Proof that Cauchy sequences of real numbers converge in R

Question

Im reading Analysis by Tao, and unless I have yet to see this it doesn't look like he ever proves that Cauchy sequences of real numbers converge to a real number.

Though its never explicitly stated, it looks like his initial definition of $\mathbb{R}$ is the field formed by equivilance classes of rational Cauchy sequences with $a_n \sim b_n$ when $\forall \epsilon>0,\epsilon \in \mathbb{Q} \ \exists N \in \ \mathbb{N} \mid \vert a_n-b_n \vert \lt \epsilon \ \forall n\geq N$, with the equivalence classes denoted as $LIM_{n\rightarrow \infty} a_n$.

Then, I believe it is implied that this field is isomorphic to the set of all real Cauchy sequences modulo the same relation (with $\epsilon$ now allowed to be a real number rather than rational). The map $\phi(LIM_{n\rightarrow \infty} a_n)=\lim_{n\rightarrow \infty} a_n$ with $\lim_{n\rightarrow \infty} a_n$ denoting the equivalence class of $a_n$ (with $a_n$ considered as a real Cauchy sequence in the image). Its not hard to see this is an injective homomorphism--proving it is surjective is less straightforward.

Surjectivity is equivalent to all real Cauchy sequences converging to a real number, so it suffices to show that $\lim_{n \rightarrow \infty} x_n =\lim_{n \rightarrow \infty} r_n$ ($=\phi(LIM_{n\rightarrow \infty} r_n)$) where $x_n \in \mathbb{R}$ and $r_n \in \mathbb{Q}$.

What I have now is: since rationals can always bisect two distinct real numbers, there must exist $r_n$ such that $x_n \lt r_n \lt x_n+\frac{1}{n}$ for all $n\geq 1$. From this, we have $\vert r_n-x_n \vert \lt \frac{1}{n}$, which is enough to show the two sequences are equivalent Cauchy sequences. Therefore, $\lim_{n \rightarrow \infty} x_n =\lim_{n \rightarrow \infty} r_n$, so the equivalence classes of rational and real Cauchy sequences are isomorphic.

I feel as if what I did was a little too simple--the result feels pretty significant (the equivalence classes of rational Cauchy sequences is not isomorphic to $\mathbb{Q}$, so there's no reason to suppose this changes when we use reals) so I would like to make sure that what I have makes sense--its only been a few days since I started learning so another pair of eyes would help. Thanks so much.

Answer 1

Reusing the following (easily understandable) notations: $$\phi\colon\quad\Bbb R:=\text{Cauchy}(\Bbb Q)/\sim\ \longrightarrow\ \text{Cauchy}(\Bbb R)/\sim',$$ I copy and paste two minor comments, before deleting them:

• "Surjectivity is equivalent to all real Cauchy sequences converging to a real number": I would rather say "to every real Cauchy sequence being ∼′ to some rational Cauchy sequence".
• "so it suffices to show that $\lim_{n \rightarrow \infty} x_n =\lim_{n \rightarrow \infty} r_n$ where $x_n∈\Bbb R$ and $r_n∈\Bbb Q$": I prefer to make the quantifiers explicit: "that $\forall(x_n)\in\text{Cauchy}(\Bbb R),\exists(r_n)\in\text{Cauchy}(\Bbb Q),\lim_{n \rightarrow \infty} x_n =\lim_{n \rightarrow \infty} r_n$".

I agree with your last comment: you "missed is proving r_n is Cauchy but that’s not very difficult to do with the triangle inequality".

Let's now get to the main point: "since rationals can always bisect two distinct real numbers, there must exist $r_n$ such that $x_n\lt r_n\lt x_n+\frac1n$ for all $n\ge1$" needs a careful justification, because our set of real numbers $\Bbb R$ here is not (or not yet) the "usual one", for which we are used to the density of the rationals.

Let us prove the following equivalent property:$$\forall x=[(x_n)]\in\Bbb R=\text{Cauchy}(\Bbb Q)/\sim,\quad\forall\epsilon\in\Bbb Q_+^*,\quad \exists r\in\Bbb Q,\quad x\le r\le x+\epsilon.$$ Given such $x=[(x_n)]$ and $\epsilon$, there exists $N$ such that $$\forall n\ge N,\quad \left|x_n-x_N\right|\le\frac\epsilon2,$$ i.e. $$\forall n\ge N,\quad x_n\le x_N+\frac\epsilon2\le x_n+\epsilon.$$ The rational number $$r:=x_N+\frac\epsilon2$$ is then suitable, i.e. satisfies $x\le r\le x+\epsilon$.