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A set-theoretic tree of height ${\omega }_2$, levels of size at most ${\aleph }_1$, and no cofinal branch. The Continuum hypothesis supplies one through the minimal-walk tree.

A set-theoretic tree of height ${\omega }_1$ with countable levels and no cofinal branch. More generally, a $\kappa$-Aronszajn tree has height a regular uncountable $\kappa$, levels of size less than $\kappa$, and no cofinal branch.

A maximal chain with nodes at unbounded heights in a set-theoretic tree. Taking predecessor closure gives one node at each level below the tree height.

A subset of a forcing order whose distinct members have no common stronger extension. A maximal forcing antichain has a compatible member for every condition. For forcing by nodes of a set-theoretic tree, ordered by extension, this coincides with a tree antichain.

For a regular uncountable cardinal number $\kappa$, a set-theoretic tree of height $\kappa$ with nonempty levels of cardinality less than $\kappa$. The regular-height convention is important for the unbounded-extension kernel of a regular tree.

A set-theoretic tree of height ${\omega }_1$ with countable levels and at least ${\aleph }_2$ distinct cofinal branches.

Let $\kappa$ be strongly inaccessible cardinal in the ground model and perform the finite Lévy collapse to omega-one. The ground full binary set-theoretic tree of height $\kappa$ has all levels of size less than $\kappa$, so these levels become countable. Its at least $({\kappa }^{+}{)}^M$ ground branches remain distinct. The chain in a partial order condition preserves $({\kappa }^{+}{)}^M$, which is the extension ${\aleph }_2$. The unchanged ground set-theoretic tree therefore witnesses the Kurepa hypothesis.

A set-theoretic tree with one root, extensions of every node at every higher level, at least two immediate successors for every node, and distinct limit-level nodes distinguished by their predecessor chains. If splitting occurs only at later levels, a continuous cofinal selection of levels can enforce immediate splitting.

For a regular uncountable kappa-tree, keep exactly the nodes whose extensions have unbounded heights:This set is predecessor-closed. At any level $\alpha$, if no node survived, the extension heights above each of its fewer than $\kappa$ nodes would be bounded. Regularity would give a single bound for their union, contradicting the height of the original set-theoretic tree. Thus every level of $T^{\ast }$ is nonempty and still has size less than $\kappa$.

If $t\in T^{\ast }$ and $\beta >\mathrm{ht}(t)$, consider its extensions at level $\beta$. If none survived, fewer than $\kappa$ bounded extension sets would again bound every extension of $t$, a contradiction. Therefore a surviving level-$\beta$ extension exists. HenceThis unbounded-extension kernel of a regular tree uses regularity essentially. If one permits singular-height set-theoretic trees in the term “$\kappa$-tree”, the unrestricted assertion is false: take fewer than $\kappa$ disjoint branches with lengths cofinal in a singular $\kappa$. The levels are small and the height is $\kappa$, but no node has unbounded extensions. A common root can be added without creating a well-pruned subtree. The usual regular-height convention is therefore the one used here.

Prune an ${\aleph }_1$-Suslin tree as in part (a), then use its nodes as forcing conditions, with extensions stronger. Two conditions are compatible exactly when comparable, so the absence of uncountable tree antichains is the forcing countable chain condition for forcing. For each $\alpha <{\omega }_1$, the set $D_{\alpha }$ of nodes of height at least $\alpha$ is dense, by well-pruned set-theoretic tree.

If $2^{{\aleph }_0}>{\aleph }_1$, full Martin's axiom includes ${\mathrm{MA}}_{{\aleph }_1}$. It would provide a filter in an ordered set meeting all these dense subsets of a forcing order. Directedness makes that filter in an ordered set a chain in a partial order, and meeting every $D_{\alpha }$ makes its heights unbounded, producing an uncountable branch. This contradicts the Suslin-tree property. A Suslin set-theoretic tree together with failure of Continuum hypothesis therefore implies failure of Martin's axiom. This is the Suslin-tree obstruction to Martin's axiom.

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.

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.

In $M$, take $T={\bigcup }_{\alpha <\kappa }{}^{\alpha }2$ ordered by extension. Strong inaccessibility gives $|T_{\alpha }|=2^{|\alpha |}<\kappa$. In $M[H]$, each of these ground levels is therefore countable, while the height $\kappa$ is ${\omega }_1$. Thus the unchanged ground set-theoretic tree is an ${\aleph }_1$-tree in the extension.

Every ground binary function of length $\kappa$ still yields a distinct cofinal branch through this set-theoretic tree. There are $(2^{\kappa }{)}^M\ge ({\kappa }^{+}{)}^M$ such branches. The $\kappa$-chain condition preserves cardinals at and above $\kappa$, so $({\kappa }^{+}{)}^M={\aleph }_2^{M[H]}$. The ground branch family still has at least that cardinality. ConsequentlyThis Kurepa tree from an inaccessible binary tree uses ground-model levels and branches; it does not claim the full binary set-theoretic tree newly computed in the extension has countable levels.