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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.

For a coloring of $(r+1)$-subsets of an ordinal, route a vertex through the least earlier vertices matching its colors on every $r$-subset of previously chosen ancestors. The path stops when the vertex itself is selected. Common path prefixes agree, yielding a set-theoretic tree; a node at level $\alpha$ is determined by its color table on $[\alpha {]}^r$. Along a branch, the color of an $(r+1)$-tuple is independent of its last entry. If the domain has size $(2^{\mu }{)}^{+}$ and $\lambda \le \mu$, levels below ${\mu }^{+}$ have size at most $2^{\mu }$; counting all those levels forces a branch of length ${\mu }^{+}$. If the domain is a regular strong-limit cardinal $\kappa$, those levels instead have size below $\kappa$, and the tree property supplies a cofinal branch when needed.

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 an infinite cardinal $\rho$ put ${\beth }_0(\rho )=\rho$ and ${\beth }_{n+1}(\rho )=2^{{\beth }_n(\rho )}$. The Erdős-Rado theorem for finite arities isThe partition relation requires a homogeneous subset of order type ${\rho }^{+}$ for every coloring of $(n+1)$-element subsets by at most $\rho$ colors. In particular it gives uncountable homogeneous sets. The case $n=0$ is the infinite pigeonhole principle: a union of $\rho$ sets each of size at most $\rho$ cannot have size ${\rho }^{+}$.

We give the induction step by a closed elementary-submodel construction of an end-homogeneous sequence. Let $\mu$ be infinite, $\theta =(2^{\mu }{)}^{+}$, and $c:[\theta {]}^{r+1}\to \rho$ with $\rho \le \mu$. Choose a sufficiently large regular $\chi$ and an elementary submodel $M\prec H_{\chi }$ with $c,\theta \in M$, all ordinals below ${\mu }^{+}$ in $M$, $|M|=2^{\mu }$, and closure under externally given sequences of length at most $\mu$. To construct it, take ${\mu }^{+}$ successive elementary Skolem-hull closures, at each step including all such sequences from the preceding stage. Cardinality stays $2^{\mu }$ because $(2^{\mu }{)}^{\mu }=2^{\mu }$. In the final union every sequence of length at most $\mu$ has all its entries in some earlier stage, by regularity of ${\mu }^{+}$, and so was included at the next stage.

Put $\beta =\sup (M\cap \theta )$. Regularity of $\theta$ gives $\beta <\theta$, and $\beta \notin M$ since otherwise its successor would contradict the definition of the supremum. Recursively, for $\alpha <{\mu }^{+}$ choose $x_{\alpha }\in M\cap \theta$ above all earlier $x_{\xi }$ and satisfyingThere are at most $\mu$ constraints. Their parameters and their colors belong to $M$, and closure puts their complete code in $M$. The point $\beta$ witnesses their simultaneous satisfiability in $H_{\chi }$; elementarity supplies a witness in $M$. This gives an increasing end-homogeneous sequence of length ${\mu }^{+}$: on it, the color of an $(r+1)$-tuple depends only on the first $r$ entries.

For the induction from arity $n$ to $n+1$, take $\mu ={\beth }_{n-1}(\rho )$ and $r=n$. Color the $n$-tuples of the resulting sequence by their common color with a later point, equivalently by their color with $\beta$. Its index order has type ${\mu }^{+}={\beth }_{n-1}(\rho {)}^{+}$, so the induction hypothesis supplies a homogeneous index set of type ${\rho }^{+}$. The original $(n+1)$-tuples on those indices have that same color. This completes the proof without substituting the theorem's name for its induction.

A useful ordinal partition-bound function isIt is total: choose an infinite cardinal $\rho \ge |\alpha |$ with ${\rho }^{+}>\alpha$; the pairs case supplies $(2^{\rho }{)}^{+}$ with the required homogeneous order type for all these color numbers. It is nondecreasing and satisfies $f(\alpha )\ge \alpha$. Start iterating $f$ and take a supremum $\kappa$ at a limit stage. If the increasing iterates do not attain that supremum, then for every $\alpha <\kappa$ some iterate $a_{\xi }$ satisfies $\alpha \le a_{\xi }<\kappa$, andIf the iterates have already stabilized, their supremum is a fixed point directly. Thus the relevant nontrivial case is an uncountable cardinal $\kappa$ closed below itself under $f$.

