New Foundations is consistent

4.2 Extensions of approximations

Definition 4.10
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We define a partial order on base approximations by setting \( \psi \leq \chi \) when \( \psi ^{E{\mathcal A}} = \chi ^{E{\mathcal A}} \) and \( \psi ^{\mathcal L}\leq \chi ^{\mathcal L}\).

Proposition 4.11 adding orbits
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Let \( \psi \) be a base approximation, and let \( L : \mathbb N \to {\mathcal L}\) be a function such that

\[ L(m) = L(n) \to L(m+k) = L(n+k) \]

for all integers \( m, n, k : \mathbb Z \). Suppose that for all \( n \), \( L(n) \notin \operatorname {\mathsf{coim}}\psi ^{\mathcal L}\). Then there is an extension \( \chi \geq \psi \) such that \( \chi ^{\mathcal L}(L(n)) = L(n+1) \) and \( \operatorname {\mathsf{coim}}\chi ^{\mathcal L}= \operatorname {\mathsf{coim}}\psi ^{\mathcal L}\cup \operatorname {\mathsf{ran}}L \).

Proof

Define the relation

\[ R = \{ (L(n), L(n+1)) \mid n : \mathbb Z \} \]

This clearly has equal image and coimage. It is injective: if \( (L_1, L_3), (L_2, L_3) \in R \), then there are \( m, n : \mathbb Z \) such that

\[ L_1 = L(m);\quad L_3 = L(m + 1);\quad L_2 = L(n);\quad L_3 = L(n + 1) \]

So \( L(m + 1) = L(n + 1) \), giving \( L_1 = L(m) = L(n) = L_2 \) by substituting \( k = -1 \) in the hypothesis. It is also coinjective by substituting \( k = 1 \) in the hypothesis. So \( R \) is permutative. Therefore, \( \psi ^{\mathcal L}\sqcup R \) is a permutative relation, so \( (\psi ^{E{\mathcal A}}, \psi ^{\mathcal L}\sqcup R) \) is an extension of \( \psi \), and it clearly satisfies the result.