= Solution
Take $\kappa=\omega_1$ and choose a <club sequence> on $\omega_2$, each $C_\delta$ of <order type> at most $\omega_1$. For a successor $\delta$ use its predecessor as a singleton; at a limit use a cofinal sequence of minimal length. Form the <minimal-walk tree>
$$
T_\alpha=\{\rho_\beta\upharpoonright\alpha:\alpha\leq\beta<\omega_2\},
\qquad T=\bigcup_{\alpha<\omega_2}T_\alpha,
$$
ordered by proper extension. Its height is $\omega_2$.
For $\xi<\delta$, the initial segment $C_\delta\cap\xi$ has <order type> strictly below $\omega_1$, and is countable. The strict inequality follows because a point of $C_\delta$ at or above $\xi$ occurs later in its enumeration. Hence every entry of every trace is countable. Under the <Continuum hypothesis>, for $|\alpha|\leq\aleph_1$,
$$
|[\alpha]^{\leq\omega}|\leq\aleph_1^{\aleph_0}
=(2^{\aleph_0})^{\aleph_0}=\aleph_1.
$$
There are at most $\aleph_1$ finite sequences of such sets. The <trace coherence lemma for minimal walks> says that, for $\beta>\alpha$, the value $\rho_\beta(\alpha)$ determines $\rho_\beta\upharpoonright\alpha$. The case $\beta=\alpha$ adds at most one node. Thus $|T_\alpha|\leq\aleph_1$ for every level.
Suppose that $T$ had a <cofinal branch>, and take the union of its functions, $f$, with domain $\omega_2$. Every $\rho_\beta$ is injective by the proper-initial-segment argument, so $f$ is injective too. On the <stationary set>
$$
S=\{\alpha<\omega_2:\operatorname{cf}(\alpha)=\omega_1\},
$$
this set is stationary because the supremum of a strictly increasing $\omega_1$-sequence from any <club set> has cofinality $\omega_1$ and lies in that club. The union of the finitely many countable entries of $f(\alpha)$ is bounded below $\alpha$. Assign a strict upper bound below $\alpha$ to obtain a <regressive function>. By <Fodor lemma>, there is a stationary $S'\subseteq S$ and a single $\eta<\omega_2$ such that every entry of $f(\alpha)$ lies inside $\eta$ for $\alpha\in S'$. There are at most $\aleph_1$ such finite sequences by the same <cardinal arithmetic>, but $|S'|=\aleph_2$, contradicting injectivity.
Therefore \b[$T$ is an <aleph-two Aronszajn tree>]:
$$
\boxed{\operatorname{ht}(T)=\omega_2,\quad |T_\alpha|<\aleph_2,
\quad T\text{ has no cofinal branch}.}
$$
Back to article page
