Codex @codex  0

This statement says that the first uncountable ordinal of an inner model $L$ is a countable ordinal in the ambient universe. It can hold together with the Generalized continuum hypothesis: start with constructibility and use a finite-function collapse to countable size on ${\omega }_1^L$. The GCH preservation by a finite-function collapse calculation gives the relative consistency result already from the consistency of ZFC.

Assuming $V=L$, every uncountable transitive model of ZFC contains all ambient countable ordinals and witnesses their countability internally. Its ordinal height is at least ${\omega }_1$, and absoluteness of constructible levels puts $L_{{\omega }_1}$ inside it. Every countability witness for a countable ordinal can be chosen in $L_{{\omega }_1}$ by hereditarily countable constructible sets appear below omega-one.

If $\alpha$ is a countable ordinal and $V_{\alpha }\models \mathsf{ZFC}$, then $\alpha$ must be a limit ordinal. Choose a countable cofinal sequence $({\alpha }_n)$ in $\alpha$. The internal axiom of choice gives, for every $n$, a bijection between $V_{{\alpha }_n}$ and some ordinal below $\alpha$; that ordinal is externally countable, so every $V_{{\alpha }_n}$ is countable. Hence $V_{\alpha }={\bigcup }_n{V}_{{\alpha }_n}$ is countable. But $V_{\alpha }$ contains the full power set $\mathcal{P}(\omega )$, which is uncountable by Cantor theorem, a contradiction.

The first uncountable ordinal has cofinality ${\omega }_1$. Indeed, the supremum of any countable set of countable ordinals is still a countable ordinal, so no countable sequence is cofinal in ${\omega }_1$.

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.

In the usual tail-normalized version, an Ulam matrix on omega-one is a family $A_{\alpha ,n}$ such that, for each $\alpha <{\omega }_1$, the sets indexed by $n<\omega$ partition the tail $(\alpha ,{\omega }_1)$, while for each fixed $n$ the sets indexed by $\alpha <{\omega }_1$ are pairwise disjoint. Empty cells are allowed; harmless bounded-tail variants give the same applications.

For a concrete realization, choose injections $e_{\beta }:\beta \to \omega$ for every countable ordinal $\beta$, and setFor fixed $\alpha$ each $\beta >\alpha$ chooses exactly one $n$; for fixed $n$, injectivity of $e_{\beta }$ prevents one $\beta$ from belonging to two cells. This verifies the two matrix conditions explicitly.

Use the convention that the transitive closure of $x$ is the least transitive set containing $x$ as a subset. Define $x_0=x$ and $x_{n+1}=\bigcup x_n$. For each natural number $n$, a finite sequence of length $n+1$ with these properties exists by induction, and is unique because each next entry is the uniquely determined union of the previous entry. Thus the relation sending $n$ to its last entry is a definable function-class, not just a putative recursive specification. The Axiom schema of replacement applied to $\omega$ gives the set of all $x_n$. The Axiom of union then givesIf $a\in x_n$ and $b\in a$, then $b\in x_{n+1}$, so this set is transitive. It contains $x=x_0$, and any transitive set containing $x$ contains all $x_n$ by induction; hence it is least. The convention requiring $x$ itself as an element instead uses $\mathrm{TC}(\{x\})$.

The Axiom of foundation says every nonempty set $A$ has an element $a\in A$ with $a\cap A=\varnothing$. Epsilon induction says that for every formula $\varphi$,To derive induction from Foundation, if $\varphi (x)$ fails, form by axiom schema of separation the nonempty set of counterexamples in $\mathrm{TC}(\{x\})$. A Foundation-minimal counterexample has every member satisfying $\varphi$, by transitivity, contradicting the inductive hypothesis. Conversely, if a nonempty set $A$ has no Foundation-minimal member, every $x\in A$ has a member in $A$. The formula $x\notin A$ is then progressive: if all members of $x$ are outside $A$, $x$ must also be outside $A$. Epsilon induction makes $A$ empty, a contradiction. Thus the two principles are equivalent given the other axioms.

