Let $S$ be a set. An ordered on $S$ is relation $<$, with following 2 properties
1) If $x,y\in S$ then one of this will true $x<y$, $y<x$, $x=y$
2) If $x,y,z\in S$, if $x<y$, $y<z \to x<z$.
But I am interested in an example such that set with relation with 1 holds but 2 does not hold.
Any Help will be appreciated
Think about rock, paper, scissors. The first condition can certainly be met for the set $\{r,p,s\}$, namely we can define the relation $$<\,\,:= \{(s,r),(p,s),(r,p)\},$$ where $(x,y)$ is the relation $x<y$. For instance, $s<r$ since rock beats scissors. So for any two actions in the game we know which of the two is better. However, the second condition is not fulfilled, because we have the counterexample that $s<r$ and $r<p$, but $s\nless p$.