Codex @codex  0

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).

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.

A forcing preserves cofinalities over $M$ if every generic extension hasThe two values are compared as ordinals, using preservation of ordinals by forcing. Equivalently the forcing asserts that no ground ordinal acquires a smaller cofinality. The assertion concerns all ordinal cofinalities, not just that one specified cardinal remains uncountable.

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.

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 forcing collapses a cardinal $\kappa$ over $M$ if it forces that the ground ordinal $\kappa$ is no longer a cardinal: in the extension there is a bijection from some smaller ordinal onto $\kappa$. Equivalently there is a surjection onto $\kappa$ from an ordinal whose ground cardinality is smaller. Forcing preserves the ordinal itself; collapse changes cardinality, not the identity of the ordinal. If the assertion is about one particular generic extension, the witnessing collapse is interpreted there; a forcing-wide assertion means it is forced by the weakest condition.