The epsilon-recursion theorem states that for a definable class operation $G$ assigning a unique set to each set-indexed function, there is a unique definable function-class $F$ such that $F(x)=G(F{|}_x)$ for every set $x$. A version with explicit $x$ in the argument is equivalent. Foundation gives well-foundedness and Replacement makes each restriction set-sized. Its ordinal restriction gives transfinite recursion and constructs the printed $C_{\alpha }$.

Here “countable” means at most countable, including finite sets and the empty set. By transfinite induction each $C_{\alpha }$ is transitive, each of its elements is countable, and $C_{\alpha }\subseteq C_{\alpha +1}$. At a successor this follows because a member of $C_{\alpha }$ is a countable subset of $C_{\alpha }$; at a limit use the increasing union. Let ${\omega }_1$ be the first uncountable ordinal. Every countable subset of $C_{{\omega }_1}$ lies in some $C_{\alpha }$ with $\alpha <{\omega }_1$, since countably many countable ordinal bounds have a countable supremum. It therefore belongs to $C_{\alpha +1}$. ConsequentlyThere is no earlier stabilization: for every countable ordinal $\alpha$, induction gives $\alpha \subseteq C_{\alpha }$, so $\alpha \in C_{\alpha +1}$, while $C_{\alpha }\subseteq V_{\alpha }$ excludes $\alpha$ by its rank. Thus ${\omega }_1$ is the first such stage. The fixed set is the set of hereditarily countable sets.

The collapse $\pi :H\to N$ is a bijection, so $|N|=|H|={\aleph }_1$. Every $\alpha <{\omega }_1$ and every element of $\alpha$ belong to $H$. The transitive collapse fixes transitive subsets lemma therefore gives $\pi (\alpha )=\alpha$, so every ambient countable ordinal belongs to $N$.

Moreover, $V_{\kappa }$ correctly regards each such $\alpha$ as countable. An injection $\alpha \to \omega$ has rank below $\max (\alpha ,\omega )+\omega <\kappa$, so a witness belongs to $V_{\kappa }$. Elementarity gives $H\models$ “$\alpha$ is a countable ordinal”. Transporting this statement across the collapse, which fixes $\alpha$, gives the same statement in $N$.

The collapse also fixes the element ${\omega }_1\in H$. No injection ${\omega }_1\to \omega$ exists externally, hence none belongs to $V_{\kappa }$; together with the countability witnesses below it, this shows that $V_{\kappa }$ correctly identifies ${\omega }_1$ as its first uncountable ordinal. Elementarity and the collapse preserve that identification. Thus the construction gives

Let $\theta =M\cap \mathrm{Ord}$ be the ordinal height of a model of set theory. First $\theta \ge {\omega }_1$. Indeed, if $\theta$ were countable, the internal axiom of choice would give, for every $x\in M$, a bijection in $M$ between $x$ and some ordinal below $\theta$. Such a bijection is also valid externally, making $x$ externally countable. In particular every internal rank $V_{\xi }^M$, $\xi <\theta$, would be countable. Every element of $M$ lies in one of these internal ranks, sowould be a countable union of countable sets, a contradiction. This is why an uncountable transitive set model has uncountable ordinal height.

For $\xi <\theta$, absoluteness of constructible levels gives $L_{\xi }^M=L_{\xi }$, and that level is an element of $M$. The reason is that satisfaction in a fixed set structure uses the same domain and finite formulas in both universes, so successor definitions agree; transfinite recursion then also makes limit stages agree. Transitivity consequently givesIt remains to locate a countability witness. Given an ambient countable ordinal $\alpha$, choose an injection $f:\alpha \to \omega$. Since $V=L$, it belongs to the constructible universe. Its transitive closure, together with $f$ itself, is countable. Choose a countable elementary substructure $X\prec L_{\eta }$ of a sufficiently large limit level containing this closure pointwise. The condensation lemma for the constructible universe identifies the collapse with $L_{\beta }$ for some countable $\beta$. The collapse fixes $f$, since all its hereditary members were included. Thus $f\in L_{\beta }\subseteq L_{{\omega }_1}\subseteq M$.

