Michael Rathjen
580 California St., Suite 400
San Francisco, CA, 94104
On the constructive Dedekind reals
2008, Logic and Analysis
https://doi.org/10.1007/S11813-007-0005-6
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Abstract
In order to build the collection of Cauchy reals as a set in constructive set theory, the only Power Set-like principle needed is Exponentiation. In contrast, the proof that the Dedekind reals form a set has seemed to require more than that. The main purpose here is to show that Exponentiation alone does not suffice for the latter, by furnishing a Kripke model of constructive set theory, CZF with Subset Collection replaced by Exponentiation, in which the Cauchy reals form a set while the Dedekind reals constitute a proper class.
References (27)
- J ∀x φ(x) iff for all r ∈ J and σ there is a J ′ containing r such that J ∩ J ′ φ(σ), and for all r ∈ J and σ there is a J ′ containing r such that J ′ φ r (σ).
- If J' ⊆ J φ then J' φ.
- If J i φ for all i then i J i φ.
- J φ iff for all r ∈ J there is a J' containing r such that J ∩ J' φ.
- For all φ, J if J φ then for all r ∈ J R φ r .
- If φ contains only ground model terms, then either R φ or R ¬φ. proof: 1. Trivial induction, as before.
- Again, a trivial induction.
- By induction. As in the previous section, for the case of →, you need to invoke the previous part of this lemma. All other cases are straightforward.
- By induction on φ. Base cases: = and ∈: Trivial from the definitions of forcing = and ∈. ∨ and ∧: Trivial induction. →: Suppose J φ → ψ and r ∈ J. We must show that R φ r → ψ r . For the first clause, suppose K ⊆ R and K φ r . If K = ∅ then K ψ r . Else let s ∈ K. Inductively, since s ∈ K φ r , R (φ r ) s . But (φ r ) s = φ r , so R φ r . Using the hypothesis on J, R ψ r , and so by part 2 above, K ψ r . For the second clause, let s ∈ R. If R (φ r ) s then R φ r . By the hypothesis on J, R ψ r , and ψ r = (ψ r ) s . ∃: If J ∃x φ(x) and r ∈ J, let J ′ and σ be such that r ∈ J ∩ J ′ φ(σ). By induction, R φ r (σ r ). σ r witnesses that R ∃x φ r (x). ∀: Let J ∀x φ(x) and r ∈ J. We need to show that R ∀x φ r (x). For the first clause, we will show that for any σ, R φ r (σ). By part 4 above, it suffices to let s ∈ R be arbitrary, and find a J ′ containing s such that J ′ φ r (σ). By the hypothesis on J, for every τ there is a J ′′ containing r such that J ′′ φ r (τ ). Introducing new notation here, let τ be shif t r-s σ, which is σ with all the intervals shifted by r -s hereditarily. So we have r ∈ J ′′ φ r (shif t r-s σ). Now shift by s -r. Letting J ′ be the image of J ′′ , note that s ∈ J ′ , the image of shif t r-s σ is just σ, and the image of φ r is just φ r . Since the forcing relation is unaffected by this shift, we have s ∈ J ′ φ r (σ), as desired. The second clause follows by the same argument. Given any s ∈ R and σ, we need to show that there is a J ′ containing s such that J ′ (φ r ) s (σ). But (φ r ) s = φ r , and we have already shown that for all σ R φ r (σ). References
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