Why is Transitivity necessary in the definition in the definition of an order on a set?

Question

In Rudin, Principles of Mathematical Analysis (ed. 3), he provides the following definition (pp. 3)

Definition: Let $S$ be a set. An order on $S$ is a relation, denoted by $<$, with the following two properties:

(1) If $x \in S$ and $y\in S$ then one and only one of the statements: $x<y$, $x=y$, or $y<x$ is true.

(2) If $x,y,z \in S$, if $x<y$ and $y<z$, then $x<z$

The question I asked myself is if property 2 must hold by definition. There answer I had come up with is yes.

Suppose that the above definition is provided with only the first condition. Now take a set, $S=(1,3,5)$. Next define the following order on $S$: $1<3$, $3<5$, $5<1$.

The above order satisfies the first property of the defintion, but not the second. Hence Transitivity as a definition is required.

Is my reasoning correct, and is there anything I missed?

Answer 1

That's a fine example how without transitivity, we would not have what we want: an ordered relation.

Can you think of an example of how transitivity alone will not suffice? (Without the first condition?).

The two properties/conditions are independent: there are relations that are transitive but not trichotomous, and there are relations that are trichotomous, but not transitive. But without having them both hold under a given relation, we would not capture, or define, what we want to define: an ordered relation.

Answer 2

By transitivity you have $1<1$ but also $1=1$, so it is a contradiction to the first condition.

Answer 3

You are correct.

What you are stating is that the properties are $independent$, in that you can find examples in which both are true or one is true and the other false.

A lot of work has been done on independence.