I'm reading this paper on topologies on trees, and I am unsure of a construction being mentioned in it.
Let $T$ be a tree. Let $C \subset T$ be a chain that is bounded above. Then a pseudo-supremum of $C$ is the set of minimal upper bounds.
A Dedekind complete tree is a tree such that all pseudo-supremum are real supremum (i.e. pseudo-supremum is a singleton). The author mentions that we can take a Dedekind completion of any tree by giving "the set of pseudo-supremum of more than one element an immediate predecessor."
Can anyone give me a simple example of this process? I'm not sure exactly what even the Dedekind completion of a basic finite height tree would look like.
Thanks.
Finite height trees don't need to be completed. They are complete, since all chains are finite, they contain a maximum which is also a supremum. You do need this when your chains do not have a maximum, though,
Consider the following tree, instead: $\Bbb N\cup\{E,O\}$, where $E$ and $O$ are the even integers and odd integers respectively, ordered as follows: $$x<_T y\iff\begin{cases}x<y \text{ and }x,y\in\Bbb N, &\text{or}\\x\in\Bbb N, y\notin\Bbb N\end{cases}$$
In other words, we put two upper bounds to the chain $\Bbb N$. This is not Dedekind-complete, but we can add a new point, $P$ which is the upper bound of $\Bbb N$, and is below $E$ and $O$.