Exercise 9.1.4

Prove that κ + κ + ( λ  times) = λ κ .

Answers

Proof. First, if λ = 0 then, by convention, we have

κ + κ + ( λ  times) = κ + κ + (0 times) = 0 = 0 κ = λ κ

regardless of what κ is. So assume in what follows that λ > 0 so λ 1 .

For each α < λ , define A α = { ( α , β ) β κ } . Consider then any α 1 < λ and α 2 < λ where α 1 α 2 . Suppose that both ( x , y ) A α 1 and ( x , y ) A α 2 . It then follows that x = α 1 and x = α 2 so that x = α 1 = α 2 , which contradicts our assumption that α 1 α 2 ! So it must be that no such ( x , y ) exists so that A α 1 and A α 2 are disjoint. Since α 1 and α 2 were arbitrary, this shows that A α α < λ are mutually disjoint sets. We also clearly have that | A α | = κ for each α < λ . It therefore follows from the definition of cardinal summation that

κ + κ + ( λ  times) = α < λ κ = | α < λ A α | .

Now we show that α < λ A α = λ × κ . First consider any ( x , y ) α < λ so that there is an α < λ where ( x , y ) A α . We then have that x = α and y κ . Therefore x = α < λ so that x λ by the definition of < for ordinal numbers. Hence x λ and y κ so that ( x , y ) λ × κ , which shows that α < λ A α λ × κ since ( x , y ) was arbitrary. Now consider any ( x , y ) λ × κ so that x λ and y κ . Let α = x λ so that α < λ . Hence ( x , y ) = ( α , y ) for α < λ and y κ , which shows that ( x , y ) A α so that clearly ( x , y ) α < λ A α . This shows that λ × κ α < λ A α since again ( x , y ) was arbitrary. Thus α < λ A α = λ × κ as desired.

Putting all this together, we have

κ + κ + ( λ  times) = α < λ κ = | α < λ A α | = | λ × κ | = λ κ

by the definition of cardinal multiplication since obviously | λ | = λ and | κ | = κ . □

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2024-07-15 11:42
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