Preorders vs partial orders - Clarification 
*

*A binary relation is a preorder if it is reflexive and transitive.

*A binary relation is a partial order if it is reflexive, transitive and antisymmetric.


Does that mean that all binary relations that are a preorder are also automatically a partial order as well?
In other words is a binary relation a preorder if its only reflexive and transitive and nothing else?
Thanks for your help.
 A: A pre-order $a\lesssim b$ is a binary relation, on a set $S,$ that is
reflexive and transitive. That is $\lesssim $ satisfies (i)  $\lesssim $ is
reflexive, i.e.,  $a\lesssim a$  for all and (ii) $\lesssim $ is transitive,
i.e., $a\lesssim b$ and $b\lesssim c$ implies $a\lesssim c,$ for all $%
a,b,c\in S.$ (A pre-ordered set may have some other properties, but these
are the main requirements.)
On the other hand a partial order $a\leq b$ is a binary relation on a set $S$
that demands $S$ to have three properties: (i) $\leq $ is reflexive, i.e., $%
a\leq a$ for all $a\in S$, (ii) $\leq $ is transitive, i.e., $a\leq b$ and $%
b\leq c$ implies $a\leq c,$ for all $a,b,c\in S$ and (iii) $\leq $ is
antisymmetric, i.e., $a\leq b$ and $b\leq a$ implies $a=b$ for all $a,b\in S$%
.
So, as the definitions go, a partial order is a pre-order with an extra
condition. This extra condition is not cosmetic, it is a distinguishing
property. To see this let's take a simple example. Let's note that $a$
divides $b$ (or $a|b)$ is a binary relation on the set $Z\backslash \{0\}$
of nonzero integers. Here, of course, $a|b$ $\Leftrightarrow $ there is a $%
c\in Z$ such that $b=ac.$
Now let's check: (i) $a|a$ for all $a\in Z\backslash \{0\}$ and (ii) $a|b$
and $b|c$ we have $a|c.$ So $a|b$ is a pre-order on $Z\backslash \{0\},$ but
it's not a partial order. For, in $Z\backslash \{0\},$ $a|b$ and $b|a$ can
only give you the conclusion that $a=\pm b,$ which is obviously not the same
as $a=b.$
The above example shows the problem with the pre-ordered set $<S$,$\lesssim
>.$ It can allow $a\lesssim b$ and $b\lesssim a$ with a straight face,
without giving you the equality. Now a pre-order cannot be made into a
partial order on a set $<S$, $\lesssim >$ unless it is a partial order, but
it can induce a partial order on a modified form of $S.$ Here's how. Take
the bull by the horn and define a relation $\sim ,$ on $<S,\lesssim >$ by
saying that $a\sim b$ $\Leftrightarrow a\lesssim b$ and $b\lesssim a$. It is
easy to see that $\sim $ is an equivalence relation. Now splitting $S$ into
the set of classes $\{[a]|$ $a\in S\}$ where $[a]=\{x\in S|$ $x\sim a\}.$
This modified form of $S$ is often represented by $S/\sim .$ Now of course $%
[a]\leq \lbrack b]$ if $a\lesssim b$ but it is not the case that $b\lesssim
a.$ Setting $[a]=[b]$ if $a\sim b$ (i.e. if $a\lesssim $ and $b\lesssim a).$
In the example of $Z\backslash \{0\}$ we have $Z\backslash \{0\}/\sim $ $%
=\{|a|$ $|$ $a\in Z\backslash \{0\}\}.$
(Oh and as a parting note an equivalence relation is a pre-order too, with
the extra requirement that $a\lesssim b$ implies $b\lesssim a.)$
A: order relations are subset of pre order relations. For instance a relation kind of "prefer or indiferent" is reflexive and transitive, but is not antisymetric, so this is an example of pre order but not order. A relation like "bigger or equal" is reflexive, transitive and also antisymetric, so this relation is pre order (since it is reflexive and transitive) but also order.
A: You have it backwards - every partial order is a preorder, but there are preorders that are not partial orders (any non-antisymmetric preorder).
For example, the relation $\{(a,a), (a, b),(b,a), (b,b)\}$ is a preorder on $\{a, b\}$, but is not a partial order.
