Example of a partially ordered set whose Hasse diagram would look like a full binary tree I am starting to study partially ordered sets and I was curious as to if there is an example of a poset whose Hasse diagram would look like a full-binary tree.

A full binary tree (sometimes proper binary tree or 2-tree) is a tree in which every node other than the leaves has two children

Also, are there any infinite posets which would have the form of a full binary tree? Would Zorn's Lemma apply to such a poset?
Thanks
 A: Let $^{< \omega}2$ be the set of all finite $0-1$ sequences. For $s,t \in ^{< \omega}2$ write $s \preceq t$ iff $s$ is an initial segment of $t$ (if you define sequences the way I do, this is equivalent to $s \subseteq t$). Then $(^{< \omega}2, \preceq)$ is a countable poset whose Hasse diagram is the full binary tree.

Note that this poset contains chains without upper bounds, e.g. $\{ (0),(0,0), (0,0,0), \ldots \}$. Therefore, Zorn's Lemma doesn't apply to this poset and since we can extend every finite $0-1$ sequence to a longer one, this poset doesn't contain a maximal element.
A: Given any directed tree $T=(V,E)$, you can define a partial order $\leq$ on $V$ by $v\leq w$ iff there is a directed path from $v$ to $w$ in $T$.  The Hasse diagram of $(V,\leq)$ is then $T$.  Indeed, if $(v,w)\in E$, then $v\leq w$, and if $v< x< w$, then concatenating the path from $v$ to $x$, the path $x$ to $w$, and the edge $(v,w)$ would give a loop contradicting that $T$ is a tree.  Conversely, if $v\leq w$ and there is nothing between them, the path from $v$ to $w$ cannot pass through any other vertices, so $(v,w)$ must be an edge of $T$
