Simplified method of building an Aronszajn tree

Question

There is a very interesting method to build an Aronszajn tree in Judith Roitman's "Introduction to Modern Set Theory", on pages 100-102. In short, we build a tree $T$ in which the nodes are well-ordered subsets of $\Bbb Q$, partially ordered so that $\sigma_0 ≤ \sigma_1$ iff $\sigma_0$ is an initial segment of $\sigma_1$. Each such node $\sigma$, as a subset of $\Bbb Q$, has supremum $\sup \sigma$, and we require this be also in $\Bbb Q$. The top level is the empty set, and then the next is the set of all 1-element subsets, and so on, with a certain technique used at limit ordinals to make sure the tree is not too wide.

It seems that something much simpler than what she is doing will also suffice. The basic idea is:

1. We start with top node the empty set.
2. At level 1, we have one node for each rational number.
3. For successor ordinal $\alpha+1$, we take all $\Bbb Q$-subsets that are nodes at $\alpha$ - and extend each in every possible way by adding one more rational at the end.

This is all her method so far without alteration. Of particular note is that, for any node $\sigma$ with supremum $\sup \sigma$, there will be nodes $n \in \Bbb N$ levels below with supremum $q + \sup \sigma$ for each possible $q > 0$. Let's call this the "arbitrarily-shifted-supremum" property.

Suppose $\alpha$ is a limit ordinal. Instead of her method, we simply do this:

4. For every node $\sigma$ previously created with some $\sup \sigma$, there will be uncountably many branches of height $\alpha$ passing through it. The union of each branch is a well-ordered subset of $\Bbb Q$ extending $\sigma$. For each $q>0$, arbitrarily choose one such branch so that its union $\tau \subset \Bbb Q$ has supremum $\sup \tau = q + \sup \sigma$, and add it as a node at level $\alpha$.

The idea is that we can't add all branches for each limit ordinal as nodes, or else we'd have uncountable width, so we choose countably many such that we "mimic" this "arbitrarily-shifted-supremum" property that we get at successor levels. Roitman's method also does this, but in a rather complex way - instead of choosing branches, she chooses cofinal $\omega$-sequences of children cofinal in some branch, and plans out the increases in suprema rather carefully within the branch.

However, either way, each level has countable width (as we have pruned at each level), no branch can have uncountable height (as its union would be a well ordered subset of order type $\omega_1$), and each node will have children arbitrarily far below in the tree - at least according to Roitman, as I note this simplified method also satisfies her (**) property.

So, have we perhaps succeeded in creating an Aronszajn tree, or is there some subtle snag requiring the extra complexity of her method? And if we have succeeded, what is the point of that extra complexity?

Answer 1

Your argument is basically Kurepa's proof from his thesis Ensembles ordonnées et ramifiés, see page 96 (a footnote has Aronszajn's construction). As noted in the comments you need to show that what you suggest in step 4 is actually possible. Kurepa had already investigated the tree of well-ordered subsets of $\mathbb{Q}$ in some depth and he could indeed `simply' assert that what you suggest is possible.