What is the best way to construct an Aronszajn Tree?

Question

What is the best definition of Aronszajn tree? And, what is the best proof that it exists?

So I write the question to learn more about Aronszajn trees, any further detail is my intention to appreciate.

Answer 1

Walks on ordinals. See Propositions 4 & 5 from here.

Answer 2

Perhaps the easiest argument is given here, in Lemmas 1.1 and 1.2. The argument for $\kappa = \omega$ is due to Koszmider and I give a generalization.

Like the classical construction, we get a system of functions $\{ f_\alpha : \alpha < \omega_1 \}$ such that $f_\alpha : \alpha \to \omega$, and any two disagree on only a finite set. But now we only get finite-to-one functions instead of one-to-one functions. It is still immediate that there is no branch by a pigeonhole argument. The advantages are (a) the construction can be done with much less "care" (especially for the analogous construction for successors above $\aleph_1$), and (b) it continues upward to produce "forests" on higher cardinals in a way that the classical construction cannot.

A few interesting questions were raised in writing my paper that I could not answer. The first one is whether it is consistent that such forests do not exist on $\aleph_\omega$. Koszmider uses $\square_{\aleph_{\omega}}$ and $\aleph_\omega^\omega = \aleph_{\omega+1}$ to continue at that stage. But instead of $\square_{\aleph_{\omega}}$ he really just uses a "Jensen matrix" which is known to be much weaker than $\square_{\aleph_{\omega}}$.

Answer 3

This is an intro to saf's answer.

Suppose we have a sequence $(r_\alpha)_{\alpha\lt\omega_1}$ where each $r_\alpha:\alpha\to\omega$. Let $T_\alpha$ consist of all $f:\alpha\to\omega$ that agree with $r_\alpha$ in all but finitely many places. (In particular, each $T_\alpha$ is countable.) On the one hand, these sets form the levels of a subtree of $\omega^{\lt\omega_1}$ iff for all $\alpha\lt\beta\lt\omega_1$, the restriction $r_\beta\upharpoonright\alpha$ agrees with $r_\alpha$ in all but finitely many places. (This is called coherence.) On the other hand, the resulting tree has no branch iff there is no $r:\omega_1\to\omega$ such that the restriction $r\upharpoonright\alpha$ agrees with $r_\alpha$ in all but finitely many places. (This is called nontriviality.)

The coherence and nontriviality requirements are seemingly at odds with each other, but it is possible to get sequences $(r_\alpha)_{\alpha\lt\omega_1}$ that are both coherent and nontrivial. This is certainly a neat way to get an Aronzajn tree. One can construct such $(r_\alpha)_{\alpha\lt\omega_1}$ by a careful transfinite recursion and this is historically how these coherent trees were first constructed. But now there is a better way: walks on ordinals... (Continued in saf's answer.)

Answer 4

For each ordinal $\alpha\lt\omega_1$ construct a countable collection $T_\alpha$ of well-ordered subsets of $\mathbb Q$ of order type $\alpha$ in such a way that, whenever $\beta\lt\alpha\lt\omega_1$, $X\in T_\beta$, and $\sup X\lt r\in\mathbb Q$, there is a set $Y\in T_\alpha$ such that $X$ is an initial segment of $Y$ and $\sup Y\lt r$.

Oops! Also make sure that, if $X\in T_\alpha$, then every proper initial segment of $X$ is in the appropriate $T_\beta$. (Thus the set $T_\alpha$ is actually level $\alpha$ of the tree.)

Answer 5

There is also Shelah's very enjoyable construction using descending sequences of infinite subsets of $\omega$, close in spirit to Aronszajn's tree of rational sequences, and described in Judith Roitman's book here: https://www.math.ku.edu/~roitman/stb3fullWeb.pdf.