Let P be a poset and let us say that a subset A of P is a down-set if: $$x \in A, y < x \implies y \in A.$$
A directed set is a poset P such that for every two elements, $a,b \in P$ we can find $c \in P$ such that $c \geq a $ and $c \geq b$. Now, I am trying to prove the following statement:
A poset P is directed if and only if for every finite down-set D we can find a $c \in P$ such that $c \geq d, \forall d \in D$.
One direction is very easy, namely, to show that if a poset is directed, we can find such an upper bound for every finite down-set. The other direction seems a bit more tricky, but I suspect there's something obvious I am missing here. How could I show the other implication?
Is the statement really true? What about say $\mathbb{Z} \amalg \mathbb{Z}$, with the usual order on each factor? Are there any non-empty finite down-sets here? For what I can see, $\mathbb{Z} \amalg \mathbb{Z}$ is not directed, yet if there are no other finite down-sets than the empty set, then it satisfies the statement above.
Your counterexample is correct and shows that the statement you're trying to prove is wrong. There's not much else to say!