11.4 Comparison of Cauchy and Dedekind reals
Let us also say something about the relationship between the Cauchy and Dedekind reals. By
\autorefRC-archimedean-ordered-field, ℝ𝖼 is an archimedean ordered field. It is also
admissible for Ω, as can be easily checked. (In case Ω is the initial
σ-frame
it takes a simple induction
, while in other cases it is immediate.)
Therefore, by \autorefRD-final-field there is an embedding
of ordered fields
ℝ𝖼→ℝ𝖽 |
which fixes the rational numbers.
(We could also obtain this from \autorefRC-initial-Cauchy-complete,\autorefRD-cauchy-complete.)
In general we do not expect ℝ𝖼 and ℝ𝖽 to coincide
without further assumptions.
Lemma 11.4.1.
If for every x:Rd there merely exists
c:∏q,r:ℚ(q<r)→(q<x)+(x<r) | (11.4.1) |
then the Cauchy and Dedekind reals coincide.
Proof.
Note that the type in (11.4.1) is an untruncated variant
of (LABEL:eq:RD-linear-order), which states that < is a weak linear order.
We already know that ℝ𝖼 embeds into ℝ𝖽, so it suffices to show that every Dedekind
real merely is the limit of a Cauchy sequence of rational numbers.
Consider any x:ℝ𝖽. By assumption there merely exists c as in the statement of the
lemma, and by inhabitation of cuts there merely exist a,b:ℚ such that a<x<b.
We construct a sequence f:ℕ→\setof(q,r)∈ℚ×ℚ|q<r by
recursion:
-
1.
Set f(0):≡(a,b).
-
2.
Suppose f(n) is already defined as (qn,rn) such that qn<rn. Define s:≡(2qn+rn)/3 and t:≡(qn+2rn)/3. Then c(s,t) decides between s<x and x<t. If it decides s<x then we set f(n+1):≡(s,rn), otherwise f(n+1):≡(qn,t).
Let us write (qn,rn) for the n-th term of the sequence f. Then it is easy to see that qn<x<rn and |qn-rn|≤(2/3)n⋅|q0-r0| for all n:ℕ. Therefore q0,q1,… and r0,r1,… are both Cauchy sequences converging to the Dedekind cut x. We have shown that for every x:ℝ𝖽 there merely exists a Cauchy sequence converging to x. ∎
The lemma implies that either countable choice or excluded middle suffice for coincidence of ℝ𝖼 and ℝ𝖽.
Corollary 11.4.3.
If excluded middle or countable choice holds then Rc and Rd are equivalent.
Proof.
If excluded middle holds then (x<y)→(x<z)+(z<y) can be proved: either x<z or ¬(x<z). In the former case we are done, while in the latter we get z<y because z≤x<y. Therefore, we get (11.4.1) so that we can apply \autoreflem:untruncated-linearity-reals-coincide.
Suppose countable choice holds. The set S=\setof(q,r)∈ℚ×ℚ|q<r is equivalent to ℕ, so we may apply countable choice to the statement that x is located,
∀((q,r):S).(q<x)∨(x<r). |
Note that (q<x)∨(x<r) is expressible as an existential statement ∃(b:𝟐).(b=0𝟐→q<x)∧(b=1𝟐→x<r). The (curried form) of the choice function is then precisely (11.4.1) so that \autoreflem:untruncated-linearity-reals-coincide is applicable again. ∎
Title | 11.4 Comparison of Cauchy and Dedekind reals |
\metatable |