For a point $(x,y)$ the points that are comparable with it are the points to the east, north-east, west and south-west of it. Therefore, a subset is ordered iff for every two points $(x,y)$ and $(x',y')$ in it such that $x\le x'$ , either $(x,y)=(x',y')$ , or $x<x'$ and $y\le y'$ . This implies two things, first, that the ordered subset has to define a function on $\mathbb{R}$ or its subset, and, second, that the function has to be weakly increasing. In fact, these two conditions are also sufficient to define an ordered subset.

Now, suppose that for a weakly increasing function $f$ , there is a “hole” in its domain $A$ , i.e. there are $x'<x<x"$ such that $x',x"\in A$ , but $x\notin A$ . Then, the set $f(A\cap(-\infty,x))$ is bounded from above by $f(x")$ , and it has the least upper bound $m$ . Similarly, the set $f(A\cap(x,+\infty))$ is bounded from below by $m$ , and it has the greatest lower bound $m'$ . Note that, $m\le m'$ , so we can add the point $(x,y)$ where $m\le y\le m'$ to $f$ , and $f$ will remain weakly increasing. This means that if $f$ has “holes” in its domain, then it does not define a maximal ordered subset. There should not be any holes.

What we have so far. A maximal ordered subset has to be a weakly increasing function defined on an interval (closed $[]$ , open $()$ , or half-open $[)$ $(]$ , finite or infinite). Now suppose that the domain of $f$ is bounded from below, and its image set is bounded from below as well. Then, there is a point that is to the left of the domain of $f$ and below its image set, and we can add the point to $f$ so that it will remain weakly increasing. Hence, at least one of the domain of $f$ and its image set should not be bounded from below. Similarly, we conclude that at least one the domain of $f$ and its image set should not be bounded from above.

This also implies that the domain of the function is an open interval, for if, for example, $f$ is defined on $[a,b)$ where $b\in\mathbb{R}$ or $b=+\infty$ , or on $[a,b]$ where $b\in\mathbb{R}$ , then the domain is bounded from below, and $f$ is bounded from below by $f(a)$ . Similarly for the domain $(a,b]$ where $a\in\mathbb{R}$ or $a=-\infty$ .

Now, suppose that $f:(a,b)\rightarrow\mathbb{R}$ is a weakly increasing function such that either $a=-\infty$ or $\lim_{x\rightarrow a+0}f(x)=-\infty$ ($f$ is not bounded from below), and either $b=+\infty$ or $\lim_{x\rightarrow b-0}f(x)=+\infty$ ($f$ is not bounded from above). Let us take a point $(x,y)$ such that either $x\notin(a,b)$ , or $x\in(a,b)$ and $f(x)\neq y$ . If $x\notin(a,b)$ , then either $x\le a$ or $x\ge b$ . In the first case when $x\le a$ , $f$ is not bounded from below, and there is some $x'\in(a,b)$ such that $x'>x$ and $f(x')<y$ . Hence, $(x,y)$ and $(x',f(x'))$ are not comparable. Similarly, the second case when $x\ge b$ . If $x\in(a,b)$ , then $(x,y)$ and $(x,f(x))$ are not comparable. We conclude that $f$ defines a maximal subset.

Summary: a subset of $\mathbb{R}^{2}$ is a maximal ordered subset iff it defines a weakly increasing function $f:(a,b)\rightarrow\mathbb{R}$ such that $a=-\infty$ or $f$ is not bounded from below, and $b=+\infty$ or $f$ is not bounded from above. Note that $f$ does not have to be continuous.

So, both $y=x^{3}$ and $y=2$ are maximal (both are weakly increasing and defined on $(-\infty,+\infty)$ ). But $y=x^{2}$ is not a maximal ordered set (it is not a weakly increasing function, so not even an ordered set), and, for example, $(0,0)$ and $(-1,1)$ are not comparable.

Other examples of maximal ordered sets defined by weakly increasing functions:

	the domain of $f$ is unbounded from above	the domain of $f$ is bounded from above
the domain of $f$ is unbounded from below	$y=\mbox{sign}x=\begin{cases} 1 & x>0\\ 0 & x=0\\ -1 & x<0 \end{cases}$	$y=-\frac{1}{x}$ , $x<0$
the domain of $f$ is bounded from below	$y=-\frac{1}{x}$ , $x>0$	$y=\tan x$ , $-\frac{\pi}{2}<x<\frac{\pi}{2}$

Note that I specifically found examples such that the function if bounded from below/above iff its domain is not bounded from below/above, and also one of these functions is not continuous, though one could have found a non-continuous function in all cases.