Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 24 / 3

The diagonal intersection of the family isMembership at $\beta$ only tests the sets indexed below $\beta$, rather than every set of the family. At a regular uncountable $\delta$, a diagonal intersection of club sets is again a club set.

Fodor lemma says that if $\lambda$ is a regular uncountable cardinal, $S\subseteq \lambda$ is stationary, and $f:S\to \lambda$ is regressive, thenHere a regressive function satisfies $f(\alpha )<\alpha$ at every nonzero $\alpha \in S$; removing $0$ has no effect on stationarity. Both regularity and stationarity are part of the hypotheses.

The club principle $♣$ asserts that there is a sequence $⟨A_{\delta }:\delta \in \mathrm{Lim}({\omega }_1)⟩$ such that $A_{\delta }\subseteq \delta$ has order type $\omega$, is cofinal in $\delta$, andIt guesses a cofinal countable subset of an uncountable set, with inclusion as the guessing requirement.

Let $⟨D_{\alpha }:\alpha \in S⟩$ witness the stationary diamond principle: for every $A\subseteq {\omega }_1$, the set of $\alpha \in S$ with $D_{\alpha }=A\cap \alpha$ is stationary. We construct a normal splitting Suslin tree; this will be a nonspecial Aronszajn tree.

Construct its levels by recursion. Start with one root, and give every node two immediate successors. At a countable limit $\alpha$, the constructed portion $T\upharpoonright \alpha$ is countable. Through each of its nodes choose a cofinal branch of that portion, and put one new node above each chosen branch at level $\alpha$. This keeps the level countable and gives every earlier node an extension. Branches are identified by their predecessor chains, so nodes at a limit level are uniquely determined by their predecessors.

At a limit $\alpha \in S$, decode $D_{\alpha }$ as a candidate tree antichain $B$ of $T\upharpoonright \alpha$. If it is maximal, choose the branches just described to meet $B$. This is possible: for any node, maximality supplies a comparable member of $B$, and normality of the already constructed portion extends the larger of those two nodes to a cofinal branch up to $\alpha$. Then every node at level $\alpha$, and every later node, lies above a member of $B$. This is antichain sealing by diamond.

Here is a precise way to handle the coding. Give the countable level $\xi$ node codes in $[\omega \xi ,\omega \xi +\omega )$. There is a club set of countable limit $\alpha$ with $\omega \alpha =\alpha$, and on this club the nodes below $\alpha$ have exactly the relevant codes below $\alpha$. Empty unused codes are ignored. Thus any subset of the entire tree has an ordinal code set to which the stationary diamond principle applies.

Let $B$ now be any maximal tree antichain of the completed tree. There is a club set of $\alpha$ such that $B\cap (T\upharpoonright \alpha )$ is maximal in $T\upharpoonright \alpha$. Indeed, choose a comparable member of $B$ for each node; closure under the heights of these witnesses gives that club. Intersect it with the coding club. Stationary correct guessing supplies an $\alpha \in S$ on this intersection at which $B$ is sealed. A member of $B$ at or above level $\alpha$ would extend a member of $B$ below $\alpha$, contradicting the tree antichain property. So $B$ is contained in the countable portion below $\alpha$. Every tree antichain extends to a maximal one, hence every tree antichain is countable.

There is no cofinal branch of length ${\omega }_1$. Otherwise, choosing at each successor level the other successor of its branch node would give an uncountable tree antichain. Thus the resulting tree is a Suslin tree. A special Aronszajn tree is a union of countably many tree antichains; here those would all be countable and could not cover the ${\aleph }_1$ nodes. Therefore

The diamond theorem in the constructible universe gives $L\models ♢$, and $L$ satisfies ZFC. Apply the preceding construction inside $L$ with $S={\omega }_1^L$. It produces a normal splitting Suslin tree.

For completeness, such a tree yields a Suslin line. Order its nodes lexicographically using the two successors at every split, treating a node itself as a position between its two successor subtrees. Each node is thus a cut point between a left and a right subtree. This gives a dense linear order without endpoints. Every nonempty interval contains a whole cone above some node: for comparable endpoints use the successor cone of the descendant endpoint directed toward the other endpoint; for incomparable endpoints use the right-successor cone of the lower endpoint. Disjoint intervals therefore supply pairwise incomparable cone roots, so the order has the countable chain condition for a linear order. A countable collection of nodes has bounded heights; a cone based above that bound contains none of them, so it is not an order-dense subset. Passing to the Dedekind completion using proper cuts, so that no endpoints are added, preserves density, the countable chain condition for a linear order, and nonseparability. For nonseparability, a countable dense set in the completion would give a countable dense set of original nodes by choosing one original node between each distinct pair of its points. This contradicts the preceding height-bound argument. The result is a Suslin line.

Thus the Suslin hypothesis fails in $L$. The constructible universe theorem is a theorem of ZFC, so this is a relative-consistency argument, without an additional assumption that a transitive model exists:

Choose ${\aleph }_2$ distinct cofinal branches of the Kurepa tree and regard them as subsets of its underlying set ${\omega }_1$. Let $\mathcal{F}$ be this family. Fix an ordinal $\alpha <{\omega }_1$. The countably many node codes below $\alpha$ have heights bounded by some $\delta <{\omega }_1$. A cofinal branch meets level $\delta$, and its node there determines all its predecessors, hence all its nodes with codes below $\alpha$. Since level $\delta$ is countable,This is the Kurepa-family hypothesis. The given compatibility between the ordinal codes and the tree order ensures in particular that the branches are consistently viewed as subsets of ${\omega }_1$; boundedness of the heights of the countably many codes is what the argument uses.

Start with the set-theoretic tree of initial characteristic functionsordered by extension. The Kurepa-family hypothesis makes each level countable, because its members are in one-to-one correspondence with the distinct traces $X\cap \alpha$. Every node extends to every higher level using an $X$ that witnesses it. Nodes at limit levels are uniquely determined by their predecessors, and the distinct $X$ give ${\aleph }_2$ distinct cofinal branches.

We can also ensure the splitting requirement in the definition of a normal set-theoretic tree. Keep only nodes through which ${\aleph }_2$ of these branches pass. There are ${\aleph }_1$ nodes in total. For a discarded node, at most ${\aleph }_1$ of the selected branches pass through it, so at most ${\aleph }_1$ branches meet any discarded node. Remove those branches; ${\aleph }_2$ branches remain, and each retained node still has ${\aleph }_2$ remaining branches through it. It therefore has two different retained extensions at some later level, and has a retained extension at every higher level.

Choose increasing countable levels $⟨{\delta }_{\xi }:\xi <{\omega }_1⟩$, starting at $0$, continuously at limits, so that all nodes at level ${\delta }_{\xi }$ split before level ${\delta }_{\xi +1}$. This is possible because each selected level is countable. Restrict to these levels and relabel them by $\xi$. The retained tree now has one root, extensions at every higher level, at least two immediate successors, and unique limits of predecessor chains. Distinct remaining branches stay distinct on this unbounded set of levels. Hence the resulting normal set-theoretic tree is a Kurepa tree, with