Metamath Proof Explorer

Theorem inf3

Description: Our Axiom of Infinity ax-inf implies the standard Axiom of Infinity. The hypothesis is a variant of our Axiom of Infinity provided by inf2 , and the conclusion is the version of the Axiom of Infinity shown as Axiom 7 in TakeutiZaring p. 43. (Other standard versions are proved later as axinf2 and zfinf2 .) The main proof is provided by inf3lema through inf3lem7 , and this final piece eliminates the auxiliary hypothesis of inf3lem7 . This proof is due to Ian Sutherland, Richard Heck, and Norman Megill and was posted on Usenet as shown below. Although the result is not new, the authors were unable to find a published proof. 

       (As posted to sci.logic on 30-Oct-1996, with annotations added.)


       Theorem:  The statement "There exists a nonempty set that is a subset
       of its union" implies the Axiom of Infinity.


       Proof:  Let X be a nonempty set which is a subset of its union; the
       latter
       property is equivalent to saying that for any y in X, there exists a z
       in X
       such that y is in z.


       Define by finite recursion a function F:omega-->(power X) such that
       F_0 = 0  (See inf3lemb .)
       F_n+1 = {y<X | y^X subset F_n}  (See inf3lemc .)
       Note: ^ means intersect, < means \in ("element of").
       (Finite recursion as typically done requires the existence of omega;
       to avoid this we can just use transfinite recursion restricted to omega.
       F is a class-term that is not necessarily a set at this point.)


       Lemma 1.  F_n subset F_n+1.  (See inf3lem1 .)
       Proof:  By induction:  F_0 subset F_1.  If y < F_n+1, then y^X subset
       F_n,
       so if F_n subset F_n+1, then y^X subset F_n+1, so y < F_n+2.


       Lemma 2.  F_n =/= X.  (See inf3lem2 .)
       Proof:  By induction:  F_0 =/= X because X is not empty.  Assume F_n =/=
       X.
       Then there is a y in X that is not in F_n.  By definition of X, there is
       a
       z in X that contains y.  Suppose F_n+1 = X.  Then z is in F_n+1, and z^X
       contains y, so z^X is not a subset of F_n, contrary to the definition of
       F_n+1.


       Lemma 3.  F_n =/= F_n+1.  (See inf3lem3 .)
       Proof:  Using the identity y^X subset F_n <-> y^(X-F_n) = 0, we have
       F_n+1 = {y<X | y^(X-F_n) = 0}.  Let q = {y<X-F_n | y^(X-F_n) = 0}.
       Then q subset F_n+1.  Since X-F_n is not empty by Lemma 2 and q is the
       set of \in-minimal elements of X-F_n, by Foundation q is not empty, so q
       and therefore F_n+1 have an element not in F_n.


       Lemma 4.  F_n proper_subset F_n+1.  (See inf3lem4 .)
       Proof:  Lemmas 1 and 3.


       Lemma 5.  F_m proper_subset F_n, m < n.  (See inf3lem5 .)
       Proof:  Fix m and use induction on n > m.  Basis: F_m proper_subset
       F_m+1
       by Lemma 4.  Induction:  Assume F_m proper_subset F_n.  Then since F_n
       proper_subset F_n+1, F_m proper_subset F_n+1 by transitivity of proper
       subset.


       By Lemma 5, F_m =/= F_n for m =/= n, so F is 1-1.  (See inf3lem6 .)
       Thus, the inverse of F is a function with range omega and domain a
       subset
       of power X, so omega exists by Replacement.  (See inf3lem7 .)
       Q.E.D.
(Contributed by NM, 29-Oct-1996)

Ref Expression
Hypothesis inf3.1 ⊢ ∃ 𝑥 ( 𝑥 ≠ ∅ ∧ 𝑥 ⊆ ∪ 𝑥 )
Assertion inf3 ω ∈ V

Proof

Step Hyp Ref Expression
1 inf3.1 ⊢ ∃ 𝑥 ( 𝑥 ≠ ∅ ∧ 𝑥 ⊆ ∪ 𝑥 )
2 eqid ⊢ ( 𝑦 ∈ V ↦ { 𝑤 ∈ 𝑥 ∣ ( 𝑤 ∩ 𝑥 ) ⊆ 𝑦 } ) = ( 𝑦 ∈ V ↦ { 𝑤 ∈ 𝑥 ∣ ( 𝑤 ∩ 𝑥 ) ⊆ 𝑦 } )
3 eqid ⊢ ( rec ( ( 𝑦 ∈ V ↦ { 𝑤 ∈ 𝑥 ∣ ( 𝑤 ∩ 𝑥 ) ⊆ 𝑦 } ) , ∅ ) ↾ ω ) = ( rec ( ( 𝑦 ∈ V ↦ { 𝑤 ∈ 𝑥 ∣ ( 𝑤 ∩ 𝑥 ) ⊆ 𝑦 } ) , ∅ ) ↾ ω )
4 vex ⊢ 𝑥 ∈ V
5 2 3 4 4 inf3lem7 ⊢ ( ( 𝑥 ≠ ∅ ∧ 𝑥 ⊆ ∪ 𝑥 ) → ω ∈ V )
6 5 1 exlimiiv ⊢ ω ∈ V