# Does every preorder induce a partial order?

A (non-strict) preorder is defined as a reflexive and transitive homogenous relation; a (non-strict) partial order is defined as a reflexive, antisymmetric, and transitive homogenous relation.

Clearly, every partial order is a preorder but not every preorder needs to be a partial order (cf. [1]). However, it seems to me that every preorder induces a partial order via the following construction:

• Every preorder induces a strict preorder (an irreflexive, transitive relation). [2]

• Every strict preorder is a strict partial order (an irreflexive, transitive, antisymmetric relation). [3]

• Every strict partial order induces a (non-strict) partial order. [4]

Is that correct or where did I go wrong? I appreciate any help you can provide.

Sources:

[3] Wikipedia on preorders: "A binary relation is a strict preorder if and only if it is a strict partial order."

[4] Wikipedia on partial orders: "Conversely, a strict partial order $$<$$ on $$P$$ may be converted to a non-strict partial order".

Here's an easy way to get a partial order:

Starting with a preorder $$R$$,

for each element $$x$$, put into the equivalence class of $$x$$ every element $$y$$ such that both $$xRy$$ and $$yRx$$.

You immediately have a partial order on the equivalence classes.

Now I refer to your steps.

Technically, every preorder induces a strict preorder unless all the elements are equivalent.

With that qualification, yes, your way works.

For illustration:

Suppose there are three elements in the preorder $$a\equiv b$$ and $$a.

The obvious way gives a partial order with two elements, $$[a,b]$$ being less than $$[c]$$.

Your way gives rise to a partial order with three elements, such that $$a$$ is less than $$c$$ and such that $$b$$ is less than $$c$$. The new partial order doesn't say anything that relates $$a$$ and $$b$$.

• Thanks, that certainly seems like a more straightforward way to get there. I suppose the partial order on the set of equivalence classes then canonically induces a partial order on the original set via $x \preceq y \quad :\Leftrightarrow \quad [x] \preceq [y]$, right? Jun 16, 2023 at 8:50
• @Maxas No it doesn't. The relation you retrieve that way is the same preorder you started with. Jun 16, 2023 at 9:08
• @amrsa Yes, I just realized $\preceq$ defined that way on the original set would not be antisymmetric. That means that the procedure sketched by Chris doesn't produce a partial order on the set itself, unlike what I presumably outlined in my original post. Or am I overlooking something? Jun 16, 2023 at 9:17
• @Maxas No, you are right. It's just a partial order on the quotient set. Jun 16, 2023 at 9:18
• @amrsa Do you know if the one in the original post works? Jun 16, 2023 at 9:19