Exercise 7.2.2

Proof. First, clearly we have by definition that $n\cdot \omega =\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{n\cdot k\mid k<\omega \right\}=\omega$. Then, since $\alpha$ is a limit ordinal, we have from Exercise 6.5.10 that $\alpha =\omega \cdot \beta$ for some ordinal $\beta$. Thus we have

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Proof. Consider any natural number $n>0$ and any ordinal number $\alpha$. We then show that $n\cdot {\aleph }_{\alpha }={\aleph }_{\alpha }$ by constructing a bijective $f:{\omega }_{\alpha }\to n\times {\omega }_{\alpha }$, which clearly shows the result by the definition of cardinal multiplication.

So consider any $\beta \in {\omega }_{\alpha }$. Then, since $n>0$, we have by Theorem 6.6.3 that there is a unique ordinal $\gamma$ and unique natural number $k<n$ such that

where $k<n$ and clearly

since $1\le n$ and $0\le k$. Hence we have $k\in n$ and $\gamma \le \beta <{\omega }_{\alpha }$ so that $\gamma \in {\omega }_{\alpha }$. We then set $f(\beta )=(k,\gamma )\in n\times {\omega }_{\alpha }$.

First we show that $f$ is injective. So consider ${\beta }_1$ and ${\beta }_2$ in ${\omega }_{\alpha }$ where $f({\beta }_1)=f({\beta }_2)$. If we have $f({\beta }_1)=(k_1,{\gamma }_1)$ and $f({\beta }_2)=(k_2,{\gamma }_2)$ then clearly this means that $k_1=k_2$ and ${\gamma }_1={\gamma }_2$. It then clearly follows that

which shows that $f$ is injective.

Now we show that $f$ is also surjective. So consider any $(k,\gamma )\in n\times {\omega }_{\alpha }$ so that $k\in n$ and $\gamma \in {\omega }_{\alpha }$. Then let $\beta =n\cdot \gamma +k$ so that clearly $f(\beta )=(k,\gamma )$. However, we must show that $\beta$ is actually in ${\omega }_{\alpha }$. To see this, first we note that since $\gamma <{\omega }_{\alpha }$ is an ordinal we have $\gamma =\delta +m$ for some limit ordinal $\delta$ and natural number $m$ by Exercise 6.5.4, where clearly $\delta \le \gamma$. Hence we have

where $n\cdot \delta =\delta$ by Lemma 1 since $\delta$ is a limit ordinal. Then since also $\delta \le \gamma <{\omega }_{\alpha }$ and $n\cdot m+k$ is a natural number we clearly have that $\beta =\delta +(n\cdot m+k)<{\omega }_{\alpha }$ as well since ${\omega }_{\alpha }$ is a limit ordinal (by Theorem 7.1.9b). Hence $\beta \in {\omega }_{\alpha }$.

Thus we have shown that $f$ is bijective so that by definition $n\cdot {\aleph }_{\alpha }=\left|n\times {\omega }_{\alpha }\right|=\left|{\omega }_{\alpha }\right|={\aleph }_{\alpha }$ as desired. □