Past exam of the logic and set theory course of the University of Oxford / 2008 / B / 5

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.

False. Use $\alpha =2$ and $\beta =\omega$. The recursive definition of ordinal multiplication givesThe first product concatenates countably many finite two-point blocks. The second concatenates two whole copies of the order type $\omega$, producing a strictly larger ordinal. Hence ordinal multiplication is not commutative.

True: ordinal multiplication on the left distributes over ordinal addition. If $\alpha =0$ both sides are zero. Otherwise fix $\alpha >0$ and $\beta$, and induct on $\gamma$. At zero the identity follows from $\alpha 0=0$. At a successor, use the recursive ordinal multiplication rule and the associativity just proved:For a nonzero limit ordinal $\lambda$, the ordinals $\beta +\xi$ are cofinal in $\beta +\lambda$. Continuity of ordinal multiplication in its right argument givesFor the last equality, $\alpha \xi$ is increasing and cofinal in $\alpha \lambda$, which is a limit ordinal because $\alpha >0$, and ordinal addition is continuous in its right argument. This establishes the limit step and the identity for all ordinals.

False. Take $\alpha =\beta =1$ and $\gamma =\omega$. ThenThus ordinal multiplication on the right does not in general distribute over ordinal addition. The side on which ordinal multiplication occurs matters; the left-distributive identity in the previous part does not imply this one.