Does every tree have an upper portion that is a non-empty chain?

Question

Given a poset $P$, call $A \subseteq P$ an upper portion iff $A^c < A$, by which I just mean that for all $a \in A^c$ and all $b \in A$ we have that $a < b$. Then every upper portion is upward closed. Proof. Suppose for a contradiction that $a \in A$ and $a \leq b$, but not $b \in A$. Then $b \in A^c$, so $b < a$, a contradiction.

Now by tree (Warning! Non-standard definition) let us mean:

• a directed poset with the property that the upward closure of any singleton is a chain.

By the upward closure of $\{x\}$ I just mean $\{y \mid y≥x\}$.

Question. Does every tree have an upper portion that is a non-empty chain?

Answer 1

No:

        ·  
        ·  
        ·  
        x  
        |\  
        x x  
        |\  
        x x  
        |\  
        x x