In Principles of Mathematical Analysis, Rudin defines an $\textit{order}$ on a set $S$ to be a relation denoted by $<$, with the following two properties:
- If $x, y \in S$ then one and only one of $x < y$, $x = y$, $x > y$ is true.
- If $x,y,z \in S$ and $x<y$ and $y<z$ then $x<z$.
He also notes that the notation $x \leq y$ indicates $x<y$ or $x=y$ without specifying which of these two is to hold.
Most definitions of an order I have come across have been the following: A relation $\leq$ is an order on a set $S$ if the following properties hold:
- $a \leq a$ for all $a \in S$
- $a \leq b$ and $b \leq a$ implies $a = b$
- $a \leq b$ and $b \leq c$ implies $a \leq c$
- For any $a, b \in S$, we have $a \leq b$ or $b \leq a$
My question is are these two orders equivalent in some way?
Best Answer
The proof of equivalence is straightforward by simply deriving the desired properties one by one, from the ones we already know:
Let $<$ be an order on $S$ in the first sense and define $x\le y\iff x<y\lor x=y$ Then
Let $\le $ be an order on $S$ in the second sense. Define $x<y\iff x\le y\land x\ne y$.