18. Large cardinals

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Case 1 Subcase 2.1 Subcase 2.2 Case 3
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Subcase 4.1 Subcase 4.2 Subcase 4.3
We claim that B is a chain in T of size κ. Suppose that t 0 , t 1 ∈ B with t 0 ≠ t 1 , and choose
x 0 , x 1 ∈ L correspondingly. Say wlog x 0 < ′′ x 1 . Now t 0 ∈ B and x 0 ≤ ′′ x 1 , so t 0 ≤ x 1 .
And t 1 ∈ B and x 1 ≤ x 1 , so t 1 ≤ x 1 . So t 0 and t 1 are comparable.
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} = ⋃
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 x ≤ ′′ a. Choose y ∈ V t with a < ′′ y and
t < y. Then t < x, t < y, and x ≤ ′′ a < ′′ y. If x = a, then t ≤ a, as desired. If x < ′′ a,
then t < a by (*).
This finishes the case in which L has a subset of order type κ. The case of order type
κ ∗ is similar, but we give it. So, suppose that L has order type κ ∗ . Define
V t ;
B = {t ∈ T : ∃x ∈ L∀a ∈ L[a ≤ ′′ x → t ≤ a]}.
We claim that B is a chain in T of size κ. Suppose that t 0 , t 1 ∈ B with t 0 ≠ t 1 , and choose
x 0 , x 1 ∈ L correspondingly. Say wlog x 0 < ′′ x 1 . Now t 0 ∈ B and x 0 ≤ x 0 , so t 0 ≤ x 0 . and
t 1 ∈ B and x 0 ≤ ′′ x 1 , so t 1 ≤ x 0 . So t 0 and t 1 are comparable.
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