Assuming we use the set-theoretic definition of a tree:
A set $T$ with a partial order $<$ is a tree if for any $t \in T$
the set $\{ s \in T : s < t \}$ is well-ordered by $<$.
https://en.wikipedia.org/wiki/Tree_(set_theory)
We know how to define discrete, dense, and continuous linear orders using cuts:
- A linear order is discrete if every cut of it is a jump;
- A linear order is dense if it contains no jumps;
- A dense linear order is continuous if it contains no gaps.
Is there a reason to not use the types of orders in trees?
Can we define discrete, dense, and continuous trees the same way?
My attempt:
A set $T$ with a partial order $\le$ is a discrete (resp. dense, continuous) tree if for any $t \in T$:
- the set $S = \{ s \in T : s \le t \}$ is totally ordered by $\le$, and
- the total order of $S$ is discrete (resp. dense, continuous).
If we add the connectivity property:
- for any two elements $x$ and $y$ of $T$ there is an element $z$ such that $z \le x$ and $z \le y$,
it looks like the new definition includes the sets of real trees:
https://en.wikipedia.org/wiki/Real_tree
Does it make sense?
Now, assuming we separated discrete and continuous trees using the order type.
Let's take a set $T = \{0\} \cup \mathbb T$ with the partial order $x < y \iff |x| < |y|$,
where $\mathbb T = \{ z \in \mathbb C : |z| = 1 \}$ is the unit circle (https://en.wikipedia.org/wiki/Circle_group),
and $0 = (0,0)$ is the center of the circle.
According to my definition, $T$ is a discrete tree.
But $T$ contains an uncountable (not discrete) subset $\mathbb T$.
What kind of tree is it?
In your example, $(\{0\}\cup\mathbb T,<)$ does not know about the elements of $\mathbb T$ forming a continuous circle in $\mathbb C$.
You'd get an isomorphic tree when taking a discrete set $X$ with cardinality $2^{\aleph_0}$ and let $T'=\{0\}\cup X$ with $0<x$ for all $x\in X$ and the elements of $X$ incomparable.