Rudin's definition of an ordered set

Question

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?

Answer 1

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

• For all $a\in S$, we have $a=a$ and hence $a\le a$
• Assume $a\le b$ and $b\le a$. If $a<b$, then by the first property of $<$, neither $b<a$ nor $b=a$, contradicting $b\le a$. Hence $a=b$.
• Assume $a\le b$ and $b\le c$. If $a=b$ or $b=c$, then trivially $a\le c$. In the remaining case $a<b$ and $b<c$, also $a<c$ and so $a\le c$
• Let $a,b\in S$. Then one of $a<b$, $a=b$, $b<a$. In the first two cases $a\le b$, in the last case $b\le a$.

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Let $\le $ be an order on $S$ in the second sense. Define $x<y\iff x\le y\land x\ne y$.

• Let $x,y\in S$. Then $x\le y$ or $y\le x$, so $x<y\lor x=y\lor y<x$. By our definition of $<$, we cannot have $x<y\land x=y$, nor $y<x\land x=y$. Remains the possibility that $x<y\land y<x$. But then $x\le y\land y\le x$ implies $x=y$, contradiction. Hence excactly one of $x<y,x=y,y<x$ holds
• Assume $x<y$ and $y<z$. Then $x\le y$ and $y\le z$, so $x\le z$. If we had $x=z$, then $y<x$, but that contradicts $x<y$. Hence $x\ne z$ and so $x<z$.