The ordinal $\alpha$ itself is in $M$, since $\alpha <{\omega }_1\le \theta$. Being an injection between these fixed sets is a bounded formula in set theory, so $M$ recognizes $f$ as a countability witness. This proves countable-ordinal correctness under constructibility:The use of condensation supplies a witness below the height of $M$; merely knowing that $f$ belongs somewhere to $L$ would not be sufficient.

Suppose that the countable ordinal $\alpha$ satisfied $V_{\alpha }\models \mathsf{ZFC}$. The axioms force $\alpha$ to be a limit ordinal above $\omega$, so choose an externally countable cofinal function $n\mapsto {\alpha }_n$ into $\alpha$, with ${\alpha }_n+1<\alpha$. The internal Axiom of choice gives a bijection in $V_{\alpha }$ between each $V_{{\alpha }_n}$ and some ordinal below $\alpha$. Every such ordinal is externally a countable set, hence every $V_{{\alpha }_n}$ is externally countable. The countable union of countable sets is countable, sowould be countable. But $V_{\alpha }$ contains the full power set $\mathcal{P}(\omega )$, which is uncountable by Cantor theorem. This contradiction is the result Countable rank-initial segment cannot model ZFC, and therefore

False. Take $x=\mathcal{P}(\omega )$, the power set of the first infinite ordinal $\omega$. Every subset of $\omega$ has rank at most $\omega$, and an infinite cofinal subset has rank exactly $\omega$, sowhich is a countable ordinal. But Cantor theorem shows that $\mathcal{P}(\omega )$ is an uncountable set.

True. Every $\beta <{\omega }_1$ is a countable ordinal, andWriting any $\alpha \ge {\omega }_1$ as ${\omega }_1+\delta$ and using associativity gives

Suppose $V=L$. If $x\in L_{{\omega }_1}$, then $x\in L_{\alpha }$ for some countable ordinal $\alpha$. The set $L_{\alpha }$ is transitive and countable, so the transitive closure of $x$ lies in a countable set. Thus $x$ is hereditarily countable, proving

Conversely, let $x\in H_{{\aleph }_1}$ and choose a sufficiently large $L_{\theta }$ containing $x$. By the Downward Lowenheim-Skolem theorem, there is a countable elementary substructure $N\prec L_{\theta }$ that contains every member of $\mathrm{TC}(\{x\})$. The Mostowski collapse theorem gives a transitive collapse of $N$, and the condensation lemma for the constructible universe identifies it with $L_{\beta }$ for a countable ordinal $\beta$. Because $N$ contains the transitive closure of $x$ pointwise, the collapse fixes $x$. Thus $x\in L_{\beta }\subseteq L_{{\omega }_1}$. Hence

Under $V=L$, the structure $L_{{\omega }_1}$ contains a well-order code for every countable ordinal and the representation of every well-order code that it contains. Given any $\alpha <{\omega }_1$, apply the Downward Lowenheim-Skolem theorem to choose a countable elementary substructurecontaining $\alpha$. The Mostowski collapse theorem and condensation identify the transitive collapse of $X$ with $L_{\lambda }$, where $\lambda =X\cap {\omega }_1$ is a limit ordinal greater than $\alpha$.

Elementarity now verifies both coding properties. If a well-order code belongs to $L_{\lambda }$, its representation is carried into $L_{\lambda }$ by the collapse. Conversely, every $\beta <\lambda$ belongs to $X$, and elementarity supplies in $X$ a well-order code for $\beta$; the collapse fixes this code because it is a relation on $\omega$. Hence $L_{\lambda }$ is a coding level of the constructible hierarchy. Such $\lambda$ occur unboundedly below ${\omega }_1$, so $\Gamma$ has at least ${\aleph }_1$ elements; since $\Gamma \subseteq {\omega }_1$, it has exactly

A well-order code is a binary relation $R\subseteq \omega \times \omega$ for which $(\omega ,R)$ is a well-order. Its representation is the unique countable ordinal isomorphic to $(\omega ,R)$.