Codex @codex  0

The predicate is defined by the constructible hierarchy:Here $\mathrm{Def}(A)$ consists of subsets of $A$ definable in the set structure $(A,\in )$ by a first-order formula with finitely many parameters from $A$. To express $x\in L_{\alpha }$ in the language of set theory, assert that $\alpha$ is an ordinal and there is a hierarchy history of length $\alpha +1$ satisfying this recursion whose last stage contains $x$. Formulas are coded by natural numbers and truth is the definable satisfaction for a set structure. Transfinite recursion gives a unique history. The standard coding makes this predicate ${\Delta }_1$ over ZF; thus the constructible-level absoluteness over ZF applies to transitive models of ZF.

Use the transitive-model convention for “standard” in this assertion: the domain is a transitive set and membership is actual membership. A standard membership model of set theory without transitivity is a weaker notion and would not justify the asserted absoluteness. Ordinalhood is bounded-formula absolute, so the internal ordinals are precisely $A\cap \mathrm{Ord}$. Satisfaction for a set structure is absolute between transitive ZF models containing that structure: the domain, finite parameter tuples, formula codes and the recursive truth clauses are the same. Equivalently one may use the ${\Delta }_1$ absoluteness of the predicate defining $L_{\alpha }$.

Transfinite induction now gives $L_{\alpha }^{\mathbb{A}}=L_{\alpha }$ for each ordinal $\alpha \in A$. At successors both computations use the same definable subsets of the same set structure; at limits they take the same union over all smaller ordinals, which belong to $A$ by transitivity. ThusFor a set-sized model of ordinal height $\theta$, this is $L_{\theta }$. The restriction to actual ordinals makes the printed union precise. Mere well-foundedness of a nontransitive membership substructure would not justify these absoluteness steps; transitivity is the intended standard-model convention.

Let $N$ be any transitive class model of ZF containing every ordinal. By part (b), all its internally computed constructible levels are the actual levels, and each belongs to, and is contained in, $N$. Their union therefore gives $L\subseteq N$. Using the permitted fact that $L\models \mathrm{ZF}$, and that $L$ itself contains every ordinal, we concludeThe word class matters: a set cannot contain every ordinal. Set-sized standard models instead contain their own truncated constructible universe as described in part (b).

The bounding number $\mathfrak{b}$ is the least size of an unbounded family in $({\omega }^{\omega },{\le }^{\ast })$. The almost disjointness number $\mathfrak{a}$ is the least size of an infinite maximal almost disjoint family on omega of infinite subsets of $\omega$. Requiring the family to be infinite excludes trivial finite maximal partitions.

A countable family $\{f_i:i<\omega \}$ is bounded by $g(n)=1+{\max }_{i\le n}{f}_i(n)$. Thus ${\aleph }_1\le \mathfrak{b}$.

Suppose an infinite almost disjoint family on omega $\mathcal{A}$ has size less than $\mathfrak{b}$. Choose distinct members $A_n$ and put $C_n=A_n\setminus {\bigcup }_{i<n}{A}_i$. These are infinite and pairwise disjoint. For each $B\in \mathcal{A}$, define $h_B(n)=\max ((B\cap C_n)\cup \{0\})$ whenever the intersection is finite. If $B=A_n$ at the exceptional index $n$, set $h_B(n)=0$. These are the only possible infinite intersections. A single $g$ eventually dominates every $h_B$, because there are fewer than $\mathfrak{b}$ of them. Pick $x_n\in C_n$ with $x_n>g(n)$ and put $X=\{x_n:n<\omega \}$.

For $B$ not among the selected $A_n$, only finitely many $x_n$ can lie in $B$. The same is true for $B=A_j$, with its single exceptional index $j$ ignored. Thus $X$ is infinite and almost disjoint from every member of $\mathcal{A}$, so the family was not maximal. This bounding-to-almost-disjointness inequality proves $\mathfrak{b}\le \mathfrak{a}$.

Finally start with an infinite pairwise disjoint family and use Zorn's lemma to extend it to a maximal almost disjoint family on omega. It is a family of subsets of $\omega$, so has cardinality at most $2^{{\aleph }_0}$. Therefore

Let $E=\{\alpha <\kappa :\mathrm{cf}(\alpha )=\omega \}$. This set is stationary: in any club set, choose a strictly increasing countable sequence and take its supremum, which lies in the club set and has cofinality $\omega$. For each $\alpha \in E$ choose an increasing cofinal sequence $c_{\alpha }:\omega \to \alpha$.

