Exercise 1.4.6

Prove that S exists for all S . Where is the assumption that S used in the proof?

Answers

Proof. Suppose that S so that there is a set X S (since sets are all that exist by the axioms). We then define the set

S = { x X x A  for every  A S } ,

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 S 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 S (i.e. S ) is a subset of every set in S . Hence, with this proof, S being nonempty is required in order for S to exist.

Note that we could have defined it and shown its existence differently and a little more generally:

Proof. Suppose that S is any set (which could possibly be empty). By the Axiom of Union, the set S exists so that the set

S = { x S x A  for every  A S } ,

exists by the Axiom Schema of Comprehension, where its uniqueness again follows immediately from the Axiom Schema of Comprehension. □

The intersection S defined this way exists even if S = , and is in fact empty in this case since S is empty. Mathematicians have chosen the former definition in which S exists only when S nonempty. The reason is that we would like S to represent the naive set

S = { x x A  for every  A S } ,

but this is problematic if S is empty because the defining property is then vacuously true for any x so that S would therefore contain every set, which we know results in a contradiction.

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2024-07-15 11:42
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