18. Large cardinals

Clearly (V κ , ∈, U) |= σ. Suppose that α < κ and (V α , ∈, V α ∩ U) |= σ. Then α is a limit
ordinal, and there is an ordinal γ < α and a function g with domain γ such that V α ∩ U
consists of all triples (γ, β, g(β)) with β < γ. (Some absoluteness is used.) Now V α ∩ U
is nonempty; choose (γ, β, g(β)) in it. Then γ = λ since it is in U. It follows that g = f.
Choose β < λ such that α < f(β). Then (λ, β, f(β)) ∈ U ∩ V α . Since α < f(β), it follows
that α has rank less than α, contradiction.
Now suppose that λ < κ ≤ 2 λ . A contradiction is reached similarly, as follows. Let f
be a function whose domain is P(λ) with range κ. Let U = {(λ, B, f(B)) : B ⊆ λ}. Let
σ be the sentence expressing the following:
For every ordinal γ there is an ordinal δ with γ < δ, U is nonempty, and there is an ordinal
µ and a function g with domain P(µ) such that U consists of all triples (µ, B, g(B)) with
B ⊆ µ.
Clearly (V κ , ∈, U) |= σ. Suppose that α < κ and (V α , ∈, V α ∩ U) |= σ. Then α is a limit
ordinal, and there is an ordinal γ < α and a function g with domain P(γ) such that
V α ∩U consists of all triples (γ, B, g(B)) with B ⊆ γ. (Some absoluteness is used.) Clearly
γ = λ; otherwise U ∩ V α would be empty. Note that g = f. Choose B ⊆ λ such that
α = f(B). Then (λ, B, f(B)) ∈ U ∩ V α . Again this implies that α has rank less than α,
contradiction.
The new equivalent of weak compactness involves second-order logic. We augment first
order logic by adding a new variable S ranging over subsets rather than elements. There
is one new kind of atomic formula: Sv with v a first-order variable. This is interpreted as
saying that v is a member of S.
Now an infinite cardinal κ is Π 1 1 -indescribable iff for every U ⊆ V κ and every secondorder
sentence σ of the form ∀Sϕ, with no quantifiers on S within ϕ, if (V κ , ∈, U) |= σ,
then there is an α < κ such that (V α , ∈, U ∩ V α ) |= σ. Note that if κ is Π 1 1-indescribable
then it is not first-order describable.
Theorem 18.15. An infinite cardinal κ is weakly compact iff it is Π 1 1 -indescribable.
Proof. First suppose that κ is Π 1 1-indescribable. By Theorem 18.14 it is inaccessible.
So it suffices to show that it has the tree property. By the proof of Theorem 18.7(iii)⇒(iv)
it suffices to check the tree property for a tree T ⊆