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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.

The Beth numbers ${\beth }_0={\aleph }_0$, ${\beth }_{n+1}=2^{{\beth }_n}$ form a strictly increasing countable sequence by Cantor's theorem. Their supremum ${\beth }_{\omega }$ has cofinality $\omega$ because no finite subset is cofinal. It exceeds $2^{{\aleph }_0}={\beth }_1$. It is also a strong limit cardinal: if $\rho <{\beth }_{\omega }$, some $n$ has $\rho <{\beth }_n$, so $2^{\rho }\le {\beth }_{n+1}<{\beth }_{\omega }$. Thus the singular cardinal example needs no Continuum hypothesis.

Every function $f:\mathbb{R}\to [\mathbb{R}{]}^{<{\omega }_1}$ has reals $x,y$ with $x\notin f(y)$ and $y\notin f(x)$. The values are countable subsets. Requiring distinct witnesses is equivalent, by first adjoining $x$ to $f(x)$. The Freiling theorem identifies this axiom with the negation of the Continuum hypothesis in ZFC.

In ZFC, the Freiling axiom of symmetry is equivalent to failure of the Continuum hypothesis. Under the hypothesis, countable initial segments of a well-order of the reals violate symmetry. When the continuum exceeds ${\aleph }_1$, the union of the countable values on an ${\aleph }_1$-sized set leaves a real outside; avoiding that real's countable value gives the two witnesses.

An inner model $W$ of ZF containing all ambient subsets of $\omega$ computes the same countable ordinals as the ambient universe. Every ambient countably infinite ordinal has a well-order code coded by a subset of $\omega$, hence that code belongs to $W$; finite ordinals are already shared. The Mostowski collapse theorem inside $W$ recovers the actual ordinal and its countability witness. Conversely any countability witness in $W$ remains one outside. If $W$ also contains the full power set of $\omega$ as a set, its Continuum hypothesis would give an ambient bijection from ${\omega }_1$ onto that same power set. Thus an ambient failure of this well-orderable formulation of the Continuum hypothesis is preserved in such an inner model. No internal axiom of choice is required for this argument.

Under the Continuum hypothesis, enumerate the meagre $F_{\sigma }$ covers by ${\omega }_1$. At stage $\alpha$ choose a new real outside all earlier covers and the previous choices. The excluded union is meagre, so the Baire category theorem makes the choice possible. Each cover then contains only countably many selected points. Arbitrary meagre sets need not themselves have only continuum many possible choices; the coded covers are what is enumerated.

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.

In the usual arithmetic under the axiom of choice of cardinal numbers, take the singular strong-limit cardinalCantor's theorem makes this sequence strictly increasing. It is countable and cofinal in its supremum, so $\mathrm{cf}(\kappa )\le {\aleph }_0$. No finite subset is cofinal in this limit ordinal, hence $\mathrm{cf}(\kappa )={\aleph }_0$. ThereforeThis construction needs no assumption such as the Continuum hypothesis. The conventional use of cardinals here includes well-orderability of these power sets, supplied by the axiom of choice. It is a separate setting from the explicit failure of well-orderability hypothesized in question 8; one must not impose choice on that later hypothesis.

Take the Beth numberCantor's theorem makes this sequence strictly increasing. It is a countable cofinal sequence in $\kappa$, while no finite set is cofinal in the limit ordinal $\kappa$. Hence $\mathrm{cf}(\kappa )={\aleph }_0$. ConsequentlyThus Beth omega is a singular strong limit above the continuum. No Continuum hypothesis is needed for the example.

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.

