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

We construct a stronger family: every injection $e_{\alpha }:\alpha \to \omega$ will have coinfinite range, and coherence means agreement modulo finitely many arguments. Begin with the empty function. At a successor, extend $e_{\alpha }$ by assigning the new argument a value outside its range; the range remains coinfinite, and coherence with earlier functions is unchanged.

At a countable limit $\alpha$, take $0={\alpha }_0<{\alpha }_1<\cdots$ cofinal in $\alpha$. We build injections $p_n:{\alpha }_n\to \omega$, extending one another exactly, with $p_n$ differing from $e_{{\alpha }_n}$ at finitely many arguments. Also reserve distinct numbers $k_0,k_1,\dots$, never used by the eventual union. After constructing $p_n$, choose $k_n$ outside its range and different from the earlier reserved numbers; its range is coinfinite since it differs finitely from that of $e_{{\alpha }_n}$.

To extend to ${\alpha }_{n+1}$, first use $e_{{\alpha }_{n+1}}$ on the new arguments while keeping $p_n$ on the old ones. Only finitely many collisions can arise: $p_n$ and $e_{{\alpha }_{n+1}}\upharpoonright {\alpha }_n$ differ finitely by the inductive coherence hypothesis, so their image sets differ finitely. The candidate is otherwise injective. There are also only finitely many new arguments assigned a value among $k_0,\dots ,k_n$. Reassign these finitely many bad arguments to distinct fresh values outside the candidate image and the reserved finite set. Infinitely many fresh values are available because the candidate image differs only finitely from the coinfinite image of $e_{{\alpha }_{n+1}}$. The resulting $p_{n+1}$ is injective, extends $p_n$, omits the reserved numbers, and differs finitely from $e_{{\alpha }_{n+1}}$.

Put $e_{\alpha }={\bigcup }_n{p}_n$. It is injective and omits all $k_n$, so its range is coinfinite. For each $\beta <\alpha$, choose $n$ with $\beta \le {\alpha }_n$. The restriction $e_{\alpha }\upharpoonright \beta$ is $p_n\upharpoonright \beta$, which differs finitely from $e_{{\alpha }_n}\upharpoonright \beta$, and hence from $e_{\beta }$. Transfinite recursion now givesThese coherent coinfinite injections into omega solve the requested problem. Reserving infinitely many omitted values is important: a union of injections with individually coinfinite ranges need not itself have coinfinite range.

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.