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A countable ordinal has a countable set as its underlying set.

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 ${\omega }_1$.

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.

Ordinal exponentiation is defined by transfinite recursion:when $\lambda$ is a nonzero limit ordinal. If $\alpha \ge 1$, each successor step and each supremum show that $\beta \mapsto {\alpha }^{\beta }$ is nondecreasing. If $\alpha \ge 2$, transfinite induction strengthens this to strict increase: ${\alpha }^{\beta }<{\alpha }^{\beta +1}$, and the value at a limit exceeds every earlier value.

Strict increase can disappear when the base changes. For example, with $\omega <\omega +1<\omega +2$,because for every positive finite $n$, both finite powers have leading term ${\omega }^n$, and taking the supremum gives ${\omega }^{\omega }$.

We next proveApply transfinite induction to $\gamma$. The zero and successor cases follow directly from the recursive definitions and the associativity of ordinal multiplication. At a limit $\lambda$, continuity of ordinal multiplication in its right argument givesThe superficially similar identity $(\alpha \beta {)}^{\gamma }={\alpha }^{\gamma }{\beta }^{\gamma }$ is false: for $\alpha =\omega$, $\beta =2$, and $\gamma =2$,

Finally,For the first equivalence, bases zero and one fail, while strict increase embeds every countable ordinal into the powers of any base $\alpha \ge 2$. For the second, $\alpha \ge {\omega }_2$ immediately gives ${\alpha }^{{\omega }_1}\ge \alpha$. Conversely, if $\alpha <{\omega }_2$, then $|\alpha |\le {\aleph }_1$. Every exponent below the first uncountable ordinal ${\omega }_1$ is countable, so each ${\alpha }^{\beta }$ has cardinality at most ${\aleph }_1$; the supremum of ${\aleph }_1$ such ordinals still has cardinality at most ${\aleph }_1$ and is therefore below the second uncountable ordinal ${\omega }_2$.

Take a transitive model $M\models \mathsf{ZFC}+2^{{\aleph }_0}={\aleph }_2$. In $M$, choose a bijection $e:{\omega }_2\to \mathcal{P}(\omega )$ and encode its graph by a set $A\subseteq {\omega }_2$, using a fixed bijection between ${\omega }_2\times \omega$ and ${\omega }_2$. The relative constructible universe $L(A)$ can decode $e$, and therefore contains every real number of $M$; being an inner model of $M$, it has no additional reals.

Models with the same reals have the same first uncountable ordinal, because their reals code exactly the same countable well-orders. If $L(A)$ satisfied the Continuum hypothesis, its bijection between ${\omega }_1$ and the reals would also belong to $M$, contradicting $M\models 2^{{\aleph }_0}={\aleph }_2$. This is the construction in relative constructible universe can violate the continuum hypothesis, and it gives

The second uncountable ordinal ${\omega }_2$ is the least ordinal whose cardinal number is strictly larger than that of the first uncountable ordinal ${\omega }_1$. Equivalently, it is the initial ordinal of the successor cardinal ${\aleph }_2$.