Isomorphism between equivalence classes of Cauchy sequences and Dedekind cuts

Question

Real numbers can be defined in terms of equivalence classes of Cauchy sequences and in terms of Dedekind cuts. It can be shown that both of these constructions give a complete ordered field. It is also known that up to isomorphism, there is only one such complete ordered field. Therefore, there should exist an isomorphism between these two constructions. My question is: How to define this isomorphism explicitly?

Answer 1

Let's build the isomorphism.

A Dedekind cut is defined by a partition of $\mathbb{Q}$ into two non-empty sets $A$ and $\complement A$, where all elements in $A$ are lower than all elements in $\complement A$, and $A$ has no greatest element (this last condition to avoid getting two different Dedekind cuts for the same number, when this number $\in \mathbb{Q}$).

If we have a Cauchy sequence of rational numbers $(a_n)_{n \in \mathbb{N}}$, we can define an equivalent Dedekind cut by its lower set $A = \left\lbrace x \in \mathbb{Q} : \exists N, \forall n \ge N, x \le a_n \right\rbrace$, its upper bound excluded.

As a Cauchy sequence is bounded, $A$ is non-empty, and $\complement A$ is non-empty too.

Any element is $\complement A$ is greater than any $a_n$ (otherwise it would be in $A$), so greater than any element in $A$.

Let's call $f$ this function from Cauchy sequences of $\mathbb{Q}$ elements to Dedekind cuts of $\mathbb{Q}$.

Reciprocally, if we have a Dedekind cut defined by its lower set $A$, we can derive an equivalent Cauchy sequence by the following process:

• $a_0$ is any element in $A$.
• $a_{2n+1}$ is the lowest of $C = B \cap \complement A$,
  where $B = \left\lbrace a_{2n}+\frac k {2^{2n}}, k \in \mathbb{N}, k \ge 0 \right\rbrace$.
  This exists because $A$ is upper bounded (by definition of a Dedekind cut) but $B$ is not, so $B \cap \complement A$ is non empty; $B$ is lower bounded by the upper bound of $A$; all elements in $B$ are isolated.
• $a_{2n+2}$ is the highest of $C' = B' \cap A$, where $B' = \left\lbrace a_{2n+1}-\frac k {2^{2n+1}}, k \in \mathbb{N}, k \ge 0 \right\rbrace$.

This defines an alternating Cauchy sequence.

Let's call $g$ this function from Dedekind cuts of $\mathbb{Q}$ to Cauchy sequences of $\mathbb{Q}$ elements.

This defines an isomorphism. To prove it, we prove that $f \circ g = $ Id.

Begin with a Dedekind cut defined by its lower set $A$.
Define a Cauchy sequence $(a_n)_{n \in \mathbb{N}}$ by $g$ as above.
Then define $A' = f((a_n)_{n \in \mathbb{N}}) = \left\lbrace x \in \mathbb{Q} : \exists N, \forall n \ge N, x \le a_n \right\rbrace$.

$\forall x \in A$, as $A$ is upper bounded but has no greatest element, $\exists y \in A, y > x$.
Let's have $N=2k$, such that $\frac 1 {2^N} < y-x$.
$\forall n > N, a_n \in [a_{N+2}, a_{N+1}]$, as the sequence is alternating.
So $a_n \ge a_{N+2} \gt a_{N+1} - \frac 1 {2^N} \gt a_{N+1} - y + x$
As $a_{N+1} \in \complement A$ and $y \in A$, $a_{N+1} > y$, so $a_n > x$. This proves $x \in A'$.

Reciprocally, $\forall x \in A', \exists k, x \le a_{2k}$.
As $a_{2k} \in A$, $\forall y \in \complement A, a_{2k} \lt y$ so $x \lt y$. This implies $x \in A$.

So we have proved $A' = A$, which proves $f \circ g =$ Id.

Answer 2

Well, a Cauchy sequence $(x_n)_n$ of rational numbers is meant to represent the real number $x=\lim_nx_n$, right? On the other hand, $x\in\mathbb R$ is represented by the Dedekind cut $x^\downarrow=\{p\in\mathbb Q|p<x\}$.

Now, a rational number $p$ is less or equal then $x=\lim_nx_n$ if and only if there is $N$ such that $p\leq x_n$, for all $n>N$. Moreover, $p=\lim_nx_n$ if and only if the constant Cauchy sequence $(p)_n$ is equivalent to $(x_n)_n$, so we can define a Dedekind cut by taking the set of rationals $p$ such that $p\leq x_n$, for all $n>N$ for some $N$, and $(p)_n$ is not equivalent to $(x_n)_n$.

You can visualize this in the following way: If we picture $(x_n)_n$ as a sequence of point that are clustering around the line $y=x=\lim_nx_n$ while $p$ represents the constant line $y=p$, then $p$ is inside the Dedekind cut if eventually all points of the sequence are above $p$ and "far" from $p$. For example, the cut associated to the sequence $(p)_n$ is the set of rationals $q<p$, which is the Dedekind cut representing $p$ inside $\mathbb R$.

You can show that this map is well-defined and injective. Surjectivity is slightly more involved, but it's ultimately an argument by "picking a rational number between $\alpha-\frac{1}{n}$ and $\alpha$, for all $n$, this gives a Cauchy sequence with associated cut equal to $\alpha$.

Answer 3

It seems reasonably simple, consider $\phi: u\mapsto A_u$ where $u = \{u(n)\}_{n\in{\mathbb N}}$ is a Cauchy sequence and \begin{equation} A_u = \left\{x\in{\mathbb Q}\mid \left(\exists k\in{\mathbb N}^*\mid \text{Card}\left(u^{-1}\left((-\infty, x + \frac{1}{k})\right)\right)<\infty\right)\right\} \end{equation} Then prove that $A_u, A_u^c$ is a Dedekind cut and that $A_u$ depends only on the equivalence class of $u$.