On changing cofinality of partially ordered sets

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Clearly, D is a dense subset of Q. So, G ∩ D ≠ ∅. Pick some q ∈ G ∩ D, q ≥ p(0). Then q is incompatible with each f p(1) (ν), ν ∈ A. Hence j 1 (q) is incompatible with p(1). This contradicts Definition 4.12(3), since the projection of p(1) to Q is weaker than p(0). Suppose now that η ∈ T (⃗p), η ≠ 〈〉. Let us show that the set Suc T (⃗p) (η) is U |η|+1 -positive. Suppose otherwise. Then there is A ∈ U |η|+1 such that for every ν ∈ A we have f p(|η|+1) (η ⌢ ν) ∉ G. Note that f p(|η|) (η) ∈ G by the definition of T (⃗p). Consider (in V ) the following set D = {q ∈ Q | ∃ν ∈ A q ≥ f p,η (ν) or ∀ν ∈ A q, f p,η (ν) are incompatible }. Clearly, D is a dense subset of Q. So, G ∩ D ≠ ∅. Pick some q ∈ G ∩ D, q ≥ f p(|η|) (η). Then q is incompatible with each f p,η (ν), ν ∈ A. Hence j |η|+1 (q) is incompatible with [f p,η ] U|η|+1 . This contradicts Lemma 4.13, Definition 4.12(3), since the projection of [f p,η ] U|η|+1 to Q is weaker than f p(|η|) (η). Definition 4.15 We call a tree T ⊆ [κ]
Definition 4.18 Let 〈η, T (⃗p)〉, 〈η ′ , T ′ (⃗p ′ )〉 ∈ P . Set 〈η, T (⃗p)〉 ≤ 〈η ′ , T ′ (⃗p ′ )〉 iff 1. η ′ ∈ T (⃗p), 2. η ′ is an end extension of η, 3. T ′ (⃗p ′ ) ⊆ T (⃗p), 4. ⃗p η ′ ≤ ⃗p ′ . Definition 4.19 Let 〈η, T 〉, 〈η ′ , T ′ 〉 ∈ P . Set 〈η, T 〉 ≤ ∗ 〈η ′ , T ′ 〉 iff 1. 〈η, T 〉 ≤ 〈η ′ , T ′ 〉, 2. η = η ′ . Lemma 4.20 Let 〈p(0), p(1), ..., p(n), ... | n < ω〉 be a good sequence and q ′ (0) ≥ p(0). Then there are 〈q(0), q(1), ..., q(n), ... | n < ω〉 such that 1. q(0) ≥ q ′ (0), 2. 〈q(0), q(1), ..., q(n), ... | n < ω〉 is a good sequence, 3. for each n < ω, q(n) ≥ p(n), Proof. Combine j ≤n (q ′ (0)) with p(n), for each n < ω, and turn the results into separated conditions. Let us argue that (6) of Definition 4.12 can be easily satisfied. Thus, we need to take care of α’s in a 0 (q ′ (0)) \ a 0 (p(0)) such that j ≤l (α) does not appear in a n(α) (p(l)), for 1 ≤ l < ω. Given such α, we consider k’s above n(α). Then j ≤k (α) ∉ a n(α) (p(k)). But then we are free to set 〈j ≤k (α), κ k 〉 ∉ ⊤ q(k),n(α) . □ Denote the result for each n < ω, by q(n). By the construction we have q(n) ≥ p(n). The following is a slightly more general statement with a similar proof. Lemma 4.21 Let 〈p(0), p(1), ..., p(n), ... | n < ω〉 be a good sequence, l < ω and q ′ (m) ≥ p(m), for every m ≤ l. Then there are 〈q(0), ..., q(n), ... | n < ω〉 such that 1. q(0) = q ′ (0), 2. q(m) ≥ q ′ (m), for every m ≤ l, 23
Page 1 and 2: On changing cofinality of partially
Page 3 and 4: as K V -the one computed in V . Jus
Page 5 and 6: It is a subset of P V (A n ) consis
Page 7 and 8: Consider (again in V ) the followin
Page 9 and 10: Force now a new ω sequence to each
Page 11 and 12: Now, for each α ∈ [κ, κ + ), w
Page 13 and 14: 3. the sequence 〈ν 0 , ν 1 , ..
Page 15 and 16: 2. n < m implies a n (q) ⊆ a m (q
Page 17 and 18: I.e. it is the set of all elements
Page 19 and 20: 4. β ∈ j ≤l,≤n (a 0 (q)) \ a
Page 21: 1. for every n, 0 < n < ω, [f p(n)
Page 25 and 26: Proof. Let σ be a statement and
Page 27 and 28: Lemma 4.24 〈P, ≤ 〉 preserves

On changing cofinality of partially ordered sets

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