Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 2 / 16H (Logic and Set Theory)

True. For $\beta >0$, multiplication $\gamma \mapsto \beta \gamma$ is strictly increasing and continuous at limit ordinals, and $\beta \gamma \ge \gamma$. Thus there is a greatest $\gamma$ with $\beta \gamma \le \alpha$: the set is bounded, and its supremum still satisfies the inequality by continuity if it is a limit. The remaining ordered interval in $\alpha$ has an ordinal order type $\delta$, so $\alpha =\beta \gamma +\delta$. If $\delta \ge \beta$, then $\beta (\gamma +1)\le \alpha$, contradicting maximality. Hence $\delta <\beta$.

False. Take $\alpha =\omega +1$, $\beta =2$. For finite $\gamma$, $\gamma \cdot 2+\delta$ with $\delta <2$ is finite. For infinite $\gamma$, $\gamma \ge \omega$ implies $\gamma \cdot 2\ge \omega \cdot 2>\omega +1$. Neither case gives the required expression.

True. This is left distributivity of ordinal multiplication. Fix $\alpha ,\beta$ and induct on $\gamma$. The zero case is immediate. The successor step uses $\alpha (\eta +1)=\alpha \eta +\alpha$ and associativity of ordinal addition. At a limit, multiplication and addition are continuous in their right argument, so taking suprema of the earlier equalities gives $\alpha (\beta +\gamma )=\alpha \beta +\alpha \gamma$.

False. For $\alpha =\beta =1$ and $\gamma =\omega$, the left side is $2\cdot \omega =\omega$, whereas the right side is $\omega +\omega >\omega$.

Under the convention that a limit ordinal is nonzero and not a successor, the assertion is false at $\alpha =0$, since $\omega \cdot 0=0$. It is true for $\alpha >0$. If $\alpha =\eta +1$, then $\omega \alpha =\omega \eta +\omega$ has no greatest predecessor. If $\alpha$ is a nonzero limit, $\omega \alpha ={\sup }_{\eta <\alpha }\omega \eta$, again a nonzero limit. If zero is counted as a limit ordinal by convention, the unrestricted assertion is true with that convention.

True. Divide a limit ordinal $\lambda$ by $\omega$ using part a: $\lambda =\omega \alpha +n$ with $n<\omega$. If $n>0$, the right side is a successor, contrary to the limit assumption. Hence $n=0$ and $\lambda =\omega \alpha$. For a nonzero limit, $\alpha >0$; if zero is allowed as a limit, it is represented by $\alpha =0$.