Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 1 / 16H (Logic and Set Theory)

The inductive definition of ordinal addition is $\alpha +0=\alpha$, $\alpha +(\beta +1)=(\alpha +\beta )+1$, and $\alpha +\lambda ={\sup }_{\beta <\lambda }(\alpha +\beta )$ for a nonzero limit ordinal $\lambda$. The synthetic definition is the order type of the order sum obtained by putting a copy of $\beta$ after a disjoint copy of $\alpha$.

These agree by transfinite induction on the second argument. Appending no points gives $\alpha$; appending one new greatest point gives a successor. At a limit stage, the order sum is exhausted by the increasing initial segments obtained by appending each $\beta <\lambda$, so its order type is their supremum. Thus the synthetic construction satisfies all three recursion clauses, and uniqueness of transfinite recursion proves equivalence.

False. By the successor and limit clauses for ordinal multiplication,Multiplication is not commutative because it counts ordered copies, rather than merely the cardinality of their union.

Always true. In the order-sum interpretation of ordinal multiplication, $\alpha (\beta +\gamma )$ consists of copies of $\alpha$ indexed first by $\beta$ and then by $\gamma$. The first portion has order type $\alpha \beta$ and the second $\alpha \gamma$, proving $\alpha (\beta +\gamma )=\alpha \beta +\alpha \gamma$. Equivalently, transfinite induction on $\gamma$ uses the successor rule, associativity of ordinal addition, and continuity of both operations in the right argument at limit ordinals.

False. Take $\alpha ={\omega }^2+1$, which satisfies the proposed lower bound. Associativity of ordinal addition and $1+{\omega }^2={\omega }^2$ giveThe final successor survives only in the second expression.

False. Take $\alpha ={\omega }_1+1$. Continuity of ordinal multiplication in the right argument gives $({\omega }_1+1){\omega }_1={\omega }_1^2$. For a precise bound, for each countable ordinal $\beta$ monotonicity givesBoth bounding expressions have supremum ${\omega }_1^2$ as $\beta$ ranges below ${\omega }_1$, because $2\beta$ is still countable and these indices are cofinal. But the successor rule gives