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

The valuation of a forcing name is defined recursively byThe recursion is on forcing name rank. Only pairs with an active condition in $H$ contribute elements, and their first coordinates are evaluated in the same filter. Thus a name is a ground-model set, while its value is a set in the generic extension.

A nice forcing name for a subset of a ground-model set $a$, often an ordinal, has the formwhere each $A_x$ is an antichain in $\mathbb{Q}$ and the entire construction belongs to $M$. Its value consists of the $x$ for which $H$ meets $A_x$. The antichains need not be maximal. Under the countable chain condition they are countable in $M$, which makes nice forcing names useful for counting possible subsets in extensions.

For a countable transitive ground model, the semantic forcing relation isThe forcing theorem identifies this with the recursively defined relation inside $M$ and makes it definable there. In particular, if $r\ge q$ is stronger, then $q⊩\phi$ implies $r⊩\phi$. The names are interpreted in each relevant generic filter, not replaced by their value in one fixed extension in the definition.

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

Start with a ground model containing a normal splitting Suslin tree $T$, and force with $\mathrm{Add}(\omega ,{\omega }_2)$, the finite partial functions from ${\omega }_2\times \omega$ to $2$. Such a starting model is relatively consistent with ZFC: the constructible universe has $◊$, and $◊$ implies the existence of a Suslin tree.

This Cohen forcing is Knaster. In an uncountable family of finite conditions, the delta-system lemma yields an uncountable family whose domains have one common root; thinning to common values on that finite root makes its members pairwise compatible. If $P$ is Knaster and $T$ is CCC, the product $P\times T$ is CCC: thin any uncountable family to pairwise compatible $P$ coordinates, then use CCC to find two compatible $T$ coordinates. Thus $P$ forces that the old $T$ is still CCC A name for an uncountable antichain would otherwise produce an uncountable antichain in this product by deciding its nodes.

Countable levels and the tree's splitting property persist, and the splitting argument from the previous part rules out a new cofinal branch in a CCC tree. Therefore the old $T$ is still Suslin. Meanwhile the forcing adds at least ${\aleph }_2$ distinct reals, with all cardinals preserved, so $2^{{\aleph }_0}\ge {\aleph }_2$.

We obtain a model with a Suslin tree and $\neg \mathrm{CH}$. Hence existence of a Suslin tree does not imply the Continuum hypothesis. This uses Knaster forcing preserves Suslin trees, not a claim that arbitrary CCC forcing preserves them.

For uncountable regular $\kappa$, let $\theta =\mathrm{otp}(C_{\zeta })$. Since $C_{\zeta }$ is a club set in $\zeta$, $\mathrm{cf}(\theta )=\mathrm{cf}(\zeta )=\kappa$, so $\theta \ge \kappa$.

If $\theta >\kappa$, its cofinality ensures $\theta >\kappa +\omega$: ordinals strictly between $\kappa$ and $\kappa +\omega$ are successors, and $\kappa +\omega$ has cofinality $\omega$. In the continuous increasing enumeration $c$ of $C_{\zeta }$, take $\gamma =c(\kappa +\omega )$. It is a limit point of $C_{\zeta }$ and has cofinality $\omega <\kappa$. Coherence gives $C_{\gamma }=C_{\zeta }\cap \gamma$, with order type $\kappa +\omega$ and size $\kappa$. This contradicts the size clause for ordinals of cofinality below $\kappa$. ThusThe printed hypotheses need this qualification. At $\kappa =\omega$, choose $C_{\zeta }=\zeta$ for every nonzero countable limit ordinal. These club sets are coherent, and the small-cofinality clause is vacuous. But at $\zeta ={\omega }^2$ their order type is ${\omega }^2$, not $\omega$. The standard square principle includes an order-type bound $\mathrm{otp}(C_{\zeta })\le \kappa$, which also repairs this countable case.

Fix $\delta <{\kappa }^{+}$ with $\mathrm{cf}(\delta )>\omega$. The limit points of $C_{\delta }$ form a club set in $\delta$. For each such point $\gamma$, coherence givesAs $\gamma$ increases through those limit points, these order types strictly increase. Hence the club set contains at most one member of a fixed fiber $E_{\xi }$. Removing an initial segment past that member leaves a club set disjoint from $E_{\xi }$. Therefore $E_{\xi }\cap \delta$ is not stationary. This proves each $E_{\xi }$ is non-reflecting, by the nonreflection of fixed order-type fibers.

For uncountable regular $\kappa$, part (a) gives type $\kappa$ at cofinality $\kappa$, and the size clause gives type less than $\kappa$ at smaller cofinalities. All nonzero limit ordinals below ${\kappa }^{+}$ fall into one of these cases. ThusThe index set $\kappa +1$ has cardinality $\kappa$, so this is a disjoint union of $\kappa$ non-reflecting subsets. Empty fibers can be omitted; if nonempty pieces are required, a fiber of size ${\kappa }^{+}$ can be split further, since nonreflection is inherited by subsets. Such a fiber exists because $\kappa$ sets of size at most $\kappa$ cannot cover ${\kappa }^{+}$ points.

At $\kappa =\omega$, the nonreflection definition is vacuous for every subset of ${\omega }_1$, since no smaller ordinal has uncountable cofinality. The partition conclusion is therefore still true by an arbitrary countable partition, although the particular fibers from the weak printed square assumptions need not cover all limit ordinals. Under the standard order-type-bounded square definition, the displayed fiber partition works in this case too.