Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 19 / 6 / iv / b / Solution

Use the displayed family to form the coherent-injection Aronszajn tree. A node at level $\beta$ is a restriction $e_{\alpha }\upharpoonright \beta$ for some $\alpha \ge \beta$. Every such node differs only finitely from $e_{\beta }$. There are countably many finite subsets of the countable domain $\beta$ and countably many assignments of natural-number values to each, so there are only countably many possible finite modifications. Hence each level is countable. It is nonempty because it contains $e_{\beta }$.

Every shorter restriction of a node is again a node, and its predecessors have order type its domain ordinal. Thus this is a set-theoretic tree of height ${\omega }_1$. If it had an uncountable chain in a partial order, its domain heights would be unbounded in ${\omega }_1$, since the levels below any countable height contain only countably many nodes. The union of that chain in a partial order would be an injection ${\omega }_1\to \omega$, impossible. ThereforeThe countable-level proof uses coherence, whereas the no-branch proof uses injectivity; the two features play different roles.