Homepage › Solution manuals › Karel Hrbáček › Introduction to Set Theory › Exercise 1.4.6
Exercise 1.4.6
Prove that exists for all . Where is the assumption that used in the proof?
Answers
Proof. Suppose that so that there is a set (since sets are all that exist by the axioms). We then define the set
noting that this exists by the Axiom Schema of Comprehension. The uniqueness of this set follows immediately from the Axiom of Extensionality. □
The non-emptiness of was used to produce a set to which the Axiom Schema of Comprehension could be applied to further limit it. This works because the intersection of all the sets in (i.e. ) is a subset of every set in . Hence, with this proof, being nonempty is required in order for to exist.
Note that we could have defined it and shown its existence differently and a little more generally:
Proof. Suppose that is any set (which could possibly be empty). By the Axiom of Union, the set exists so that the set
exists by the Axiom Schema of Comprehension, where its uniqueness again follows immediately from the Axiom Schema of Comprehension. □
The intersection defined this way exists even if , and is in fact empty in this case since is empty. Mathematicians have chosen the former definition in which exists only when nonempty. The reason is that we would like to represent the naive set
but this is problematic if is empty because the defining property is then vacuously true for any so that would therefore contain every set, which we know results in a contradiction.