How to prove that $\mathbb R$ is a complete metric space without Bolzano-Weierstrass theorem Consider $\mathbb R$ defined as the completion of $\mathbb Q$ (with Cauchy sequence approach).
Wikipedia : https://en.wikipedia.org/wiki/Construction_of_the_real_numbers#Construction_from_Cauchy_sequences
How to prove that $(\mathbb R, \vert \,.\vert)$ is a complete metric space without using the Bolzano-Weierstrass theorem? Otherwise, how to prove that any Cauchy sequence $\{x_n\}_{n\in\mathbb N} \in\mathbb Q^\mathbb{N}$ converges to the real $\overline{x}$ which represents the sequence ?
I've started with : $\forall\varepsilon>0 \;\mathrm{s.t.}\; \varepsilon\in\mathbb Q \;,\; \exists n_0\in\mathbb N \;:\; \forall(p,q)\in\mathbb N^2 \;,\; p,q \geqslant n_0 \Rightarrow \vert x_p - x_q\vert \leqslant \varepsilon$ then, let $p\geqslant n_0$ fixed, $\overline{x_p - \varepsilon} \leqslant \overline{x} \leqslant \overline{x_p+\varepsilon}$, can I write $p\geqslant n_0 \Rightarrow \vert x_p - \overline{x}\vert\leqslant \varepsilon$ if I've defined an isomorphism $\varphi : \mathbb Q \to \mathbb R'$ ($\mathbb R'$ a subfield of $\mathbb R$) such that $\varphi(q) = \overline{q}$ ? Does it possible ?
Thanks in advance.
 A: It appears that you want to take somewhat harder route to prove completeness of real numbers. If we define a real number as an equivalence class of Cauchy sequences of rational numbers then we must show that a Cauchy sequence of real numbers converges to a specific real number.
Here is the simplest and most natural way to do it. Assume that $\{x_{n} \} $ is a Cauchy sequence of real numbers. Each term $x_{i}$ here is an equivalence class of Cauchy sequences of rationals and we pick one representative member say $\{x_{i, n}\} $ of this class. Let $y_{n} =x_{n, n} $ then you need to show the following:


*

*$\{y_{n} \}$ is a Cauchy sequence of rationals and let $y$ be the real number corresponding to it i.e. $y$ is an equivalence class of Cauchy sequences of rationals one of whose members is the sequence $y_{n} $. 

*$x_{n} \to y$ as $n\to\infty$
Proving the above statements is not hard and you should be able to do it. Also note that while proving the statements above is not hard, it is definitely boring and difficult to type because of the symbolism involved and involves reasonable amount of $\epsilon$ gymnastics.
A slightly different approach is to prove the following statements:


*

*$\mathbb{Q}$ is dense in $\mathbb{R}$ i.e. for any given $\epsilon > 0$ and any $x \in \mathbb{R}$ there is a $y \in \mathbb{Q}$ such that $|x - y| < \epsilon$.

*If $\{y_{n}\}$ is a Cauchy sequence of rational numbers then it converges to a real number $y$ where $\{y_{n}\}$ is a member of the equivalence class $y$.
If $\{x_{n}\}$ is a Cauchy sequence of real numbers then using denseness of $\mathbb{Q}$ in $\mathbb{R}$ we can find rationals $y_{n}$ such that $|x_{n} - y_{n}| < 1/n$ and then show that $\{y_{n}\}$ is a Cauchy sequence of rational numbers which tends to real number $y$ (and $\{y_{n}\}$ is a member of equivalence class $y$). Because $|x_{n} - y_{n}| < 1/n$ it follows that $x_{n}$ also tends to the same limit as $y_{n}$ and thus $x_{n}$ converges to $y$.
A more preferable and easy approach to define reals as Dedekind cuts of rationals and then establish completeness of real numbers. 
A: I think we take the abstraction of the real number line $\mathbb{R}$ for granted.  It assumes for example we can evaluate numbers to arbitrary precision (which you can't).  So there is always room for error and fuzziness.  
Sacrificing our connection to physical reality, you may therefore consider all possible quotients of numbers and where they fit on the real number line.  Some of them will be bigger than $\sqrt{2}$ and some of them will be less than $\sqrt{2}$ more precisely some of them will have:


*

*$ x^2 = \frac{p^2}{q^2} < 2$

*$ x^2 = \frac{p^2}{q^2} > 2$

*$ x^2 = \frac{p^2}{q^2} \neq 2$ (let's just make up a number for this)


This is called a Dedekind Cut.  And in this way we show that the rational numbers $\mathbb{Q}$ are completed by the real number line $\mathbb{R}$. [1]

This is the only "picture" I could find of a Dedekind cut.  Let's try one out:
$$ 577^2 = 332929 > 3322928 =  2 \times 408^2$$
Therefore it should go on the $\color{#4433FF} {\textrm{blue}}$ side.
