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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.

If $T$ is a normal splitting Suslin tree and $P$ is Knaster, then $P\times T$ is CCC, so $P$ forces that the ground tree is still CCC Countable levels and splitting persist. A new cofinal branch would give an uncountable antichain by splitting, so none is added. This permits adding many Cohen reals while retaining a Suslin tree, and distinguishes Knaster preservation from an unjustified preservation claim for every CCC forcing.

There exists an ${\omega }_1$-tree with countable levels and at least ${\aleph }_2$ distinct cofinal branches.

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

The tree of restrictions ${\rho }_{\beta }\upharpoonright \alpha$, ordered by extension. For a club sequence on ${\omega }_2$ with club order types at most ${\omega }_1$, the Continuum hypothesis bounds each level by ${\aleph }_1$. Trace injectivity and Fodor lemma exclude a cofinal branch.

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:

There is a cofinal branch. We give a proof that does not require distinct limit-level nodes to have different predecessor chains.

The set $S=\{\alpha <\lambda :\mathrm{cf}(\alpha )=\kappa \}$ is stationary. Indeed a strictly increasing continuous $\kappa$-sequence in any club set has supremum in that club set, below $\lambda$, of cofinality $\kappa$. At each $\alpha \in S$, there are fewer than $\kappa$ pairs of nodes on $T_{\alpha }$. For every pair whose predecessor chains below $\alpha$ differ, choose a height where they differ. Since $\mathrm{cf}(\alpha )=\kappa$, all these heights are bounded by some ${\beta }_{\alpha }<\alpha$. Consequently nodes in $T_{\alpha }$ having the same predecessor at ${\beta }_{\alpha }$ have identical predecessor chains below $\alpha$.

The Fodor lemma states that a regressive function on a stationary subset of a regular uncountable cardinal is constant on a stationary subset. Applying it here, ${\beta }_{\alpha }$ has a constant value $\beta$ on a stationary subset $S^{\prime }$. Choose $t_{\alpha }\in T_{\alpha }$ for each $\alpha \in S^{\prime }$. The level $T_{\beta }$ has fewer than $\kappa <\lambda$ nodes. Partitioning $S^{\prime }$ according to the predecessor of $t_{\alpha }$ at height $\beta$, one fiber $S^{\prime \prime }$ is stationary, since the union of fewer than $\lambda$ nonstationary sets is nonstationary. Let its common predecessor be $t$.

For $\alpha <{\alpha }^{\prime }$ in $S^{\prime \prime }$, the predecessor of $t_{{\alpha }^{\prime }}$ at level $\alpha$ and $t_{\alpha }$ have the same predecessor $t$ at level $\beta$. Their chains below $\alpha$ therefore agree. For each $\nu <\lambda$, choose $\alpha \in S^{\prime \prime }$ above $\nu$ and let $b_{\nu }$ be the predecessor of $t_{\alpha }$ at level $\nu$. The preceding comparison makes $b_{\nu }$ independent of that choice. The nodes $b_{\nu }$ form a chain through every level:This proves the uniformly narrow regular-height tree branch theorem. The uniform bound below the smaller regular $\kappa$ is stronger than merely bounding each level below $\lambda$.

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.

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.

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.

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.

A Kurepa tree is a set-theoretic tree of height ${\omega }_1$, with every level countable, possessing at least ${\aleph }_2$ distinct cofinal branches. The predecessors of each node are well ordered; their order type is the node's height. A cofinal branch is a maximal chain with nodes at unbounded heights below ${\omega }_1$, equivalently one node at every level after taking its predecessor closure.

Take $\kappa ={\omega }_1$ and choose a club sequence on ${\omega }_2$, each $C_{\delta }$ of order type at most ${\omega }_1$. For a successor $\delta$ use its predecessor as a singleton; at a limit use a cofinal sequence of minimal length. Form the minimal-walk treeordered by proper extension. Its height is ${\omega }_2$.

