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.
On
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 $b\lesssim a.)$
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.