Assume the Continuum hypothesis and enumerate $\mathbb{R}=\{r_{\alpha }:\alpha <{\omega }_1\}$. DefineEvery value is countable. For any two indices, say $\alpha \le \beta$, one has $r_{\alpha }\in f(r_{\beta })$, so the two required nonmembership conditions cannot both hold. Including the endpoint in each initial segment also rules out taking the two points equal. Therefore the free-pair assertion implies the negation of Continuum hypothesis. Cantor theorem and choice already give $2^{{\aleph }_0}\ge {\aleph }_1$, so

The Gödel constructible universe theorem gives $\mathrm{Con}(\mathrm{ZFC})\Rightarrow \mathrm{Con}(\mathrm{ZFC}+\mathrm{CH})$. The forcing independence theorem for the Continuum hypothesis gives $\mathrm{Con}(\mathrm{ZFC})\Rightarrow \mathrm{Con}(\mathrm{ZFC}+\neg \mathrm{CH})$. For example, the latter can be obtained by first passing to the constructible universe and then adding sufficiently many Cohen reals. By the equivalence in part (b), these are respectively models of the negation and affirmation of the free-pair assertion. Thus if ZFC is consistent, the assertion is independent of ZFC. The consistency qualification is essential: an inconsistent theory proves every sentence. These are syntactic relative-consistency implications, not a claim that bare consistency supplies a countable transitive model.

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.

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:

Use diamond-sequence forcing. A condition is a sequenceordered by end extension. Its countable descending chains have lower bounds: take their union and, if their lengths approach a new limit, add an arbitrary subset at that last index. Hence it is countably closed and preserves ${\omega }_1$. The union of the generic filter supplies $⟨A_{\xi }:\xi <{\omega }_1⟩$.

To prove the diamond principle, let a condition force that $\dot{X}\subseteq {\omega }_1$ and that $\dot{C}$ is a club set. Below any such condition build $p_n$ and strictly increasing countable ordinals ${\delta }_n$ so that ${\delta }_n>{\delta }_{p_n}$, $p_{n+1}$ forces ${\delta }_n\in \dot{C}$, decides $\dot{X}\cap {\delta }_n$, and has length past ${\delta }_n$. Unboundedness supplies ${\delta }_n$; countable closure allows deciding all the bits below it. The construction can be carried out in $M$.

Let $\delta ={\sup }_n{\delta }_n={\sup }_n{\delta }_{p_n}$. The union of the conditions has entries exactly below $\delta$ and decides a ground-model set $x=\dot{X}\cap \delta$. Extend that union by setting $A_{\delta }=x$. This condition forces $\delta \in \dot{C}$ by closure, and $A_{\delta }=\dot{X}\cap \delta$. The conditions giving a correct guess inside any named club set are therefore dense. ThusNo ground-model Continuum hypothesis is needed; this forcing is allowed to collapse higher cardinals.

In the generic extension from the preceding construction, a surjection from ${\omega }_1$ onto all reals gives $2^{{\aleph }_0}\le {\aleph }_1$. Cantor theorem gives the reverse inequality, so the Continuum hypothesis holds.

For the exact consistency-strength statement, existence of a countable transitive model is stronger than bare consistency of ZFC. One can formalize the forcing proof syntactically, or use the constructible universe theorem: ZFC proves that its constructible universe satisfies ZFC and the Generalized continuum hypothesis. If there is any model of ZFC, its internal constructible universe is therefore a model of ZFC with CH. By the Godel completeness theorem,This deduction does not silently replace consistency by existence of a transitive model.

The Freiling theorem givesFirst assume the Continuum hypothesis and fix a well-order $(r_{\alpha }{)}_{\alpha <{\omega }_1}$ of the real numbers. Define $f(r_{\alpha })=\{r_{\beta }:\beta \le \alpha \}$. Each value is countable. For any pair, one index is at most the other, so at least one of $x\in f(y)$ or $y\in f(x)$ holds. The same holds for $x=y$, since $x\in f(x)$. This violates the Freiling axiom of symmetry.

Conversely, assume $|\mathbb{R}|>{\aleph }_1$ and let $f$ assign a countable set to each real. Choose $X\subseteq \mathbb{R}$ of cardinality ${\aleph }_1$. By infinite cardinal arithmetic,Choose $y$ outside this union. Since $f(y)$ is countable and $X$ is uncountable, choose $x\in X\setminus f(y)$. Then $y\notin f(x)$ and $x\notin f(y)$, with $x\ne y$. This proves the Freiling axiom of symmetry and completes both implications.

