# Baby Rudin Definition of Order

At the start of Baby Rudin (Walter Rudin's Principles of Mathematical Analysis), Rudin defines an order $$<$$ on a set $$S$$ as being any relation which obeys the following two axioms:

(1) One and only one of $$x is true. (2) If $$x and $$y then $$x.

My question is, to what notion of order (in order theory) does this correspond? Is it a total order (which is defined as a relation which is reflexive, transitive, antisymmetric, and strongly connected)?

• It is not a total order. Clearly we do not have $x<x$. Mar 12 at 21:29
• @beeclu Of course, I should have seen that! I've come across the notion of "strict total order". Is this correct? The reason I ask is that Rudin later proves that $\mathbb{R}$ is the unique complete ordered field (up to isomorphism) and I want to ensure I understand what the "ordered" caveat there means.
– EE18
Mar 12 at 21:30
• I am not familiar with order theory, but from the definition of "strict total order" on wiki these seem to be equivalent notions. Mar 12 at 21:32

It is an alternative description of a total order.

Usually an order on a set is defined as relation $$\le$$ which has suitable properties. One can then define $$x < y$$ iff $$x \le y$$ nd $$x \ne y$$.

If $$\le$$ is a total order, then $$<$$ clearly satisfies Rudin's two axioms. Conversely, given Rudin's relation $$<$$, we can define $$x \le y$$ iff $$x < y$$ or $$x = y$$. It is an easy exercise to verify that $$\le$$ is a total order.

• Thank you for your answer! I agree that $\leq$ is a total order but Rudin is working with $<$ as the primary order of interest. Would it be possible to comment on $<$ in particular?
– EE18
Mar 13 at 0:39
• It is a total order. Note that Rudin uses $=$ as well. By using $=$ in the first axiom, he's informally defining $\leq$ without ever using the symbol. Mar 13 at 0:54
• @EE18 Working with relations $<$ or $\le$ is a matter of taste. As I wrote, they can be transformed into eacb other without loss of information. Also see en.wikipedia.org/wiki/Partially_ordered_set and math.stackexchange.com/questions/364289/strict-partial-order. Mar 13 at 8:54
• I see, thank you!
– EE18
Mar 13 at 17:05