Such closure implies that $\kappa$ is a strong limit. Suppose $\lambda <\kappa$ is infinite and $\kappa \le 2^{\lambda }$. Take distinct binary strings of length $\lambda$, indexed in order by $\kappa$, and color a pair by its first differing coordinate. No three strings are homogeneous: three bits cannot be pairwise different at the same coordinate. Closure would give $f(\lambda +1)<\kappa$, and its restriction would have a homogeneous subset of order type $\lambda +1$ in $\lambda$ colors, a contradiction. Hence $2^{\lambda }<\kappa$; finite exponents are harmless for an uncountable $\kappa$.

Use the usual tree property: every set-theoretic tree of height $\kappa$ with nonempty levels of cardinality below $\kappa$ has a cofinal branch. For an uncountable cardinal satisfying this all-set-theoretic trees convention, $\kappa$ must be regular. If it were singular with cofinality $\tau <\kappa$, attach to one root $\tau$ chains whose lengths increase cofinally to $\kappa$. Every level has at most $\tau$ nodes, the height is $\kappa$, and no branch is cofinal. This contradiction proves regularity. Combined with the preceding closure argument it proves that the cardinal under discussion is strongly inaccessible.

For any $c:[\kappa {]}^2\to \nu$, $0<\nu <\kappa$, make a separated set-theoretic tree of functions. Its level $\alpha$ consists of the functions $t:\alpha \to \nu$ realized as $t(\xi )=c(\{\xi ,\beta \})$ for some $\beta \ge \alpha$, with $\beta <\kappa$. Order these functions by restriction. Levels are nonempty and have size at most ${\nu }^{|\alpha |}<\kappa$, using the strong-limit property. The tree property gives a branch $b:\kappa \to \nu$. Recursively select $x_{\eta }$ for $\eta <\kappa$, larger than all earlier selections, whose coloring function agrees with $b$ through a level above those earlier selections; regularity allows that level to remain below $\kappa$, and its node has a realizing point. Then $c(\{x_{\xi },x_{\eta }\})=b(x_{\xi })$ for $\xi <\eta$. One of the fewer-than-$\kappa$ colors occurs on $\kappa$ selected points, by regularity. These points are homogeneous, proving $\kappa \to (\kappa {)}_{\nu }^2$ for every $\nu <\kappa$. ConsequentlyA strictly increasing countable iteration has a supremum of countable cofinality, so it cannot meet this regular set-theoretic tree-property condition; no continuity of $f$ has been assumed.

The final printed assertion needs its convention specified. The usual tree property alone does not imply strong inaccessibility; it can consistently hold at successor cardinals. In the argument just given, strong inaccessibility follows from function closure together with that set-theoretic tree property. There is also a stronger branching form of the tree property for which the assertion holds literally: require a cofinal branch in every separated rooted height-$\kappa$ set-theoretic tree whose nodes each have fewer than $\kappa$ immediate successors, without restricting level sizes. The singular-chain example proves regularity under this convention too. If $2^{\lambda }\ge \kappa$ for $\lambda <\kappa$, take the full binary set-theoretic tree through level $\lambda$, choose $\kappa$ distinct nodes on that level, and attach to its $\xi$th chosen node a chain of length $\xi$, for $\xi <\kappa$. It has height $\kappa$ and at most two successors per node, but every branch has length below $\kappa$. Thus this branching property implies the strong-limit condition and hence strong inaccessibility. At a strongly inaccessible cardinal it agrees, for separated set-theoretic trees, with the usual level-size convention: regularity bounds the union of fewer than $\kappa$ successor sets, and strong-limit arithmetic bounds the possible predecessor sequences at limit levels.

Represent the arena by a pruned set-theoretic tree $T$ of legal finite positions. Players I and II alternate; every position has a legal continuation. The usual terminal-position convention can also be handled by adding terminal wins and losses at the initial stage. Give the branches the product topology generated by finite-position cylinders, and let $A$ be I's winning open set.