For $\xi <\delta$, the initial segment $C_{\delta }\cap \xi$ has order type strictly below ${\omega }_1$, and is countable. The strict inequality follows because a point of $C_{\delta }$ at or above $\xi$ occurs later in its enumeration. Hence every entry of every trace is countable. Under the Continuum hypothesis, for $|\alpha |\le {\aleph }_1$,There are at most ${\aleph }_1$ finite sequences of such sets. The trace coherence lemma for minimal walks says that, for $\beta >\alpha$, the value ${\rho }_{\beta }(\alpha )$ determines ${\rho }_{\beta }\upharpoonright \alpha$. The case $\beta =\alpha$ adds at most one node. Thus $|T_{\alpha }|\le {\aleph }_1$ for every level.

Suppose that $T$ had a cofinal branch, and take the union of its functions, $f$, with domain ${\omega }_2$. Every ${\rho }_{\beta }$ is injective by the proper-initial-segment argument, so $f$ is injective too. On the stationary setthis set is stationary because the supremum of a strictly increasing ${\omega }_1$-sequence from any club set has cofinality ${\omega }_1$ and lies in that club. The union of the finitely many countable entries of $f(\alpha )$ is bounded below $\alpha$. Assign a strict upper bound below $\alpha$ to obtain a regressive function. By Fodor lemma, there is a stationary $S^{\prime }\subseteq S$ and a single $\eta <{\omega }_2$ such that every entry of $f(\alpha )$ lies inside $\eta$ for $\alpha \in S^{\prime }$. There are at most ${\aleph }_1$ such finite sequences by the same cardinal arithmetic, but $|S^{\prime }|={\aleph }_2$, contradicting injectivity.

Therefore $T$ is an aleph-two Aronszajn tree:

Let $⟨D_{\alpha }:\alpha \in S⟩$ witness the stationary diamond principle: for every $A\subseteq {\omega }_1$, the set of $\alpha \in S$ with $D_{\alpha }=A\cap \alpha$ is stationary. We construct a normal splitting Suslin tree; this will be a nonspecial Aronszajn tree.

Construct its levels by recursion. Start with one root, and give every node two immediate successors. At a countable limit $\alpha$, the constructed portion $T\upharpoonright \alpha$ is countable. Through each of its nodes choose a cofinal branch of that portion, and put one new node above each chosen branch at level $\alpha$. This keeps the level countable and gives every earlier node an extension. Branches are identified by their predecessor chains, so nodes at a limit level are uniquely determined by their predecessors.

At a limit $\alpha \in S$, decode $D_{\alpha }$ as a candidate tree antichain $B$ of $T\upharpoonright \alpha$. If it is maximal, choose the branches just described to meet $B$. This is possible: for any node, maximality supplies a comparable member of $B$, and normality of the already constructed portion extends the larger of those two nodes to a cofinal branch up to $\alpha$. Then every node at level $\alpha$, and every later node, lies above a member of $B$. This is antichain sealing by diamond.

Here is a precise way to handle the coding. Give the countable level $\xi$ node codes in $[\omega \xi ,\omega \xi +\omega )$. There is a club set of countable limit $\alpha$ with $\omega \alpha =\alpha$, and on this club the nodes below $\alpha$ have exactly the relevant codes below $\alpha$. Empty unused codes are ignored. Thus any subset of the entire tree has an ordinal code set to which the stationary diamond principle applies.

Let $B$ now be any maximal tree antichain of the completed tree. There is a club set of $\alpha$ such that $B\cap (T\upharpoonright \alpha )$ is maximal in $T\upharpoonright \alpha$. Indeed, choose a comparable member of $B$ for each node; closure under the heights of these witnesses gives that club. Intersect it with the coding club. Stationary correct guessing supplies an $\alpha \in S$ on this intersection at which $B$ is sealed. A member of $B$ at or above level $\alpha$ would extend a member of $B$ below $\alpha$, contradicting the tree antichain property. So $B$ is contained in the countable portion below $\alpha$. Every tree antichain extends to a maximal one, hence every tree antichain is countable.

