# Hasse Diagrams of trees with height $>\omega^2$

I'm looking to improve my intuition and visualization of what large countable trees look like and I've ran into the issue that I have no understanding of what a tree of height, say, $\omega^2$ looks like. I'll include pictures of the kind of thing I'm looking for, and where my understanding breaks down.

An $\omega$ tree has a subtree that's isomorphic to something that looks like this, An $\omega+1$ subtree has a subtree isomorphic to something that looks like this, Repeating this process (of adding a node at the top) and taking unions means I've got some handle on what a tree of height $\omega . 2$ looks like, it's a union of things like this.

I can understand how to keep going with this process and my imagination (maybe) stretches far enough as $\omega.3$, but I really can't get any further with it.

My question is, is there any nice ways of visualizing a tree of height $\omega ^2$ as a Hasse diagram like this? Or is trying to use your intuition on a problem like this kind of futile?

• I like this question, but it may actually be a better fit at math.stackexchange.com. – Joel David Hamkins Aug 20 '13 at 15:10
• And by the way, height isn't really the right term for your usage here, since in set theory when we speak of a tree of height $\alpha$, we usually mean that it can be stratified into $\alpha$ many levels, with the predecessors of a node on lower levels. For example, a Suslin tree is a kind of tree of height $\omega_1$. What you are concerned with would usually be called the rank of a well-founded tree. – Joel David Hamkins Aug 20 '13 at 16:00
• So a tree of height $\omega_1$ might not be well founded (meaning it might have an $\omega^{-1}$ path within it), right? – Eran Aug 21 '13 at 14:33
• If you use height in my sense (the usual sense), then the trees grow upward, and the root is minimal. Any node on level $\omega$ or higher has a linear $\omega$-chain below it. So this is a very different meaning than the OP's. – Joel David Hamkins Aug 21 '13 at 19:01

One can construct explicit trees of quite large rank, using the following two rules:

• Given a tree whose of rank $\beta$, simply add a new root atop the old root to have a tree of rank $\beta+1$.

• Given trees $T_n$ of rank $\alpha_n$, join them together with a new root to have a tree of rank $\sup_n(\alpha_n+1)$.

Applying these rules, one can easily construct trees of any countable ordinal height. In particular, to make a tree of rank $\omega\cdot 3$, first make trees of rank $\omega\cdot 2+n$ by adding a chain of $n$ nodes on top of your tree of rank $\omega\cdot 2$, and then join them together to realize $\sup_n\omega\cdot 2+n=\omega\cdot 3$.

This methods gets you up to $\omega^2$, $\omega^3$, $\omega^\omega$, even $\epsilon_0$ and beyond.

• The first rule is actually a special case of the second rule, where one has only one tree. – Joel David Hamkins Aug 20 '13 at 15:09
• My only issue is that this doesn't provide much insight as to what these things look like or any characterizing features of them. The definition makes sense and this process makes sense it just lacks insight to me. – Robadob Aug 21 '13 at 14:46
• Following the rules, you can easily draw them, for quite large ordinals. Perhaps it would help your intuition to draw pictures of the ordinals themselves in this range. – Joel David Hamkins Aug 21 '13 at 15:34