# $<$ in a preorder

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?

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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
This is discussed on Wikipedia – Henry Jul 13 '12 at 22:46
@azarel That is one reason for this and a related question. – Code-Guru Jul 13 '12 at 23:17
@Henry Thanks for the link. I'll study that in more detail as I can. – Code-Guru Jul 13 '12 at 23:18

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 $[x]=[y]$.
(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.