Homepage › Solution manuals › Karel Hrbáček › Introduction to Set Theory › Exercise 6.3.1
Exercise 6.3.1
Let be a property such that for every there is at most one for which holds. Then for every set there is a set such that, for all , if holds for some , then holds for some .
Answers
Proof. Define a property such that holds if and only if
- 1.
- holds, or
- 2.
- and there is not a such that holds.
Clearly this property is such that for every there is a unique for which holds.
Now consider any set . Then by the Axiom Schema of Replacement there is a set such that, for every , there is a for which holds. Consider any . Then by the above there is a such that holds. Now suppose that holds for some . Then option 2 above cannot be the case so that, holds (option 1) since does. Thus holds for some as we were required to show. □