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A set-theoretic tree is a partial order for which the strict predecessors of every node are well-ordered by the tree order. The height of a node is that predecessor order type, and $T_{\alpha }$ is the set of nodes of height $\alpha$. Under the ordinary height-and-width definition, a kappa-tree satisfies

For an arbitrary cardinal in this definition one must distinguish it from additional conventions such as being well-pruned, normal or splitting. Splitting means that every node has two incompatible extensions; it is not implied by the height-and-width clauses. This distinction is material in Question 5(ii)(a).

A well-pruned set-theoretic tree of height $\kappa$ has the property that every node $t$ extends to every higher level below $\kappa$: if $\mathrm{ht}(t)<\beta <\kappa$, there is $u$ of height $\beta$ with $t{<}_{T}u$. Equivalently, the heights of extensions of every node are unbounded in $\kappa$, since taking predecessors then gives an extension at any prescribed intermediate level. This is stronger than merely having no terminal nodes. For a kappa-tree we use the usual regular uncountable height cardinal and levels of size less than that cardinal.

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.

A tree is splitting if every node has two incompatible extensions. This property is additional to height and level-size bounds in a kappa-tree. A cofinal branch through a splitting tree of regular uncountable height produces an equally large antichain by taking off-branch extensions and proceeding beyond their divergence heights. Thus a splitting ${\omega }_1$-tree with countable antichains has no uncountable chains; the nonsplitting chain ${\omega }_1$ itself is a counterexample to omitting splitting.

For a kappa-tree, retain nodes whose extension heights are unbounded in $\kappa$. The retained nodes are predecessor-closed. Each level remains nonempty: otherwise regularity would bound the union of fewer than $\kappa$ bounded extension sets. The same argument above a retained node ensures retained extensions at every later level. Singular height invalidates this proof and the unrestricted conclusion.