Exercise 5.1.3

Show that 1 κ = 1 for all κ and 0 κ = 0 for all κ > 0 .

Answers

Proof. Suppose that κ = | A | for a set A .

Note first that if κ = 0 then by Exercise 5.1.2 it follows that

1 κ = 1 0 = 1 .

In the case where κ > 0 we claim that there is a unique f 1 A so that clearly then

1 κ = | 1 A | = 1 .

For existence define f : A 1 by f ( x ) = for all x A so that clearly f 1 A . For uniqueness consider any f 1 , f 2 1 A . Since 1 = { } it has to be that f 1 ( x ) = f 2 ( x ) = for all x A . Hence f 1 = f 2 .

Now suppose also that κ > 0 so that A . Hence there is an a A . We claim that in this case that A = so that

0 κ = | A | = | | = 0 .

So suppose that A so that there is an f A . Then it’s true that for every a A there is a unique b such that f ( a ) = b . But since there is an a A this implies that there is also a b , which is a contradiction. Hence there can be no f A so that A = . □

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2024-07-15 11:42
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