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Exercise 1.3.7
The Axiom of Pair, the Axiom of Union, and the Axiom of Power Set can be replaced by the following weaker versions.
Weak Axiom of Pair: For any and , there is a set such that and .
Weak Axiom of Union: For any , there exists such that if and , then .
Weak Axiom of Power Set: For any set , there exists such that implies .
Prove the Axiom of Pair, the Axiom of Union, and the Axiom of Power Set using these weaker versions. [Hint: Use also the Comprehension Schema.]
Answers
All the below proofs follow the same basic structure:
- 1.
- Establish a property that exactly matches the defining property of the set guaranteed to exist by the normal axiom.
- 2.
- Instantiate the set that exists by the weak form of the axiom.
- 3.
- Show that for all .
- 4.
- This proves that the set exists by the Axiom Schema of Comprehension, which is exactly the set asserted to exist by the normal axiom.
The Axiom of Pair
Proof. Suppose that and are sets and define the property by or . Now let be a set guaranteed to exist by the Weak Axiom of Pair such that and . Now suppose that holds for , hence or . If then since . Similarly, in the other case in which then also since . Hence, in both cases so that implies that . Therefore, the set = exists by the Axiom Schema of Comprehension. Of course is exactly the set guaranteed to exist by the (normal) Axiom of Pair, proving the result. □
The Axiom of Union
Proof. Let be any set and define the property as for some . Now let be a set guaranteed to exist by the Weak Axiom of Union so that, for every , implies that also. Suppose that holds for so that there is a particular such that . Then, since and , it follows from the Weak Axiom of Union that . Hence, we have established that implies that . Therefore, by the Axiom Schema of Comprehension, the set exists, which is exactly the set guaranteed to exist by the (normal) Axiom of Union. □
The Axiom of Power Set
Proof. Let be any set and define the property by . Now let be a set that exists by the Weak Axiom of Power Set so that implies that for all . Next, suppose that holds for so that . Then, by the above, we have that so that clearly implies that . Then the set exists by the Axiom Schema of Comprehension. Of course is exactly the set guaranteed to exist be the (normal) Axiom of Power Set, proving our result. □