Covering relation

Remarks

Order the positive integers by divisibility. Then $2 \lessdot 6$: the only divisors of $6$ that $2$ divides are $2$ and $6$, so nothing lies strictly between. In general $m \lessdot n$ exactly when $m \mid n$ and $n / m$ is prime. The pair $2 < 8$ is not a cover, since $2 < 4 < 8$, so the second condition fails. In a finite poset, $x < y$ holds if and only if there is a finite sequence $x = z_0 \lessdot z_1 \lessdot \cdots \lessdot z_k = y$ with $k \ge 1$. So the covering relation determines a finite partial order. This fails for infinite posets: in $\Q$ the covering relation is empty.