The construction gives a Lusin set $\boxed{A=\{x_{\alpha }:\alpha <{\omega }_1\}}$ with all $x_{\alpha }$ distinct.

Every meagre set is contained in a meagre $F_{\sigma }$ set: replace each nowhere dense set in a countable covering by its closure. There are at most $2^{{\aleph }_0}$ such covers, since each closed subset of $\mathbb{R}$ is determined by a countable rational basis for its open complement, and a countable sequence of such codes is again coded by a real. Under the Continuum hypothesis, enumerate all meagre $F_{\sigma }$ sets as $(M_{\alpha }{)}_{\alpha <{\omega }_1}$, allowing repetitions.

By transfinite recursion, chooseFor each $\alpha <{\omega }_1$, the excluded union is a meagre set: it is a countable union of meagre sets and singletons. The Baire category theorem ensures that it does not cover $\mathbb{R}$, so the choice is possible. The resulting $A$ is uncountable. If $M$ is meagre, choose $\gamma$ with $M\subseteq M_{\gamma }$. Every $x_{\alpha }$ with $\alpha \ge \gamma$ avoids $M_{\gamma }$, hencewhich is countable. Thus $A$ has precisely the Lusin set property. Enumerating covers rather than all meagre sets avoids an unjustified claim that there are only continuum many arbitrary meagre subsets.

A witness to failure of the Axiom of power set is $\boxed{{\omega }_1\in H_{{\omega }_2}}$. Every subset $A\subseteq {\omega }_1$ belongs to $H_{{\omega }_2}$, since its transitive closure has cardinality at most ${\aleph }_1$. If some $p\in H_{{\omega }_2}$ were the internal power set of ${\omega }_1$, transitivity and bounded subsethood would force $p$ to contain every ambient subset of ${\omega }_1$, and only those subsets. Thus it would be the full $\mathcal{P}({\omega }_1)$.

By the Cantor theorem,Its transitive closure contains all its members, so $|\mathrm{tc}(\{p\})|\ge |p|\ge {\aleph }_2$. This contradicts $p\in H_{{\omega }_2}$. ThereforeThe power-set failure in hereditarily small sets needs no assumption about the Continuum hypothesis.

The assumed beth number equality ${\beth }_1={\aleph }_2$ means $2^{{\aleph }_0}={\aleph }_2$ in the ambient universe. Put $W=L(X)$ with $X=\mathcal{P}(\omega )$, using the relative constructible hierarchy defined above.

The base $X$ is transitive because every member of a member of $X$ is a finite ordinal, hence itself a subset of $\omega$. Moreover $X\subseteq W$ and $X\in W$, and $W$ is an inner model of ZF. The latter standard fact follows from the hierarchy and reflection: bounded-rank witnesses give pairing and union; reflection gives separation and replacement; ambient replacement bounds the stages of all relatively constructible subsets of any fixed set, giving power set. It does not require axiom of choice inside $W$. Since no new subsets of $\omega$ can appear in an inner class,

Since inner models with all reals preserve omega-one, it suffices to check that their real-code argument applies to $W$. Every ambient countably infinite ordinal has a well-order code on $\omega$, coded by a subset of $\omega$ in $X$. Finite ordinals and $\omega$ are already shared. This code belongs to $W$, is well-founded there, and its Mostowski collapse theorem interpretation inside $W$ is the same ordinal as outside. Hence the ordinal is countable in $W$. Conversely, any countability witness in $W$ remains a witness in the ambient universe. Thus ${\omega }_1^W={\omega }_1$.

If $W$ satisfied the Continuum hypothesis in its usual well-orderable formulation, it would contain a bijection from ${\omega }_1^W$ onto its power set of $\omega$. The same bijection would exist externally from ${\omega }_1$ onto $X$, contradicting $|X|={\aleph }_2$. ThereforeThis proof uses agreement on all reals and on ${\omega }_1$, not an unwarranted assertion that $L(X)$ satisfies axiom of choice or computes every higher cardinal number correctly.