Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 19 / 4

The partition relation $\kappa \to (\lambda {)}_{\beta }^{\alpha }$ means that for every coloring $c:[\kappa {]}^{\alpha }\to \beta$, there is $H\subseteq \kappa$ with $|H|=\lambda$ such that $c$ is constant on $[H{]}^{\alpha }$. Here $[X{]}^{\alpha }$ denotes the subsets of $X$ of cardinality $\alpha$, and the subscript is the number of colors. The domain consists of subsets of size $\alpha$, not ordered tuples, and the definition also makes sense for infinite arity.

The stationary-indexed club principle ${♣}_S$ asserts a sequence $⟨A_{\delta }:\delta \in S\cap \mathrm{Lim}({\omega }_1)⟩$ with each $A_{\delta }\subset \delta$ cofinal of order type $\omega$, such that for every uncountable $X\subseteq {\omega }_1$ some $\delta \in S$ satisfies $A_{\delta }\subseteq X$. It predicts a cofinal ladder contained in the target, rather than predicting the entire initial segment of the target as diamond does. For a stationary $S$ with successor members, only its stationary limit-ordinal part is relevant; countable limit ordinals have cofinality $\omega$. This is the restricted version of the club principle.

A tree with unique limits has no distinct nodes at a limit level with the same history. Precisely, for a limit ordinal $\delta$ and $s,t\in T_{\delta }$,Here $s\upharpoonright \beta$ denotes the unique predecessor of $s$ of height $\beta$. This is uniqueness of a limit node when it exists, not a requirement that every cofinal chain in a partial order below a limit level acquire a limit node.

Use the standard normal set-theoretic tree convention: a unique root, extensions at every higher level, splitting into at least two successors, and tree with unique limits. The small-level and height assumptions already give an ${\omega }_1$-tree, while the given tree antichain condition gives the countable chain condition for forcing. We only need to exclude an uncountable branch.

If such a branch existed, its heights would be unbounded, since each initial segment contains only countably many nodes. Fill in predecessors to obtain its node $b_{\alpha }$ at every level. At each successor step choose a successor $s_{\alpha }$ of $b_{\alpha }$ different from $b_{\alpha +1}$. For $\alpha <\beta$, the node $s_{\beta }$ extends the branch successor $b_{\alpha +1}$, and so is incompatible with $s_{\alpha }$. Thus $\{s_{\alpha }:\alpha <{\omega }_1\}$ is an uncountable tree antichain, a contradiction.

Therefore the set-theoretic tree is ${\aleph }_1$-Suslin. The splitting part of normality matters: a single chain in a partial order of height ${\omega }_1$ would satisfy the tree antichain condition but not the conclusion if one used a weakened definition of normality allowing no splitting.

Fix a diamond principle sequence $D_{\alpha }\subseteq \alpha$. Construct a normal splitting set-theoretic tree of height ${\omega }_1$ with countable levels. At successors give every node two successors. At a countable limit stage $\alpha$, the set-theoretic tree below $\alpha$ is countable. Choose countably many cofinal branches through it covering all its nodes, and put one node at level $\alpha$ above each distinct chosen branch. This preserves extension to all higher levels and tree with unique limits.

Arrange a coding of each level into the ordinal block $[\omega \alpha ,\omega (\alpha +1))$. On the club set of limit fixed points of $\alpha \mapsto \omega \alpha$, the nodes coded below $\alpha$ are exactly the nodes of height below $\alpha$. At a limit stage, if $D_{\alpha }$ codes a maximal tree antichain of the current set-theoretic tree below $\alpha$, require every chosen branch to meet it. This is possible: for any starting node $t$, maximality provides a comparable tree antichain member; if above $t$, first extend to it, and if below $t$, it has already been met. Then extend along a sequence of heights cofinal in $\alpha$. If the prediction is not a maximal tree antichain, use the ordinary covering branches. Thus every level is countable and the construction remains normal.

Here is the full chain-condition verification. Let $A$ be a maximal tree antichain in the final set-theoretic tree. For every node $t$, choose a witness $a_t\in A$ comparable with $t$. There is a club set of countable limit stages $\alpha$ closed under these witness choices: starting from any bound, repeatedly bound the heights of witnesses for all the countably many nodes below the current stage, and take the supremum after countably many steps. At such an $\alpha$, $A\cap T_{<\alpha }$ is already maximal in $T_{<\alpha }$.

View $A$ as a subset of ${\omega }_1$ through the coding. Diamond gives stationarily many stages with $D_{\alpha }=A\cap \alpha$. Choose one also in the witness-closure club set and the coding club set. The construction at that stage seals this very tree antichain: every node of level $\alpha$ extends one of its members below $\alpha$, and so does every node at a later level. No such node can itself belong to $A$, since it is comparable with an earlier member of $A$. HenceEvery tree antichain extends to a maximal one, so the set-theoretic tree has no uncountable tree antichain. Its normal splitting also excludes uncountable branches by part (ii). It is therefore a Suslin tree. By the standard Suslin-tree characterization of Suslin hypothesis, diamond implies failure of Suslin hypothesis. The decisive step is antichain sealing by diamond, with maximality below a correctly guessed club set stage verified explicitly.

A kappa-filtration is an increasing continuous sequence with union $A$ and $|A_{\alpha }|<\kappa$ at every stage. Intersect $S$ with the club set of nonzero limit ordinals. For each remaining $\alpha$, continuity gives $A_{\alpha }={\bigcup }_{\beta <\alpha }{A}_{\beta }$, so choose $r(\alpha )<\alpha$ with $f(\alpha )\in A_{r(\alpha )}$. This is a regressive function. Fodor lemma gives a stationary subset $S_1$ and a fixed $\beta$ such that all these values lie in $A_{\beta }$.

Since $|A_{\beta }|<\kappa$, partition $S_1$ into fewer than $\kappa$ fibers of $f$. If every fiber were nonstationary, choose a club set avoiding each one. Their intersection is club set by regularity, contradicting stationarity of $S_1$. Thus one fiber is stationary, andThis proves the filtration form of Fodor lemma; continuity at limit stages, the small size of each stage, and regularity of $\kappa$ are all used.