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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.

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up vote 1 down vote accepted

Your counterexample is correct and shows that the statement you're trying to prove is wrong. There's not much else to say!

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