According to Munkres' book Topology, an order relation on $A$ is defined as follows: A relation $C$ on a set $A$ is called an order relation if it has the following properties:

  1. (Comparability) for every $x$ and $y$ in $A$ for which $x\neq y,$ either $xCy$ or $yCx.$

  2. (Nonreflexivity) for no $x$ in $A$ does the relation $xCx$ hold.

  3. (Transitivity) If $xCy$ and $yCz,$ then $xCz.$

With the definition above, I have tried to figure out the following statement:

Let function $f: \mathbb{R}\to\mathbb{R}.$ For $P_0=(x_0,y_0), P_1=(x_1,y_1)\in\mathbb{R}^2,$ define \begin{gather*} P_0<P_1 :=\big(y_0-f(x_0)<y_1-f(x_1)\big) \lor \big(y_0-f(x_0)=y_1-f(x_1)\land x_0<x_1\big). \end{gather*} Then $<$ is an order relation on $\mathbb{R}^2.$

And I have given a proof of the statement.

Proof $\quad$ Let $f: \mathbb{R}\to\mathbb{R}.$ (1) Comparability. Let $P_i=(x_i,y_i), i=1,2.$ Assume that $P_1\neq P_2.$ If $y_1-f(x_1)<y_2-f(x_2),$ then by definition we have $P_1<P_2.$ If $y_2-f(x_2)<y_1-f(x_1), $ then $P_2<P_1.$ If $y_1-f(x_1)=y_2-f(x_2),$ then, since $P_1\neq P_2,$ we arrive at $x_1\neq x_2,$ for otherwise we would have, from $y_1-f(x_1)=y_2-f(x_2),$ $y_1=y_2,$ and so $P_1=P_2,$ a contradiction. Thus, if $x_1<x_2,$ then $P_1<P_2,$ and while $x_2<x_1,$ we have $P_2<P_1.$

(2) Nonreflexivity. Let $P_i=(x_i,y_i), i=1,2.$ If $P_1<P_2,$ then we have $P_1\neq P_2,$ for if $P_1=P_2,$ then $x_1=x_2,y_1=y_2,$ which would imply that $y_1-f(x_1)=y_2-f(x_2),$ and $x_1=x_2,$ a contradiction to the hypothesis of $P_1<P_2.$

(3) Transitivity. Let $P_i=(x_i, y_i), i=1,2,3.$ Then \begin{align*} &(P_1<P_2)\land (P_2<P_3)\\ \iff &\left[\big(y_1-f(x_1)<y_2-f(x_2)\big)\lor\big(y_1-f(x_1)=y_2-f(x_2)\land x_1<x_2\big)\right]\\ &\land \left[\big(y_2-f(x_2)<y_3-f(x_3)\big)\lor\big(y_2-f(x_2)=y_3-f(x_3)\land x_2<x_3\big)\right]\\ \iff &\left[\big(y_1-f(x_1)<y_2-f(x_2)\big)\land \big(y_2-f(x_2)<y_3-f(x_3)\big)\right]\\ &\lor \left[\big(y_1-f(x_1)<y_2-f(x_2)\big)\land \big(y_2-f(x_2)=y_3-f(x_3)\land x_2<x_3\big)\right]\\ &\lor \left[\big(y_1-f(x_1)=y_2-f(x_2)\land x_1<x_2\big)\land y_2-f(x_2)<y_3-f(x_3)\right]\\ &\lor \left[\big(y_1-f(x_1)=y_2-f(x_2)\land x_1<x_2\big)\land \big(y_2-f(x_2)=y_3-f(x_3)\land x_2<x_3\big)\right]\\ \Rightarrow& \big(y_1-f(x_1)<y_3-f(x_3)\big)\lor \big(y_1-f(x_1)=y_3-f(x_3)\land x_1<x_3\big)\iff P_1<P_3. \end{align*} Hence $<$ is transitive. Therefore $<$ is indeed an order relation on $\mathbb{R}^2.$

My question is: (1) Is my proof correct? (2) If (1) is true, what kind of order relation is this, or What are properties of this order relation? (3) Is there any reference about order relation like this?

  • $\begingroup$ Proof looks OK (I didn't go through all the transitive cases). It seems to be a variation on a "lexicographic order". $\endgroup$ – coffeemath Aug 29 '16 at 2:04
  • $\begingroup$ Note that what Munkres is here calling an order relation is usually called a strict linear order or a strict total order. And yes, the proof is correct. $\endgroup$ – Brian M. Scott Aug 29 '16 at 19:43

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