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 $x<y, x=y, y<x$ is true. (2) If $x<y$ and $y<z$ then $x<z$.

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)?

  • $\begingroup$ It is not a total order. Clearly we do not have $x<x$. $\endgroup$
    – beeclu
    Mar 12 at 21:29
  • $\begingroup$ @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. $\endgroup$
    – EE18
    Mar 12 at 21:30
  • $\begingroup$ I am not familiar with order theory, but from the definition of "strict total order" on wiki these seem to be equivalent notions. $\endgroup$
    – beeclu
    Mar 12 at 21:32

1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – EE18
    Mar 13 at 0:39
  • $\begingroup$ 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. $\endgroup$
    – Git Gud
    Mar 13 at 0:54
  • $\begingroup$ @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. $\endgroup$
    – Paul Frost
    Mar 13 at 8:54
  • $\begingroup$ I see, thank you! $\endgroup$
    – EE18
    Mar 13 at 17:05

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