Define the winning-position attractor in an infinite game by transfinite recursion. Start withAt a successor stage retain $W_{\alpha }$ and add I-positions with at least one child in $W_{\alpha }$, and II-positions with every child in $W_{\alpha }$. At a limit stage take the union. The increasing sequence stabilizes because the positions form a set: strict increases at arbitrarily many stages would select too many distinct positions. Write $W$ for the stable set. The rank of a position in $W$ is its first entry stage; any positive such rank is a successor ordinal.

If the initial position belongs to $W$, at every I-position of positive rank choose a child that was present at the preceding stage. Every move by II also leads to a child of smaller rank, by the rule for II-positions. There is no infinite strictly decreasing sequence of ordinals, so every play following this strategy in an infinite game reaches $W_0$ after finitely many moves. Its complete branch then belongs to $A$. Hence this is a winning strategy in an infinite game for I.

If the initial position is outside $W$, every move of I stays outside $W$. At a II-position outside $W$ at least one child is outside $W$: otherwise the entry ranks of its set of children have a common ordinal bound, and that position would enter at the next stage. Choose such a child. This gives II a strategy in an infinite game that never enters $W_0$. A resulting branch cannot belong to $A$, because $A$ is an open set and would therefore have a finite prefix whose whole cylinder lies in $A$, namely a member of $W_0$. Thus II wins. This proves the Gale-Stewart theorem for an arbitrary set-sized arena, in the ordinary metatheory with choice. Countability of the move set was never used.

The restriction on the payoff is essential. Under the axiom of choice, already for binary moves there is an undetermined game. There are $\mathfrak{c}=2^{{\aleph }_0}$ strategies for each player, since a strategy is a binary-valued map on a countable set of positions. List I's strategies ${\sigma }_{\alpha }$ and II's strategies ${\tau }_{\alpha }$ for $\alpha <\mathfrak{c}$. Each strategy is compatible with $\mathfrak{c}$ different plays, obtained by varying all the opponent's binary moves.

At stage $\alpha$, choose a fresh play $x_{\alpha }$ following ${\sigma }_{\alpha }$ and then a different fresh play $y_{\alpha }$ following ${\tau }_{\alpha }$, avoiding every previously chosen play. Fewer than $\mathfrak{c}$ plays have been used at that stage, even when $\mathfrak{c}$ is singular, so these choices are possible. Let $B=\{y_{\alpha }:\alpha <\mathfrak{c}\}$. No I-strategy wins $B$, because its selected $x_{\alpha }$ is outside $B$; no II-strategy wins, because its selected $y_{\alpha }$ is inside $B$. Future choices avoid all earlier $x_{\alpha }$, so they never spoil the first assertion. This is the choice diagonalization of an undetermined game. Consequently determinacy for all payoff sets is incompatible with the axiom of choice. The attractor still makes sense for a nonopen payoff, but an avoiding play may then enter the payoff without ever having a winning finite prefix.

The exponent-two Erdős-Rado theorem for pairs says that for every infinite cardinal $\lambda$,In particular, a finite or countable coloring of pairs from $(2^{{\aleph }_0}{)}^{+}$ has an uncountable monochromatic subset. We prove the general statement by an end-homogeneous tree construction.

Let $c:[\mu {]}^2\to \nu$ be any coloring on an ordinal $\mu$. Route a vertex $\beta$ through an increasing sequence of preceding vertices as follows. Start with the least vertex. After ancestors $a_{\eta }$, $\eta <\xi$, have been selected, choose the least remaining candidate $\gamma \le \beta$, above those ancestors, satisfyingThe terminal candidate $\beta$ always qualifies, so the procedure eventually reaches it. Candidate sets shrink and their least elements increase. If an intermediate candidate is $\gamma$, its route is precisely the earlier part of the route to $\beta$, because all earlier color tests agree. Consequently the routes define a set-theoretic tree on the vertices.

Along a branch, the color from an ancestor to every later branch vertex is constant. Also a node of height $\alpha$ is determined by its sequence of $\alpha$ colors to its ancestors: reconstruct the ancestors as successive least candidates with those prescribed colors, then reconstruct the node as the least final candidate. Thus its level has at most ${\nu }^{|\alpha |}$ nodes.