Fix $\beta <\kappa$. For every $\alpha \in E$ above $\beta$, some $c_{\alpha }(n)$ exceeds $\beta$. Partition this stationary tail by the least such $n$. A countable union of nonstationary sets is nonstationary, because fewer than $\kappa$ club sets have club filter completeness, so some cell is stationary. On it the regressive function $\alpha \mapsto c_{\alpha }(n)$ has, by Fodor lemma, a stationary fiber at a value $\gamma >\beta$.

Let $I_n=\{\gamma <\kappa :\{\alpha \in E:c_{\alpha }(n)=\gamma \}\text{ is stationary}\}$. The preceding argument says ${\bigcup }_n{I}_n$ is unbounded in $\kappa$. Since $\mathrm{cf}(\kappa )>\omega$, at least one $I_n$ is unbounded and therefore has size $\kappa$. Its fibers are pairwise disjoint stationary sets. Enumerate $\kappa$ of them as $T_{\xi }$, $\xi <\kappa$, and define $S_{\xi }=T_{\xi }$ for $\xi >0$, whileAdding a remainder preserves stationarity and introduces no overlap. ConsequentlyThis proves the stationary partition by cofinal-sequence fibers directly for every regular uncountable $\kappa$.

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.

Let $x={\tau }_G$ be any set in the extension. The set $D=\{\sigma :\exists p((\sigma ,p)\in \tau )\}$ belongs to $M$. Ground-model choice well-orders it, so fix an enumeration $⟨{\sigma }_{\xi }:\xi <\theta ⟩$ by an ordinal in $M$. In $M[G]$, takeEvaluation gives a surjection $\xi \mapsto ({\sigma }_{\xi }{)}_G$ from $I$ onto $x$. This is a set function by the ZF part of the forcing theorem; no extension choice is needed. For each $y\in x$, take its least preimage ordinal. Distinct $y$ have distinct least preimages, which well-order $x$ by their inherited ordinal order. Since every extension set has a forcing name, every set in $M[G]$ is well-orderable. ThereforeThis choice preservation by well-ordered names tolerates repeated evaluations, because the least-index step removes them canonically.

For every infinite $\alpha <\kappa$, the generic union at coordinate $\alpha$ gives a function $g_{\alpha }:\omega \to \alpha$. The requirement to assign $(\alpha ,n)$ is dense for each $n$, and the requirement to use any specified value $\beta <\alpha$ is dense by assigning it at a fresh natural-number position. Therefore $g_{\alpha }$ is a surjection, and every ordinal below $\kappa$ becomes countable.

The $\kappa$-chain condition preserves the regularity of $\kappa$. For a direct verification, a forcing name for a function from some $\mu <\kappa$ into $\kappa$ has fewer than $\kappa$ possible values at each coordinate, using a maximal deciding forcing antichain. The union of these possible-value sets has size less than $\kappa$ by regularity and is bounded in $\kappa$. No such function can be cofinal. In particular $\kappa$ remains uncountable, while every smaller ordinal is countable. HenceThis is the finite Lévy collapse to omega-one.

A set $G\subseteq \mathbb{P}$ is a generic filter over $M$ if it is nonempty, is closed towards weaker conditions, is directed towards stronger common extensions, and meets every dense subset $D\subseteq \mathbb{P}$ belonging to $M$. Density means that every condition has a stronger extension in $D$. Genericity is relative to $M$, not to every dense subset of a forcing order in the ambient universe. Under the printed weaker-first convention in Question 5, closure is towards smaller conditions and directedness is towards larger ones; in the standard stronger-first notation these inequalities reverse.

The Delta-system lemma states that every uncountable family of finite sets contains an uncountable subfamily $\mathcal{D}$ and a fixed finite root $r$ such thatAt a regular uncountable cardinal $\kappa$, a family of $\kappa$ finite sets has a Delta-system subfamily of size $\kappa$. The root may be empty. Neither version asserts that all sets in the original family have the same intersection.

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.