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Exercise 9.1.4
Prove that .
Answers
Proof. First, if then, by convention, we have
regardless of what is. So assume in what follows that so .
For each , define . Consider then any and where . Suppose that both and . It then follows that and so that , which contradicts our assumption that ! So it must be that no such exists so that and are disjoint. Since and were arbitrary, this shows that are mutually disjoint sets. We also clearly have that for each . It therefore follows from the definition of cardinal summation that
Now we show that . First consider any so that there is an where . We then have that and . Therefore so that by the definition of for ordinal numbers. Hence and so that , which shows that since was arbitrary. Now consider any so that and . Let so that . Hence for and , which shows that so that clearly . This shows that since again was arbitrary. Thus as desired.
Putting all this together, we have
by the definition of cardinal multiplication since obviously and . □