There is no cofinal branch of length ${\omega }_1$. Otherwise, choosing at each successor level the other successor of its branch node would give an uncountable tree antichain. Thus the resulting tree is a Suslin tree. A special Aronszajn tree is a union of countably many tree antichains; here those would all be countable and could not cover the ${\aleph }_1$ nodes. Therefore

Start with the set-theoretic tree of initial characteristic functionsordered by extension. The Kurepa-family hypothesis makes each level countable, because its members are in one-to-one correspondence with the distinct traces $X\cap \alpha$. Every node extends to every higher level using an $X$ that witnesses it. Nodes at limit levels are uniquely determined by their predecessors, and the distinct $X$ give ${\aleph }_2$ distinct cofinal branches.

We can also ensure the splitting requirement in the definition of a normal set-theoretic tree. Keep only nodes through which ${\aleph }_2$ of these branches pass. There are ${\aleph }_1$ nodes in total. For a discarded node, at most ${\aleph }_1$ of the selected branches pass through it, so at most ${\aleph }_1$ branches meet any discarded node. Remove those branches; ${\aleph }_2$ branches remain, and each retained node still has ${\aleph }_2$ remaining branches through it. It therefore has two different retained extensions at some later level, and has a retained extension at every higher level.

Choose increasing countable levels $⟨{\delta }_{\xi }:\xi <{\omega }_1⟩$, starting at $0$, continuously at limits, so that all nodes at level ${\delta }_{\xi }$ split before level ${\delta }_{\xi +1}$. This is possible because each selected level is countable. Restrict to these levels and relabel them by $\xi$. The retained tree now has one root, extensions at every higher level, at least two immediate successors, and unique limits of predecessor chains. Distinct remaining branches stay distinct on this unbounded set of levels. Hence the resulting normal set-theoretic tree is a Kurepa tree, with

Choose ${\aleph }_2$ distinct cofinal branches of the Kurepa tree and regard them as subsets of its underlying set ${\omega }_1$. Let $\mathcal{F}$ be this family. Fix an ordinal $\alpha <{\omega }_1$. The countably many node codes below $\alpha$ have heights bounded by some $\delta <{\omega }_1$. A cofinal branch meets level $\delta$, and its node there determines all its predecessors, hence all its nodes with codes below $\alpha$. Since level $\delta$ is countable,This is the Kurepa-family hypothesis. The given compatibility between the ordinal codes and the tree order ensures in particular that the branches are consistently viewed as subsets of ${\omega }_1$; boundedness of the heights of the countably many codes is what the argument uses.

The finite binary-function Cohen forcing $\mathrm{Fn}(\omega ,2,<\omega )$ is countable in $M$. It preserves ${\omega }_1$, and the countable levels, height, and normal extensions of the ground-model tree remain unchanged.

If the extension contained an uncountable tree antichain, apply the ground-model uncountable subset lemma for countable forcing to obtain an uncountable $X\in M$ contained in it. Incomparability in the fixed ground-model tree is absolute, so $M$ already regards $X$ as an uncountable tree antichain, contradicting that $T$ is Suslin in $M$. Similarly, a new cofinal branch has an uncountable ground-model subset. Comparability is absolute, so this would be an uncountable chain in the ground-model tree, again impossible.

Therefore countable forcing preserves Suslin trees, and in particular

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.

A well-pruned set-theoretic tree that is an Aronszajn tree and a Suslin tree gives a forcing with the countable chain condition for forcing: stronger nodes extend weaker ones. The ${\omega }_1$ dense subsets of a forcing order of nodes at or above each level cannot all be met by a filter in an ordered set, since that would produce a cofinal branch. Thus ${\mathrm{MA}}_{{\aleph }_1}$ fails. When the continuum exceeds ${\aleph }_1$, full Martin axiom includes this instance.