Understanding proof that $\mathbb{R}$ is Cauchy complete

They seem to saying the following:

Suppose $(x_n)_n$ is a Cauchy sequence of rationals. For each $n$, consider the real number $\alpha_n$ represented by the constant (trivially Cauchy) sequence $[(x_n,x_n,\cdots,x_n,\cdots)]$. That is, $\alpha_n$ is the image of $x_n$ under the natural embedding $\iota:\Bbb Q\hookrightarrow\Bbb R$.

We want to show $\lim_n\alpha_n=x$ for some $x$. If we take $x$ to be the real number represented by $(x_1,x_2,\cdots,x_n,\cdots)$, then we can show $\lim_n\alpha_n=x$ as desired.

Therefore, any Cauchy sequence of rationals has a limit in $\Bbb R$.

There are some points here. The zero$^{th}$ point is that the definition of the absolute value function in terms of $\lim_n|x_n|$ is nonsense if we haven't yet established that all Cauchy sequences converge. $[(x_n,\cdots)]-[(x_1,\cdots)]$ converges to zero because for all rational $\epsilon>0$ there exists an $N\in\Bbb N$ such that $m,n\ge N$ implies $|x_n-x_m|<\epsilon$, so by definition of the ordering in $\Bbb R$, I know $|[(x_n,\cdots,)]-[(x_1,\cdots)]|<\iota(\epsilon)$ for $n\ge N$. Then it follows for any real $\epsilon>0$ that there is $N$, $n\ge N$ implies $|[(x_n,\cdots)]-[(x_1,\cdots)]|<\epsilon$.

Firstly, we must make sure that the notion of limit and the notion of "Cauchy sequence" is independent of the choice of representative elements.

Secondly, we must satisfy ourselves that a sequence of "rationals" $(\iota(x_n))_n\in\Bbb R$ is Cauchy in $\Bbb R$ iff. it is Cauchy in $\Bbb Q$. Really, we would like to know that $\iota$ is an isometric embedding of metric spaces (though, 'metric space' involves the notion of $\Bbb R$, so this is possibly circular language). Because then there is no ambiguity when referring to "Cauchy sequences of rationals".

Thirdly, we must check that any Cauchy sequence of reals can be well approximated by a Cauchy sequence of rationals, in the sense that a limit of the latter sequence will also be a limit of the former sequence. Then what the authors have shown will show $\Bbb R$ is complete.

I'll address the third point, I recommend you have a think about the first two points. The strategy I present is a more general one and is an over complication in this case. The simpler way: if $\beta_n$ is a Cauchy sequence of reals then rational numbers $\alpha_n\in(\beta_n,\beta_n+2^{-n}$ and $\gamma_n\in(\beta_n-2^{-n},\beta_n$ can be chosen by density, and $\alpha,\gamma$ will be Cauchy sequences too. They will then converge (by the first lemma) to have the same limit, and by squeezing $\beta$ has this limit too. So the limit exists!

The general way:

Let $(\beta_n)_n$ be a Cauchy sequence of reals. Let $(\alpha_{n,m})_m$ be a sequence of rationals representing $\beta_n$. With no loss of generality, we can take $|\alpha_{n,m}-\alpha_{n,m+1}|<2^{-(1+m)}$ for all $n,m$.

Then we can show the following:

• The sequence $(\alpha_{n,n})_n$ is Cauchy (in $\Bbb R$ or in $\Bbb Q$)
• As a Cauchy sequence of rationals, it has a limit $x\in\Bbb R$
• $x$ is also a limit for the sequence $\beta_n$

Concluding the proof. This template works in general for constructing the completion of any metric space.

Fix a (rational) $\epsilon>0$. There is an $N$, s.t. $i,j\ge N$ implies $|\beta_i-\beta_j|<\epsilon/3$. There is an $M$, $2^{-M}<\epsilon/3$. Then whenever $n,m\ge\max(N,M)$ we have: $$|\alpha_{n,n}-\alpha_{m,m}|\le|\alpha_{n,n}-\alpha_{n,k}|+|\alpha_{n,k}-\alpha_{m,k}|+|\alpha_{m,k}-\alpha_{m,m}|<\epsilon$$

Where $k\ge M$ is chosen from $|\beta_n-\beta_m|<\epsilon/3$, implying $|\alpha_{n,j}-\alpha_{m,j}|<\epsilon/3$ for sufficiently large $j$. We also used the simplifying assumption $|\alpha_{a,b}-\alpha_{a,b+1}|<2^{-(1+b)}$ for all $a,b$ which implies by geometric series that $|\alpha_{a,c}-\alpha_{a,b}|<2^{-\min(b,c)}$.

Hence the diagonal sequence is Cauchy.

I claim the sequence in $\beta$ converges to $[(\alpha_{1,1},\alpha_{2,2},\cdots)]$ now that we have established it is Cauchy and thus represents a real. We need to show, for all $\epsilon>0$ there is $N$, $n\ge N$ implying $|\alpha_{n,m}-\alpha_{m,m}|<\epsilon$ for all large $m$, where “large” depends on $n$. There is $K$, $n,m\ge K$ implying $|\alpha_{n,n}-\alpha_{m,m}|<\epsilon/2$ and there is $M$, $2^{-M}<\epsilon/2$. $N:=\max(M,K)$ suffices as: $$|\alpha_{n,m}-\alpha_{m,m}|\le|\alpha_{n,m}-\alpha_{n,n}|+|\alpha_{n,n}-\alpha_{m,m}|$$So for $n\ge N$ and “large $m\ge N$”, the intended inequality is satisfied.