# Order relation and the same order type

Definition 1: 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 $xCz$.

Remark: As the tilde, $\sim$, is the generic symbol for an equivalence relation, the "less than"symbol, $<$, is commonly used to denote an order relation. Stated in this notation, the properties of an order relation become

(1) If $x\neq y$, then either $x<y$ or $y<x$.

(2) If $x<y$, then $x\neq y$.

(3) If $x<y$ and $y<z$, then $x<z$.

Definition 2: Suppose that $A$ and $B$ are two sets with order relations $<_A$ and $<_B$ respectively. We say that $A$ and $B$ have the same order type if there is a bijective correspondence between them that preserves order; that is, if there exists a bijective function $f:A\to B$ such that $$a_1<_A a_2 \Rightarrow f(a_1)<_B f(a_2).$$

Example: The interval $(-1,1)$ of real numbers has the same order type as the set $\mathbb{R}$ of real numbers itself, because the function $f:(-1,1)\to \mathbb{R}$ given by $f(x)=\dfrac{x}{1-x^2}$ is an order-preserving bijective correspondence.

After reading this paragraph I have two questions (I was thinking on them but no approach).

1) Why the conditions (2) in the Definition 1 and Remark are the same?

2) It's obvious to me that $f(x)=\dfrac{x}{1-x^2}$ is bijective mapping of $(-1,1)$ to reals. But how to prove rigorously that it preserves order? Namely, if $a_1<_A a_2 \Rightarrow f(a_1)<_B f(a_2).$ The symbols $<_A, <_B$ are confusing me a bit.

I was trying in this way: If $a_1<_A a_2$ i.e. $(a_1,a_2)\in C$ where $C$ - relation on $A=(-1,1)$. Then we need to show that $(f(a_1),f(a_2))\in R$, where $R$ - relation on $\mathbb{R}$. But i am not sure in this and this seems to me weird.

I have never worked with this topic before so I will be very grateful for help!

• 1) is just contraposition. In 2) both $<_A$ and $<_B$ are $<$ - the usual order of real numbers. So you have to show that for $x,y\in(-1,1)$, $x<y$ implies $x/(1-x^2)<y/(1-y^2)$. – SMM Sep 2 '18 at 20:11

For 2: assume $x<y$ and subtract $x/(1-x^2)$ from $y/(1-y^2)$ you'll get $$\frac{(1+xy)(y-x)}{(1-x^2)(1-y^2)}$$ which is positive.
$$f'(x) = \frac{(1-x^2)\cdot 1 - x \cdot (-2x)}{1-x^2)^2} = \frac{1+x^2}{(1-x^2)^2} > 0$$
so $f$ is monotonically increasing by standard calculus/analysis results.