Exercise 7.2.6

Proof. We have

If $\lambda$ is finite then clearly $\lambda \cdot 2$ is finite so that $\kappa +\lambda \le \lambda \cdot 2<{\aleph }_{\alpha }$. On the other hand if $\lambda$ is infinite then $\lambda ={\aleph }_{\beta }$ where $\beta <\alpha$ since ${\aleph }_{\beta }=\lambda <{\aleph }_{\gamma }$. Thus we have $\kappa +\lambda \le \lambda \cdot 2=2\cdot \lambda =2\cdot {\aleph }_{\beta }={\aleph }_{\beta }<{\aleph }_{\alpha }$ by Corollary 7.2.2. Hence in either case $\kappa +\lambda <{\aleph }_{\alpha }$ as desired. □

Proof. First, we clearly have that ${\omega }_{\alpha }-X$ and $X$ are disjoint and $({\omega }_{\alpha }-X)\cup X={\omega }_{\alpha }$ so that, by the definition of cardinal addition, we have

Now suppose that $\left|{\omega }_{\alpha }-X\right|<{\aleph }_{\alpha }$ Since ${\omega }_{\alpha }-X\subseteq {\omega }_{\alpha }$ and $X\subseteq {\omega }_{\alpha }$ are both sets of ordinals, they are clearly both well ordered by $<$. Thus $\left|{\omega }_{\alpha }-X\right|\le \left|X\right|$ or $\left|X\right|\le \left|{\omega }_{\alpha }-X\right|$ by Lemma ??. Since we also have $\left|{\omega }_{\alpha }-X\right|<{\aleph }_{\alpha }$ and $\left|X\right|<{\aleph }_{\alpha }$, in either case it follows from Lemma 1 that

which is clearly a contradiction. Thus it must be that $\left|{\omega }_{\alpha }-X\right|\nless {\aleph }_{\alpha }$ so that ${\aleph }_{\alpha }\le \left|{\omega }_{\alpha }-X\right|$ by Corollary ??. Since ${\omega }_{\alpha }-X\subseteq {\omega }_{\alpha }$ we clearly also have that $\left|{\omega }_{\alpha }-X\right|\le \left|{\omega }_{\alpha }\right|={\aleph }_{\alpha }$. Hence $\left|{\omega }_{\alpha }-X\right|={\aleph }_{\alpha }$ by the Cantor-Bernstein Theorem. □