Equivalence of All Constructions of R

 Equivalence of All Constructions of R

 Having made the effort to abstract the properties of the integers, the rationals and the real numbers, we are now in a position to reap their fruits: we can show that all constructions of R are isomorphic, that is are in a one-toone correspondence which preserves their algebraic properties (addition and multiplication and their associated properties), their order properties, and their least upper bound properties. That is if R and R 0 are two constructions,

 then for all a, b ∈ R and a 0 , b0 ∈ R 0 with a ↔ a 0 and b ↔ b 0 we have a + b ↔ a 0 + b 0 , ab ↔ a 0 b 0 and a ≤ b ⇐⇒ a 0 ≤0 b 0 , 

while for all S ⊆ R and S 0 ⊆ R 0 with S ↔ S 0 we have sup S ↔ sup S 0 . Formally, Theorem 4.1 Every complete totally ordered field is both isomorphic and order-isomorphic to R, so in this sense all constructions of R are equivalent.

 Proof : In what follows, suppose R is any construction of the reals satisfying the axioms given in Construction 1 of above, and let (F, ≤0 , −0 , −10 , +0 , · 0 , 1 0 , 0 0 ) be any other construction of R, by which we here mean any complete totally ordered field, 

so that F is an ordered field that satisfies the least upper bound property. By basic properties of rings, we know there a monomorphism (injective ring homomorphism) 

f : Q ,→ F which is also an order-embedding. We can extend this function to an embedding of R into F as follows: for each r ∈ R let Dr = {q ∈ Q | q < r} be the associated Dedekind cut. 

Since Dr is nonempty and bounded above in Q, we have that f(Dr) is nonempty and bounded above in F, so applying the assumed least upper bound property of F

 we define the function g : R → F by g(r) = sup f(Dr) Then g is also a monomorphism and order-embedding,

 since if r, s ∈ R, then g(r + s) = sup f(Dr+s) = sup f(Dr) +0 sup f(Ds) = g(r) +0 g(s) g(rs) = sup f(Drs) = sup f(Dr) · sup f(Ds) = g(r)g(s) r ≤ s =⇒ g(r) = sup f(Dr) ≤ sup f(Ds) = g(s) g(r) = g(s) =⇒ r = s and clearly g is an extension of f. 

We’ll prove the second equation, since the other three follow similarly: if x ≤ 0 sup f(Drs) 25 then x ≤0 a for any a ∈ f(Drs) u , so in particular x ≤0 bc for all b ∈ f(Dr) u , c ∈ f(Ds) u , 

whence x ≤ 0 sup f(Dr) · 0 sup f(Ds) 

Conversely if x ≤ 0 sup f(Dr) · 0

 sup f(Ds) then x ≤0 bc for all b ∈ f(Dr) u , c ∈ f(Ds) u . Now, for any a ∈ f(Drs) u \{sup f(Drs)} there exist such b and c that also satisfy bc ≤ a, as follows: if a ∈ f(Drs) u \{sup f(Drs)} then sup f(Drs) <0 a, so by the order denseness of Q we can always find a p ∈ Q such that sup f(Drs) < 0 b = f(p) < 0 a and then we can pick c to be c = sup f(Drs) + a 2b Then we have sup f(Drs) <0 bc <0 a. Because we can always find such b and c we must have x ≤ 0 sup f(Drs) We have just shown that g(R) ⊆ F and R ∼= g(R). If we can show that g(R) ⊃ F, we will have proved that g is surjective, and therefore bijective, and so an isomorphism (both a ring isomorphism and an orderisomorphism), and we will have thereby finished the proof. Towards this end, note first that F also satisfies the Archimedean property, which follows via T ?? from the fact that R is order-embedded into F and R satisfies the Archimedean property. Consequently, for any k ∈ F there exists an n ∈ N, and so a g(n) ∈ F, such that −g(n) < k < g(n). Let Dk = {r ∈ R | g(r) < k} and note that because g(R) ∼= R we have that we have that Dk and g(Dk) are both nonempty and bounded above, so that ∃ sup Dk ∈ R, whence ∃g(sup Dk) ∈ F, and also ∃ sup g(Dk) ∈ F, and since g(R) ⊆ F we have that sup g(Dk) ≤ g(sup Dk) (4.1) 26 by the least upper bound property of F. We claim that sup g(Dk) = g(sup Dk) (4.2) Suppose not, that is suppose sup g(Dk) < g(sup Dk). Then, by the Archimedean property of F, there is some n ∈ N such that 0 = g(sup Dk) − g(sup Dk) < g(n) −1 < g(sup Dk) − sup g(Dk) (4.3) This implies that sup g(Dk) < g(sup Dk) − g(n) −1 < g(sup Dk) (4.4) But then by (4.1), the second inequality in (4.4), and the definition of g(sup Dk) and sup g(Dk), the first of which implies that g −1 (g(sup Dk) − g(n) −1 ) = sup Dk − n ∈ Dk, we have g(sup Dk) − g(n) −1 < sup g(Dk) which by the first inequality of (4.4) implies the contradiction sup g(Dk) < sup g(Dk) Hence we must have sup g(Dk) = g(sup Dk), and so g(k) ∈ g(R), whence g(R) ⊃ F and consequently g(R) = F