Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 2 / 16F (Logic and Set Theory)

The inductive definition of ordinal addition isfor nonzero limit ordinals $\lambda$. The synthetic definition is the order type of a disjoint copy of $\alpha$ followed by a disjoint copy of $\beta$, retaining the order within each block and declaring every element of the first block smaller than every element of the second.

These are equivalent by transfinite induction in the second argument. An empty second block leaves $\alpha$; a successor block appends one final point; and a limit block is the increasing union of its proper initial blocks, whose order types have the indicated supremum. Thus the synthetic construction satisfies exactly the inductive recursion, which determines the operation uniquely. It is not commutative: $1+\omega =\omega$, whereas $\omega +1>\omega$.

This assertion is always true. Synthetically, either bracketing places three disjoint blocks of order types $\alpha ,\beta ,\gamma$ in that order. The identity on these blocks is an order isomorphism, so both sums have the same order type. This proves associativity of ordinal addition for successor and limit ordinals alike.

This assertion is false, even for nonzero limit ordinals. Take $\alpha =\omega$ and $\beta ={\omega }^2$. Then $\omega +{\omega }^2={\omega }^2$, while ${\omega }^2+\omega >{\omega }^2$. The first identity follows from the limit recursion, since ${\sup }_{n<\omega }(\omega +\omega n)={\omega }^2$; the second sum strictly extends the initial block of type ${\omega }^2$.

This assertion is false. Set $\alpha =1$ and $\beta ={\omega }_1$. The limit recursion givesEvery term is countable, and they are cofinal in the first uncountable ordinal. Yet $\alpha =1$ is neither zero nor ${\omega }_1$.

This assertion is always true. Ordinal addition obeys $\alpha +\beta \ge \alpha$ and $\alpha +\beta \ge \beta$. Thus both summands are at most ${\omega }_1$. If $0<\beta <{\omega }_1$ and $\alpha <{\omega }_1$, both ordinals are countable, and their disjoint ordered union is countable, so their sum cannot be ${\omega }_1$. If instead $\alpha ={\omega }_1$ and $\beta >0$, appending a nonempty second block gives a sum strictly larger than ${\omega }_1$. These exhaust the possibilities with $0<\beta <{\omega }_1$; hence $\beta =0$ or $\beta ={\omega }_1$.