Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 19 / 5 / ii

Under the ordinary height-and-width definition of an ${\aleph }_1$-tree, the printed claim is false. The chain $({\omega }_1,\le )$ has one node at each level, every antichain has size at most one, and it is itself an uncountable chain.

Here is the intended argument under the additional splitting convention. Any uncountable chain in a tree with countable levels is unbounded in height, because a bounded set of levels below ${\omega }_1$ has only countably many nodes. Its predecessor closure therefore gives a cofinal branch $B$. At each node on $B$, splitting supplies an extension off $B$, incompatible with a later node on $B$. Recursively for $\xi <{\omega }_1$, choose such an off-branch node $s_{\xi }$, then move sufficiently far along $B$ that all subsequent choices are above a branch node incompatible with $s_{\xi }$. At a limit stage the previous countably many heights are bounded below ${\omega }_1$, so the recursion continues. The $s_{\xi }$ are pairwise incompatible, an uncountable antichain.

Thus a splitting ${\aleph }_1$-tree with only countable antichains has no uncountable chains. A splitting convention must be stated; the raw tree hypothesis alone does not suffice.

Use the equivalent tree formulation of the Suslin hypothesis: there is no normal, well-pruned Suslin tree of height ${\omega }_1$, with countable levels and no uncountable chains or antichains. This is equivalent to the linear-order formulation that every complete dense order without endpoints satisfying the countable chain condition for a linear order is separable.

If such a tree $T$ existed, use its nodes as forcing conditions, with a higher extension stronger. It is CCC because its antichains are countable. For every $\alpha <{\omega }_1$, the set $D_{\alpha }=\{t:\mathrm{ht}(t)\ge \alpha \}$ is dense because the tree is well-pruned. The assertion ${\mathrm{MA}}_{{\aleph }_1}$ is precisely that a CCC forcing and a family of at most ${\aleph }_1$ dense sets admit a filter meeting all of them.

Apply it to these dense sets. A directed filter in a tree is a chain: two compatible nodes are comparable, since both lie among the well-ordered predecessors of a common extension. Meeting every $D_{\alpha }$ makes this chain cofinal, contrary to the defining absence of uncountable chains in a Suslin tree. Consequently