# 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?

• A partial order is a preorder that is also antysymmetric. Oct 30, 2016 at 23:24

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.

• Ah yes, my bad! Thank you. Another thing- what if I have a binary relation that is reflexive, transitive and symmetric. Would this binary relation be considered a preorder? Oct 30, 2016 at 23:08
• @MarkJ Yes, that would be a preorder. Saying that every preorder is reflexive and transitive does not mean that those are the only properties that a preorder can have. Oct 30, 2016 at 23:11
• This is exactly what I was looking for. Thank you for your help! Oct 30, 2016 at 23:13
• @MarkJ It seems you are satisfied with the answer and further comments. Then why didn't you accept the answer? It's just a click to acknowledge the time that Noah Schweber took to clarify your confusion :) Oct 31, 2016 at 9:55

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 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 $$,$$\lesssim >.$$ It can allow $$a\lesssim b$$ and $$b\lesssim a$$ with a straight face, without giving you the equality. Now a pre-order cannot be made into a partial order on a set $$, $$\lesssim >$$ 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 $$$$ by saying that $$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 $$b\lesssim a.)$$