Some PST Theorems
- 1. The Russell class is a proper class.
- Proof. cannot be a set by Russell's paradox. ∎
- 2. The empty class is a set.
- Proof. Suppose (towards a contradiction) that is a proper class. By (A4), must be equinumerous with, in which case is empty. Let i be an infinite set, and consider the class . It is not equinumerous with, thus it is a set. It is finite, but its single element is infinite, thus it cannot be an element of itself. Therefore, it is an element of . This contradicts that is empty. ∎
- 3. The singleton class is a set.
- Proof. Suppose that is a proper class. Then by (A4), every proper class is a singleton. Let i be an infinite set and consider the class . It is neither a proper class (because it is not singleton) nor an element of itself (because it is neither empty nor infinite). Thus holds by definition, so has at least two elements, and . This contradicts the initial assumption that proper classes are singletons. ∎
- 4. is infinite.
- Proof. Let . Suppose that this class is a set. Then either or . In the first case, the definition of implies that, from, which it follows that, a contradiction. In the second case, the definition of implies either and hence, a contradiction, or . But cannot be empty because it has at least one element, namely . ∎
- 5. Every finite class is a set.
- Proof. Let X be a proper class. By (A4), there exists an such that F is a bijection. This contains a pair, and for each member r of, a pair . Let and . By (A4), both of these classes exist. Now, is a bijection. Thus by (A4), is a proper class, too. Clearly, and . Now, another application of (A4) shows that there exists a bijection . This proves that X is infinite. ∎
Once the above facts are settled, the following results can be proved:
- 6. The class V of sets consists of all hereditarily countable sets.
- 7. Every proper class has the cardinality .
- Proof. Let i be an infinite set, in which case the class has cardinality . By (A4), all proper classes have cardinality . ∎
- 8. The union class of a set is a set.
PST also verifies the:
- Continuum hypothesis. This follows from (5) and (6) above;
- Axiom of replacement. This is a consequence of (A4);
- Axiom of choice. Proof. The class Ord of all ordinals is well-ordered by definition. Ord and the class V of all sets are both proper classes, because of the Burali-Forti paradox and Cantor's paradox, respectively. Therefore there exists a bijection between V and Ord, which well-orders V. ∎
The well-foundedness of all sets is neither provable nor disprovable in PST.
Read more about this topic: Pocket Set Theory