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. Possible extensions