covering relation question.

Question

How can i show that "a partial order on a finite set is uniquely determined by its covering relation"?

Answer 1

Take the reflexive and transitive closure of the cover relation.

Answer 2

Let $C$ be a finite set equipped with a partial order $\leq$.

Define the relation $<$ on $C$ by: $a<b\iff a\leq b\wedge a\neq b$.

Define the relation $\lessdot$ on $C$ by: $a\lessdot b\iff a<b$ and no $c\in C$ exists with $a<c\wedge c<b$.

$a<b$ implies the existence of arrays $(c_0,\dots,c_n)$ with $c_0=a$, $c_n=b$ and $c_{i-1}<c_i$ for $i=1,\dots,n$.

All $c_i$ are distinct and the set containing them is finite, so we can take $n$ maximal.

It is evident that in that situation $c_{i-1}\lessdot c_i$ for every $i$ (if not then the maximality of $n$ is contradicted).

This proves that $<$ is a subset of the transitive closure of $\lessdot$.

Conversely $\lessdot$ is a subset of $<$ and $<$ is transitive.

We conclude that the transitive closure of $\lessdot$ is also a subset of $<$.

Proved is now that $<$ coincides with the transitive closure of $\lessdot$.

Of course $<$ completely determines $\leq$ (and vice versa), so we are ready.