$<$ in a preorder

Question

The author of the book I am studying defines $<$ for a poset as

If $x, y \in X$, where $X$ is a poset, then we shall write $x < y$ to mean that $x \le y$ and $x \ne y$.

From this, I can conceive of two definitions for $<$ for a preorder:

1) If $x, y \in X$, where $X$ is a preorder, then we shall write $x < y$ to mean that $x \le y$ and $x \ne y$.

or

2) If $x, y \in X$, where $X$ is a preorder, then we shall write $x < y$ to mean that $x \le y$ and $y \not\le x$.

Which of these is more appropriate?

The second one. As it may happen that $x \leq y $, $y\leq x$ and $x\ne y$. — azarel, Jul 13 '12 at 22:42
@Henry Thanks for the link. I'll study that in more detail as I can. — Code-Guru, Jul 13 '12 at 23:18

score 3 · Answer 1 · answered Jul 13 '12 at 22:44

(1) is incorrect, since if $\leq$ is only a pre-order, then it can have $x\leq y\leq x$ with $x\neq y$, and this will mean that the relation $\lt$ defined in (1) is not transitive, since you will have $x\lt y\lt x$, but $x\not\lt x$. So in (1), you won't have a partial order.

(2) is the strict partial order commonly associated with a pre-order; it is transitive and irreflexive.

score 3 · Accepted Answer · edited Jul 23 '12 at 17:35

3

You may know by now that if you have a preorder then you can take a quotient by the equivalent relation $x\sim y\iff x\leq y\land y\leq x$ and have a poset.

The definition should be such that it carries over to the quotient, so the second definition is more appropriate. In the first one we can have $x\neq y$ and $x\leq y\land y\leq x$, but $x\sim y$ so in the induced poset $[x]=[y]$.

edited Jul 23 '12 at 17:35

Code-Guru

2,235

answered Jul 13 '12 at 22:44

Asaf Karagila

405,794

$<$ in a preorder

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