All tree orders are lattice orders? Say that a set is tree ordered if the downset $\downarrow a =\{b:b\leq a\}$ is linearly ordered for each $a$. In a comment, Keinstein says that such sets are also semi-lattices, provided they are connected. This doesn't seem true to me.
Consider e.x. this tree:
   a    c    e
    \  / \  /
      b    d

$b\wedge d$ does not exist, so it is not a lattice. 
Am I missing something?
 A: Your example is not tree-ordered since ${\downarrow}c$ is not linearly ordered.
A: Suppose $L$ is tree-ordered - downward sets are linear - and connected. Let $a,b\in L$. Connectedness tells us there is a path betweeen $a$ and $b$: by using transitivity such a path can be simplified into a "zig-zag" path just as you have in your example (since $u< v< w\Rightarrow u< w$), and so in particular we can take a zig-zag path of minimal length between $a$ and $b$. This minimal path cannot contain an upright triangle $x< y> z$: taking the downward set of $y$ we see $x$ and $z$ are comparable and hence this triangle can be simplified to either $x< z$ or $x>z$ contradicting minimality. This leaves only two possibilities: $a$ and $b$ are directly comparable, or $a>c<b$ is the entire path, so the set of common lower bounds for $a$ and $b$ is nonempty. Since this is the intersection of two linearly ordered sets it is itself linearly ordered and has a maximum, so there is a maximum lower bound on $a,b$.
