Theorem·T02
There is no set of all sets
Russell's paradox, tamed: no set contains everything.
In words
It is not the case that there is a set V such that every set x is a member of V: there is no universal set.
Never needed: F05 · F10 · F11 · F12 · F13 · A02 · A03 · A04 · A05 · A07 · A08 (computed from the citation graph, not asserted).
Proof
- 1Suppose, for contradiction, that is a set with (negation introduction: we derive falsum from this assumption).
- 2By Separation applied to with the property , the set exists.
- 3Since contains every set, . Hence, by the defining property of : .
- 4A proposition equivalent to its own negation yields falsum: if then , contradiction; if then , contradiction. Either way (excluded middle provides the two cases, disjunction elimination combines them), we reach (falsum). So no such exists.
∎
Remarks
This is Russell's paradox converted from a catastrophe into a theorem. In naive set theory, unrestricted comprehension lets one form
directly, and the contradiction of step 4 destroys the whole system. ZFC's Separation only carves subsets out of already given sets, so the argument no longer produces a paradox; instead it proves that the collection of all sets is not itself a set. Collections too big to be sets (the universe, the ordinals) are called proper classes. Regularity gives an independent proof:
would be a membership cycle, contradicting L05; but the Russell argument needs no regularity at all.