Apply this with $\mu =(2^{\lambda }{)}^{+}$ and $\nu =\lambda$. If every node had height less than ${\lambda }^{+}$, each level would have size at most ${\lambda }^{\lambda }\le 2^{\lambda }$, and there would be at most ${\lambda }^{+}$ such levels. Their total size would be at most $2^{\lambda }$, contradicting $|\mu |=(2^{\lambda }{)}^{+}$. Hence some route has at least ${\lambda }^{+}$ ancestors. For each ancestor mark its constant color to all later route points. Since there are only $\lambda$ colors, one occurs on ${\lambda }^{+}$ ancestors. Those ancestors form the required monochromatic set. This proves the theorem.

To make the ordinal function precise, one useful ordinal partition-bound function isThe target here is ordinal order type. For infinite $\alpha$, the proved theorem with $\lambda =|\alpha |$ supplies a bound $(2^{\lambda }{)}^{+}$: its homogeneous set has order type at least ${\lambda }^{+}>\alpha$, and it handles all the required color counts. Finite targets are handled by finite Ramsey theorem bounds. Thus $f$ is total, nondecreasing and $f(\alpha )\ge \alpha$.

Suppose a sequence of transfinite iterates ${\alpha }_{\xi +1}=f({\alpha }_{\xi })$, with suprema at limit stages, has supremum an uncountable cardinal $\kappa$ at a limit stage, with every earlier iterate below $\kappa$. Thenchoose an earlier iterate at least $\theta$ and use the next one as a bound. If the iteration has already stabilized, it has already found a fixed point instead.

This closure forces $\kappa$ to be a strong limit cardinal. For any infinite $\lambda <\kappa$, color pairs of distinct binary strings of length $\lambda$ by their first differing coordinate. This uses $\lambda$ colors and has no monochromatic triangle: three binary digits at one coordinate cannot be pairwise different. It gives a coloring on $2^{\lambda }$ with no homogeneous target of type $\lambda +1$. Since $\lambda$ colors are allowed when the target is the ordinal $\lambda +1$, we haveFinite exponents satisfy the strong-limit requirement automatically for uncountable $\kappa$.

The needed additional condition is the tree property: every tree of height $\kappa$, with nonempty levels of size less than $\kappa$, has a cofinal branch. In the present convention it forces $\kappa$ to be a regular cardinal. If $\kappa$ were singular, take a cofinal sequence of smaller cardinals of length $\mathrm{cf}(\kappa )$ and form the disjoint union of chains of those lengths. This tree has height and total size $\kappa$, levels of size at most $\mathrm{cf}(\kappa )<\kappa$, but no branch of length $\kappa$, a contradiction.

For any $\nu <\kappa$, apply the preceding color-routing tree to a coloring $[\kappa {]}^2\to \nu$. If a route already has length $\kappa$, use it. Otherwise its levels are indexed below $\kappa$ and each has sizefor infinite exponents, by the strong-limit property; finite cases satisfy the same needed bound directly. The tree has $\kappa$ nodes, so regularity forces its height to be $\kappa$. The tree property supplies a cofinal branch. The ancestor-color partition of that branch has one class of size $\kappa$, since $\nu <\kappa$ and $\kappa$ is regular. Thus $\kappa \to (\kappa {)}_{\nu }^2$ for every $\nu <\kappa$, andWe have proved that an uncountable function-closed cardinal with the tree property is regular and strong limit, hence strongly inaccessible, and that it is the requested fixed point.

The paper's final sentence, read as a claim about the tree property alone, is false. Already ${\aleph }_0$ has it by the König infinity lemma but is not uncountable; even with uncountability stipulated, there are relative-consistency models with the tree property at ${\aleph }_2$. The latter counterexample is documented in the research account of Mitchell forcing. The rigorous conclusion in the iterated-function context is the one just proved: function closure supplies strong limit, and the tree property supplies regularity and a branch. Strong limit and regularity follow from distinct hypotheses in this context.

An Aronszajn tree is a set-theoretic tree of height ${\omega }_1$, with countable levels and no uncountable chain in a partial order, equivalently no cofinal branch. The strict predecessors of a node are well-ordered, and its height is their order type. Such trees exist in ZFC, as the following rationally labelled Aronszajn tree construction shows.

