Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 24 / 2

A Suslin line is a dense linear order without endpoints that is order-complete, has the countable chain condition for a linear order, and is not separable in its order topology. Thus every disjoint family of nonempty open intervals is countable, but there is no countable order-dense subset. The completeness condition says that every nonempty bounded-above subset has a supremum. These requirements distinguish a Suslin line from the real line, which has the countable order-dense subset $\mathbb{Q}$.

A Kurepa tree is a set-theoretic tree of height ${\omega }_1$, with every level countable, possessing at least ${\aleph }_2$ distinct cofinal branches. The predecessors of each node are well ordered; their order type is the node's height. A cofinal branch is a maximal chain with nodes at unbounded heights below ${\omega }_1$, equivalently one node at every level after taking its predecessor closure.

A cofinal map $f:\delta \to \alpha$ has a range cofinal in $\alpha$:For a limit $\alpha$, this is equivalent to $\sup f[\delta ]=\alpha$. For a successor $\alpha$, its largest element must occur in the range. Monotonicity is an additional property and is not part of this definition. The cofinality of $\alpha$ is the least ordinal that admits such a cofinal map into $\alpha$.

For clarity, each ${\rho }_{\beta }(\alpha )$ is an ordered finite sequence. Its minimal walk along a club sequence is finite because its nodes strictly decrease until reaching $\alpha$, and an infinite strictly decreasing sequence of ordinals cannot exist. At a step from $\delta >\alpha$, unboundedness of $C_{\delta }$ supplies $\min (C_{\delta }\setminus \alpha )$ in $[\alpha ,\delta )$.

Compare the two minimal walks along a club sequence to $\xi$ and to $\alpha$. As long as their current node is the same $\delta >\alpha$, their next nodes agree exactly when $C_{\delta }$ has no point in $[\xi ,\alpha )$. At the first such point, the walk to $\xi$ moves towhereas the walk to $\alpha$ stays at or above $\alpha$. If this does not occur before the second walk terminates, the first walk reaches $\alpha$ with it and then makes its next step below $\alpha$.

Thus there is an index $j$, possibly the terminal index of the walk to $\alpha$, withIt is unique: after this step the first walk stays below $\alpha$, so it can never again coincide with a node of the walk to $\alpha$. This is the first-divergence lemma for minimal walks.

Let $j$ be the common-prefix index from the first-divergence lemma for minimal walks. If $j=n$, the walk to $\xi$ has followed the entire walk to $\alpha$ and must make at least one further step, contradicting the assumed equality of lengths. Hence $j<n$.

The current nodes agree, so put $\delta ={\beta }_j^{\xi }={\beta }_j^{\alpha }$. Then $C_{\delta }\cap \xi$ is an initial segment of $C_{\delta }\cap \alpha$. It is proper because ${\beta }_{j+1}^{\xi }\in C_{\delta }\cap [\xi ,\alpha )$. ThereforeIn particular, each function ${\rho }_{\beta }$ is injective: equality of two trace sequences would give equal lengths and then contradict this proper-initial-segment conclusion.

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: