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Exercise 5.1.3
Show that for all and for all .
Answers
Proof. Suppose that for a set .
Note first that if then by Exercise 5.1.2 it follows that
In the case where we claim that there is a unique so that clearly then
For existence define by for all so that clearly . For uniqueness consider any . Since it has to be that for all . Hence .
Now suppose also that so that . Hence there is an . We claim that in this case that so that
So suppose that so that there is an . Then it’s true that for every there is a unique such that . But since there is an this implies that there is also a , which is a contradiction. Hence there can be no so that . □