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Lemma·L06

ω is a transitive set

Every member of a natural number is a natural number.

is a transitive set:
In words
Omega is transitive: for every n, if n is a natural number then every member of n is itself a natural number.
Rests onF02F09F14A01A06
Never needed: F03 · F04 · F05 · F06 · F08 · F10 · F11 · F12 · F13 · A02 · A03 · A04 · A05 · A07 · A08 (computed from the citation graph, not asserted).

Proof

  1. 1
    Let , a set by Separation. We show by induction.
  2. 2
    Base: holds vacuously (D002, D001), so .
  3. 3
    Step: let , so . A member of (D006) is either a member of , hence in because , or equals itself, which is in . So , giving .
  4. 4
    By T05, : every satisfies , which is transitivity of .
∎

Remarks

Concretely: the members of are natural numbers, and likewise for every natural. This closure property is used silently whenever an argument passes from a natural to one of its members: the member is again a natural, so induction hypotheses and definitions on apply to it. Together with L07 it makes the membership order on the naturals well behaved.
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Used by

L15T20
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