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The tree of restrictions ${\rho }_{\beta }\upharpoonright \alpha$, ordered by extension. For a club sequence on ${\omega }_2$ with club order types at most ${\omega }_1$, the Continuum hypothesis bounds each level by ${\aleph }_1$. Trace injectivity and Fodor lemma exclude a cofinal branch.

There is a useful explicit reconstruction of a shorter trace from a longer one. Supposeand fix $\xi <\alpha$. Let $j<n$ be the first index with $S_j\setminus \xi \ne \varnothing$, if it exists. Before that index, the walk to $\xi$ follows the walk to $\alpha$. At index $j$ it moves to $\delta =\min (S_j\setminus \xi )<\alpha$. ThusWhen $\delta =\xi$, the appended trace is empty. If no such $j$ exists, the walk first reaches $\alpha$ and then continues toward $\xi$, givingBoth formulas depend only on the displayed sequence, $\alpha ,\xi$, and the fixed club sequence. Equal traces ${\rho }_{\beta }(\alpha )={\rho }_{\gamma }(\alpha )$ consequently reconstruct the same trace at every $\xi <\alpha$. This proves the trace coherence lemma for minimal walks:

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 treeordered 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 $|\alpha |\le {\aleph }_1$,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 $|T_{\alpha }|\le {\aleph }_1$ 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 setthis 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 $S^{\prime }\subseteq S$ and a single $\eta <{\omega }_2$ such that every entry of $f(\alpha )$ lies inside $\eta$ for $\alpha \in S^{\prime }$. There are at most ${\aleph }_1$ such finite sequences by the same cardinal arithmetic, but $|S^{\prime }|={\aleph }_2$, contradicting injectivity.

Therefore $T$ is an aleph-two Aronszajn tree: