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A Von Neumann ordinal $x$ is finite exactly when $x$ and every member of $x$ are either zero or successor ordinals. Every finite ordinal has that property. Conversely, an infinite ordinal is either $\omega$, which is a nonzero limit, or is greater than $\omega$ and contains that nonzero limit ordinal. The test can be expressed by a bounded first-order formula: ordinality uses transitivity and membership comparability, and a nonempty ordinal is a successor exactly when it has a greatest member. This defines the class of natural numbers without quantifying over arbitrary infinite inductive sets.

A finite ordinal is the order type of a finite well-order; in the von Neumann representation it is one of $0,1,2,\dots$. They are exactly the ordinals $x$ such that $x$ and each member of $x$ are either zero or successor ordinals. If an ordinal is infinite, it either is the nonzero limit $\omega$ or contains $\omega$ as a member, so it fails this test. Thus the finite ordinals give the usual representation of the natural numbers.

Fix $m$ and induct on $n$. The initial equality $m+0=0+m$ follows from the defining zero clause and (ii). Assuming $m+n=n+m$, the successor clause and (iii) giveThis provesThe extension to arbitrary ordinals is false. By the limit clause for ordinal addition,whereas $\omega +1$ is the successor ordinal of $\omega$ and is strictly greater than it. ThereforeFinite addition is commutative; ordinal addition records the order of concatenated well-orders and need not be commutative.

First give the recursive definitions, keeping the right argument as the recursion variable. For ordinals $\alpha ,\beta$ and nonzero limit ordinals $\lambda$,Here $\xi +1$ is the successor ordinal, and the supremum of a set of ordinals is its union. These are ordinal arithmetic, defined by transfinite recursion, and differ from cardinal sum and ordinal multiplication.

Associativity of ordinal addition is true. Fix $\alpha ,\beta$ and use transfinite induction on $\gamma$. The case $\gamma =0$ is immediate. In the successor step the recursive definitions and the induction hypothesis giveFor a nonzero limit ordinal $\lambda$, the ordinals $\beta +\xi$ for $\xi <\lambda$ form an increasing cofinal subset of the limit ordinal $\beta +\lambda$. The limit clause for ordinal addition, restricted to this cofinal subset, givesCofinal restriction is valid because ordinal addition by a fixed ordinal on the left is increasing and continuous in its right argument, directly from its recursive definition. This completes the induction.

False. Take $\alpha =1$ and $\beta =\omega$. By the limit clause for ordinal addition,The first order type is one point followed by a countably infinite sequence, which is still of type $\omega$. The second has a greatest element after that infinite sequence, so its order type is the successor ordinal $\omega +1$. Thus ordinal addition is not commutative, even though the two resulting sets have equal cardinality.

The cumulative hierarchy is defined by transfinite recursion:The rank of a set is $\mathrm{rk}(x)={\sup }_{y\in x}(\mathrm{rk}(y)+1)$, equivalently the least $\alpha$ such that $x\subseteq V_{\alpha }$. The standard hierarchy properties give $\mathrm{rk}(V_{\alpha })=\alpha$, so every ordinal occurs as a set rank; the ordinal $\alpha$ itself is another example of rank $\alpha$.

Let $x$ be transitive of rank $\alpha$, and fix $\beta <\alpha$. The supremum formula gives an element of $x$ whose rank is at least $\beta$. Choose such an element $y$ of least possible rank. If $\mathrm{rk}(y)>\beta$, the same supremum formula gives $z\in y$ with rank at least $\beta$. Transitivity gives $z\in x$, and membership strictly decreases rank, contradicting the choice of $y$. Thus $x$ contains an element of every rank below $\alpha$.

A nonempty finite set has rank the maximum of finitely many successor ordinals, hence a successor ordinal; the empty set has rank zero. Conversely, for any $\beta$, the singleton $\{V_{\beta }\}$ has rank $\beta +1$. ThusA finite transitive set of rank $\alpha$ contains, by the proved property, distinct elements of every rank $\beta <\alpha$. Finiteness therefore forces $\alpha <\omega$. Every finite ordinal is itself a finite transitive set of that rank. Consequently

True. If $x$ is a nonempty finite set, thenThis maximum is a successor ordinal, whereas the empty set has rank $0$. Hence a set of nonzero limit ordinal rank cannot be finite and must be an infinite set.

The inductive definition of ordinal addition isfor a limit ordinal $\lambda$.

The synthetic definition takes the order type of the disjoint union of a copy of $\alpha$ followed by a copy of $\beta$: every point of the first copy precedes every point of the second, and each copy retains its original total order.

Fix $\alpha$ and use transfinite induction on $\beta$. The synthetic sum with the empty order is $\alpha$. Appending a greatest element produces the successor ordinal rule. At a limit ordinal $\lambda$, the second copy is the union of its initial segments of types $\beta <\lambda$, so the whole order has type
${\sup }_{\beta <\lambda }(\alpha +\beta )$. Thus the synthetic operation satisfies the inductive recursion; uniqueness in transfinite recursion proves the definitions equivalent.