We build countable nonempty levels $T_{\alpha }$ for all $\alpha <{\omega }_1$. A node $t\in T_{\alpha }$ is a strictly increasing function $t:\alpha \to \mathbb{Q}\cap (0,1)$, and the tree order is proper initial-segment extension. Every restriction $t\upharpoonright \beta$ must lie in $T_{\beta }$. Give nonempty $t$ the rational labeland give the empty root label $0$. We require labels to increase strictly along the tree. We also maintain the bounded rational extension property:Keeping room below every rational cap is what makes the limit step possible.

Start with $T_0=\{\varnothing \}$. At a successor stage, for every $s\in T_{\alpha }$ and every rational $r$ with $\ell (s)<r<1$, includein $T_{\alpha +1}$. Its label is $r$, strictly larger than its predecessor's. This level is countable. To extend an earlier node below a prescribed cap $q$, first extend to $T_{\alpha }$ below $q$ by the inductive property, if necessary, then choose a new rational label between that label and $q$. Thus the extension property is preserved.

Now let $\lambda <{\omega }_1$ be a nonzero limit. The whole tree below $\lambda$ is countable. For each pair $(s,q)$ with $s\in T_{\beta }$, $\beta <\lambda$, and $\ell (s)<q<1$, construct one node at height $\lambda$ extending $s$ below $q$. There are only countably many such pairs.

Choose a rational $r$ with $\ell (s)<r<q$ and an increasing sequenceSet $s_0=s$. Given $s_n\in T_{{\alpha }_n}$ with $\ell (s_n)<r$, setThis is rational and satisfies $\max \{\ell (s_n),r-1/(n+1)\}<r_n<r$. The successor rule supplies $u_n=s_n^{⌢}{r}_n$ at height ${\alpha }_n+1$. If ${\alpha }_{n+1}={\alpha }_n+1$, take $s_{n+1}=u_n$; otherwise use the already established extension property below $\lambda$ to extend $u_n$ to $s_{n+1}\in T_{{\alpha }_{n+1}}$ with label below $r$.

The union $t={\bigcup }_n{s}_n$ has domain $\lambda$ and extends $s$. It is strictly increasing, and every shorter restriction is already in its prescribed level. All its values are below $r$, while it contains the values $r_n>r-1/(n+1)$. HenceEvery predecessor's label is below the label of some $s_n$, and hence below $r$. Thus strict increase of labels survives at the limit. Include one such $t$ for each pair $(s,q)$ in $T_{\lambda }$, discarding duplicates. This gives a countable level and preserves every bounded extension requirement. In particular applying it to the root makes the level nonempty. The choices here are permitted by the ambient axiom of choice; transfinite recursion through the set ${\omega }_1$ completes the construction.

Put $T={\bigcup }_{\alpha <{\omega }_1}{T}_{\alpha }$. A node of domain $\alpha$ has exactly the restrictions of domains $\beta <\alpha$ as predecessors, so its height is $\alpha$. Therefore $T$ has height ${\omega }_1$ and countable levels. Since $\ell :T\to \mathbb{Q}$ is strictly increasing on comparable nodes, an uncountable chain would inject into the countable set $\mathbb{Q}$, which is impossible. Each label fiber is a tree antichain, so the tree is even a special Aronszajn tree:

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).

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.

Use the displayed family to form the coherent-injection Aronszajn tree. A node at level $\beta$ is a restriction $e_{\alpha }\upharpoonright \beta$ for some $\alpha \ge \beta$. Every such node differs only finitely from $e_{\beta }$. There are countably many finite subsets of the countable domain $\beta$ and countably many assignments of natural-number values to each, so there are only countably many possible finite modifications. Hence each level is countable. It is nonempty because it contains $e_{\beta }$.

Every shorter restriction of a node is again a node, and its predecessors have order type its domain ordinal. Thus this is a set-theoretic tree of height ${\omega }_1$. If it had an uncountable chain in a partial order, its domain heights would be unbounded in ${\omega }_1$, since the levels below any countable height contain only countably many nodes. The union of that chain in a partial order would be an injection ${\omega }_1\to \omega$, impossible. ThereforeThe countable-level proof uses coherence, whereas the no-branch proof uses injectivity; the two features play different roles.