Description: Our Axiom of Infinity ax-inf 9103 implies the standard Axiom of Infinity.
The hypothesis is a variant of our Axiom of Infinity provided by
inf2 9088, 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 9105 and zfinf2 9107.) The main proof is provided by
inf3lema 9089 through inf3lem7 9099, and this final piece eliminates the
auxiliary hypothesis of inf3lem7 9099. 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 9090.)
F_n+1 = {y<X | y^X subset F_n} (See inf3lemc 9091.)
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 9093.)
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 9094.)
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 9095.)
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 9096.)
Proof: Lemmas 1 and 3.
Lemma 5. F_m proper_subset F_n, m < n. (See inf3lem5 9097.)
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 9098.)
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 9099.)
Q.E.D.
(Contributed by NM, 29-Oct-1996.) |
