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Given coherent coinfinite injections into omega, use all restrictions $e_{\alpha }\upharpoonright \beta$, where $\beta \le \alpha <{\omega }_1$, ordered by proper extension. Every level is countable because its nodes differ finitely from a fixed function on a countable domain. An uncountable chain in a partial order would have unbounded domain heights, and its union would inject ${\omega }_1$ into $\omega$, which is impossible.

With $\kappa$ an uncountable regular cardinal, finite partial functions on $\kappa \times \omega$ assigning a value below the first coordinate collapse every ground ordinal below $\kappa$ to countable size. The Delta-system lemma gives the $\kappa$ chain in a partial order condition; the possible-values lemma for chain-condition forcing then preserves the regularity of $\kappa$. Therefore $\kappa$ becomes the extension ${\omega }_1$. Strong inaccessibility is enough but is not needed for this identification.

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.

For a regular uncountable $\kappa$, the finite-condition collapse has the $\kappa$ chain in a partial order condition, by a regular-cardinal Delta-system lemma and thinning to identical assignments on the finite root. Thus every maximal forcing antichain has ground-model size less than $\kappa$. There is no fixed size: for any nonzero cardinal $\mu <\kappa$, assigning all values below $\mu$ at the one coordinate $(\mu ,0)$ is a maximal forcing antichain of size $\mu$.

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.

A tree with unique limits has no distinct nodes at a limit level with the same history. Precisely, for a limit ordinal $\delta$ and $s,t\in T_{\delta }$,Here $s\upharpoonright \beta$ denotes the unique predecessor of $s$ of height $\beta$. This is uniqueness of a limit node when it exists, not a requirement that every cofinal chain in a partial order below a limit level acquire a limit node.

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.

Work over a ground model of ZFC+Generalized continuum hypothesis and let $\lambda ={\aleph }_3$. Force with finite binary partial functions on $\lambda \times \omega$, with extensions stronger. The Delta-system lemma thins any uncountable family of finite domains to an uncountable family with one common root. Only finitely many binary assignments on that root occur, so two conditions agree there and their union is a common extension. Hence the forcing has the countable chain condition for forcing and preserves cardinals and cofinalities.

The generic union yields $\lambda$ distinct reals. Totality at each coordinate is dense, and for two different coordinates it is dense to assign different values at some unused natural-number position. Thus the extension satisfies $2^{{\aleph }_0}\ge \lambda$.

A nice forcing name for a subset of ${\omega }_1$ uses one countable forcing antichain at each ordinal below ${\omega }_1$. Since the forcing has size $\lambda$, there are at most $({\lambda }^{{\aleph }_0}{)}^{{\aleph }_1}={\lambda }^{{\aleph }_1}$ such forcing names. Ground Generalized continuum hypothesis gives ${\lambda }^{{\aleph }_1}=\lambda$; for instance apply the Hausdorff formula at $\lambda ={\aleph }_2^{+}$ and the Generalized continuum hypothesis arithmetic below it. Therefore $2^{{\aleph }_1}\le \lambda$ in the extension. Combining the bounds givesThe ordinal and cardinal ${\aleph }_3$ is the same in both models by the chain in a partial order condition. The forcing theorem formalizes this construction as the requested relative-consistency implication. A countable transitive ground model is a convenient presentation, not an additional consequence silently derived from mere consistency. This is the Cohen forcing two-level continuum plateau.

Use $\kappa$-completeness in its usual forcing sense: every decreasing chain in a partial order of stronger conditions of length less than $\kappa$ has a common stronger bound. Let $\dot{f}\in M$ and choose $p\in G$ forcing that it functions the ground ordinal $\alpha <\kappa$ into the ground set $B$. Below any stronger condition $r$, recursively decide each value of $\dot{f}$ in order. At a successor step the deciding conditions are dense, and at a limit stage use $\kappa$-completeness. After all $\alpha$ steps take another common bound. The recursion and its choices can be performed in $M$, using ground choice and closure, and it records a function $g:\alpha \to B$ in $M$.

Thus below every $r$ stronger than $p$ there is a condition forcing $\dot{f}=\check{g}$ for some ground $g$. The set $D$ of such whole-function deciding conditions belongs to $M$ and is dense below $p$. Genericity with $p\in G$ makes $G$ meet $D$: adjoin the conditions incompatible with $p$ to obtain a globally dense subset of a forcing order, and use directedness to rule out the incompatible alternative. A condition in $G\cap D$ then gives $f=g\in M$.

The reverse inclusion follows because ground functions remain functions with the same domain and values. ThereforeThis closed forcing adds no short ground-valued sequences argument needs density of complete decisions. A single arbitrarily constructed lower bound need not belong to $G$, and would not by itself prove the claim.

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.

Zorn lemma states that a nonempty partially ordered set in which every chain in a partial order has an upper bound has a maximal element. Here is a proof from the axiom of choice and Hartogs theorem.

Suppose there is no maximal element. Every chain $C$ has a strict upper bound: choose an upper bound $b$ and then an element strictly greater than $b$. The family of sets of strict upper bounds is a set of nonempty subsets of the underlying set $P$. The axiom of choice is used here to choose, simultaneously for every chain $C$, one strict upper bound $b(C)$.

By Hartogs theorem there is an ordinal $\kappa$ admitting no injection into $P$. Transfinite recursion defines $x_{\alpha }$ for $\alpha <\kappa$ byAt each stage the previous elements form a chain, and $x_{\alpha }$ is strictly above every one of them. Hence $\alpha \mapsto x_{\alpha }$ is an injection $\kappa \to P$, a contradiction. Therefore a maximal element exists.