Rudin’s definition of an ordered set

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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

  • 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 $b\le a$.

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 $x<z$.
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