18. Large cardinals

Now let α < κ; we show that B has an element of height α. For each t of height α let
V t = {x ∈ L : t ≤ x}. Then
{x ∈ L : ht(x) ≥ α} = ⋃
V t ;
ht(t)=α
since there are fewer than κ elements of height less than κ, this set has size κ, and so there
is a t such that ht(t) = α and |V t | = κ. We claim that t ∈ B. To prove this, take any
x ∈ V t such that t < x. Suppose that a ∈ L and a ≤ ′′ x. Choose y ∈ V t with y < ′′ a and
t < y. Then t < x, t < y, and y < ′′ a ≤ ′′ x. If a = x, then t < a, as desired. If a < ′′ x,
then t < a by (*).
Theorem 18.7. For any uncountable cardinal κ the following conditions are equivalent:
(i) κ is weakly compact.
(ii) κ is inaccessible, and it has the linear order property.
(iii) κ is inaccessible, and it has the tree property.
(iv) For any cardinal λ such that 1 < λ < κ we have κ → (κ) 2 λ .
Proof. (i)⇒(ii): Assume that κ is weakly compact. First we need to show that κ is
inaccessible.
To show that κ is regular, suppose to the contrary that κ = ∑ α