a-1,0)<(a,x)<(a+1,0)$, we get
$$
f(a-1,0)<f(a,x)<f(a+1,0)
$$
and so $f(a-1,0)$ is a lower bound for $S_a$, while $f(a+1,0)$ is an upper bound for $S_a$.
Therefore $S_a$ has a greatest lower bound $l_a$ and a least upper bound $u_a$. Note that the sets $S_a$ have neither a minimum nor a maximum, so $S_a\subseteq(l_a,u_a)$.
Can you now find a contradiction?

For other orders on $\mathbb{R}^2$, an injective order preserving map may exist or not. For instance, take a bijection $f\colon\mathbb{R}^2\to\mathbb{R}$ and order $\mathbb{R}^2$ by
$$
\mathbf{x}<\mathbf{y} \text{ if and only if } f(\mathbf{x})<f(\mathbf{y})
$$
Then, by definition, $f$ is an order preserving map.
