## Literature on total strict preorders [closed]

I am considering a set $C$ and want to impose a binary relation $\prec$ on it that is

• transitive, i.e. if $a\prec b\land b\prec c$ then $a\prec c\ \forall a,b,c\in C$
• irreflexive, i.e. $\neg a \prec a \ \forall a \in C$
• total, i.e. $a \prec b \lor b \prec a\ \forall a,b\in C$ edit: this should read $a \prec b \lor b \prec a \lor a \sim b\ \forall a,b\in C$ to be consistent, where $\sim$ denotes equivalence of $a$ and $b$

and by consulting wikipedia (http://en.wikipedia.org/wiki/Strict_weak_ordering#Total_preorders) it appears that this binary relation is a strict, total preorder. However, Fishburn (http://www.jstor.org/stable/168680) seems to imply that it is sufficient to consider a strict, total order because he assumes that for $a,b\in C$, $\neg a \prec b \land \neg b \prec a$ implies equivalence and not equality.

I have not had much luck finding classic literature on the topic and would be very thankful for suggestions as well as a clarification of whether antisymmetry of a strict total order implies equivalence or equality.

Thank you!

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Your axioms are contradictory, since by totality, in the case $a=b$, you would need $a\prec a$, which violates irreflexivity. Usually, the pre-order terminology is used for reflexive transitive relations, but for which one might have $a\leq b\leq a$ without $a=b$. – Joel David Hamkins Jan 23 2012 at 13:56
Unless $C$ has no elements. – Joel David Hamkins Jan 23 2012 at 14:08
Yes, agreed. The axioms are contradictory and should be modified to read $a\prec b \lor b\prec a \lor a\sim b$ where $\sim$ denotes equivalence of $a$ and $b$ with respect to the ordering. – Till Hoffmann Jan 23 2012 at 14:41
No. If you want just something resembling totality, you need to start with "for a and b with a not equal to b ...". Otherwise you need not just the equivalence relation, but some guarantee that your antisymmetric order respects that relation, otherwise you might have c less than a equiv b less than c, and other messy things which make transitivity useless. Gerhard "Ask Me About System Design" Paseman, 2012.01.23 – Gerhard Paseman Jan 23 2012 at 14:59
Ordinarily, the equivalence relation associated with a pre-order $\leq$ is the relation $a\sim b\iff a\leq b\leq a$. But since your relation $\prec$ is not reflexive, this will not be an equivalence relation for you. I am voting to close this question, and I suggest you post your question at math.stackexchange.com, where you will find someone who will straighten out these elementary issues about pre-orders. – Joel David Hamkins Jan 23 2012 at 15:09