Lemma·L05
No membership cycles
No set belongs to itself, and no two sets belong to each other.
(i)
; (ii)
.
In words
No set is a member of itself, and there are no two sets each of which is a member of the other.
Never needed: F05 · F08 · F09 · F10 · F11 · F12 · F13 · F14 · A01 · A02 · A03 · A04 · A05 · A06 · A07 (computed from the citation graph, not asserted).
Proof
- 1(i) Suppose for some . The singleton exists by pairing and is non-empty, so Regularity applies: there is such that no member of belongs to . But forces , and has the member (assumption), which does belong to : contradiction (negation). So .
- 2(ii) Suppose and . Apply Regularity to the non-empty pair of D003: some has no member inside . If : then is a member of lying in , contradiction. If : then is a member of lying in , contradiction. Both cases of the disjunction fail, so no such exist.
∎
Remarks
Regularity in fact forbids membership cycles of every finite length and infinite descending membership chains; the two shortest cases proved here are the ones the arithmetic development needs. Part (i) is what makes the order on the naturals irreflexive, and part (ii) gives the injectivity of the successor in the Peano axioms. Note that (i) yields another proof that there is no universal set (T02): such a set would contain itself.