A QUESTION ON INDISCERNIBLES

The question is considered, whether for some limit ordinal α, Lα has an infinite set of indiscernibles. This is true if α is an ω-Erdos cardinal. Whether the hypothesis can be weakened is a question of interest. AMS Subj. Classification: 03e55


Introduction
Let II denote the statement: for some limit ordinal α, L α has an infinite set of indiscernibles (ordinals equipped with their natural order).It is well-known that if there is an ω-Erdos cardinal (a cardinal κ such that κ → (ω) <ω ) then II holds (see theorem 9.3 of [2]).In particular ¬II is a very strong statement, implying that ω-Erdos cardinals do not exist.
It is a question of interest whether II be deduced from a weaker hypothesis than the existence of an ω-Erdos cardinal.It is also of interest what properties α must have for L α to have indiscernibles.
It is also of interest whether II L holds.Since α → L α and the satisfaction predicate are absolute, II L holds iff there as a limit ordinal α and a set I ∈ L such that I is a set of indiscernibles for L α .
Theorem 1.If II L holds then II holds.
Proof.This follows by the remarks preceding the theorem.Since theorem 9.3 of [2] holds in L, II L holds if there is an ω-Erdos cardinal in L, and this holds if there is an ω-Erdos cardinal (theorem 9.15 of [2]),

Basic Facts
It is well-known (see [1]) that there is a collection of function definitions {h φ } such that h φ defines a Skolem function for φ in L α for any limit ordinal α.The function defined in L α will be denoted h Lα φ , or h φ if there is no danger of confusion.The Skolem hull of S ⊆ L α will always be taken using these functions, and denoted H(S).
Let I be a set of indiscernibles for L α .For S ⊆ L α the transitive collapse of H(S) is isomorphic to L α for some α; the composition j : L α → L α of the isomorphism with inclusion is an elementary embedding.Consequently, j −1 [S] is a set of indiscernibles for L α.
Theorem 2. If II holds then there is a countable α such that L α has an infinite set of indiscernibles I, and such that L α = H(I).
Proof.Let J be a set of indiscernibles for L β .Let S be the first ω elements of J. Let L α be the transitive collapse of H(S).
Theorem 3. If II holds then it is not provable in ZFC that II implies the existence of inaccessible cardinals.
Proof.By theorem 2 and absoluteness, if II holds then it holds in V κ where κ is the smallest inaccessible.If it were provable that II implied that an inaccessible cardinal existed, then an inaccessible cardinal would exist in V κ , which is a contradiction.Theorem 4. If II L holds then there is an α < ω L 1 such that L α has an infinite set of indiscernibles I ∈ L, and such that L α = H(I).
Proof.The proof of theorem 2 is an argument in ZFC.Note that by absoluteness H(S) is the same in L and V .
Theorem 5.If II L holds then it is not provable in ZFC that II L implies the existence of inaccessible cardinals in L.
Proof.As in the proof of theorem 3, if II L holds then it holds in L κ where κ is the smallest inaccessible in L.

F n -Indiscernibles
Let F be the class of augmented formulas in the language of set theory expanded by symbols for the Skolem functions, where an augmented formula φ(x 1 , . . ., x n ) is a formula φ together with a sequence x 1 , . . ., x n of variables, which includes the free variables of φ.For C ⊆ F and α a limit ordinal, a subset I ⊆ α is said to be a set of C-indiscernibles for L α if for all φ(x 1 , . . ., x n ) ∈ C, and sequences Let F n denote the formulas of F , where the variable sequence has length at most n.For a cardinal κ and an integer n let IE(κ, n) be defined by the recursion: IE(κ, 0) = κ, IE(κ, n + 1) = 2 IE(κ,n) .Theorem 6.For an integer n > 0, L κ has a set of F n -indiscernibles of order type (2 ℵ 0 ) + where κ = IE(ℵ 0 , n) + .

Atomic Formulas
Let A be the set of atomic formulas of F , and let A n be the set of atomic formulas of F n .Theorem 7. A set of A-indiscernibles for L α is a set of F -indiscernibles.A set of A n -indiscernibles for L α is a set of F n -indiscernibles.
Proof.Let I be a set of A-indiscernibles.By induction on the formation of φ, I is a set of indiscernibles for φ.This follows by hypothesis for atomic formulas.The induction step for a propositional connective is straightforward.For φ = ∃yψ(y, x), inductively I is a set of indiscernibles for ψ(h ψ ( x), x), and hence for φ.