Exercise 6.3.1

Let 𝐏 ( x , y ) be a property such that for every x there is at most one y for which 𝐏 ( x , y ) holds. Then for every set A there is a set B such that, for all x A , if 𝐏 ( x , y ) holds for some y , then 𝐏 ( x , y ) holds for some y B .

Answers

Proof. Define a property 𝐑 ( x , y ) such that 𝐑 ( x , y ) holds if and only if

1.
𝐏 ( x , y ) holds, or
2.
y = and there is not a z such that 𝐏 ( x , z ) holds.

Clearly this property is such that for every x there is a unique y for which 𝐑 ( x , y ) holds.

Now consider any set A . Then by the Axiom Schema of Replacement there is a set B such that, for every x A , there is a y B for which 𝐑 ( x , y ) holds. Consider any x A . Then by the above there is a y B such that 𝐑 ( x , y ) holds. Now suppose that 𝐏 ( x , z ) holds for some z . Then option 2 above cannot be the case so that, 𝐏 ( x , y ) holds (option 1) since 𝐑 ( x , y ) does. Thus 𝐏 ( x , y ) holds for some y B as we were required to show. □

User profile picture
2024-07-15 11:42
Comments