# Proof verification - Cauchy completeness of $\mathbb{R}$

Let $(x_n)$ be a Cauchy sequence in $\mathbb{R}$. We want to show that $(x_n)$ converges in $\mathbb{R}$. I'm presenting my proof below. I've used standard $\epsilon-N$ technique, triangle inequality and Bolzano-Weierstrass theorem. I think that this scheme, while keeping it intuitive, has lengthened the proof considerably. I'd appreciate if anyone checks the proof, whether or not there is any hole in it (I checked, but still) and most importantly, if you can suggest me some other scheme to attack the problem, which may give us a quick and short proof. Thank you.

Let $\epsilon>0$. Since $(x_n)$ is Cauchy, there exists $N=N(\epsilon) \in \mathbb{N}$ such that $$m,n \geq N \implies |x_n-x_m|<\frac{\epsilon}{3}$$ In particular, choosing $m=N$, we have $$n \geq N \implies |x_n-x_N|<\frac{\epsilon}{3}$$ It is clear that, for all $n \in \mathbb{N}$, $$\min\{x_1,\ldots,x_{N-1},x_N-\frac{\epsilon}{3}\} \leq x_n \leq \max\{x_1,\ldots,x_{N-1},x_N+\frac{\epsilon}{3}\}$$ Hence the sequence $(x_n)$ is bounded. So we are in a position to employ the powerful Bolzano-Weierstrass theorem. There exists a subsequence $(x_{n_k})$ that converges in $\mathbb{R}$. Let $x_{n_k} \to x$ as $k \to \infty$. Hence, there exists $K_1 \in \mathbb{N}$ such that $$k \geq K_1 \implies |x_{n_k}-x|<\frac{\epsilon}{3}$$ This gives us a point $x$ of attack, a "guesstimate" for where the sequence $(x_n)$ might converge. Now, there exists $K_2 \in \mathbb{N}$ such that $$k \geq K_2 \implies n_k \geq N \implies |x_{n_k}-x_N|<\frac{\epsilon}{3}$$ Choose $K_0=\max\{K_1,K_2\}$. Then in particular, $$|x_{n_{K_0}}-x|<\frac{\epsilon}{3},\,\,\,\,|x_{n_{K_0}}-x_N|<\frac{\epsilon}{3}$$ Finally, We combine all these estimates by triangle inequality to write $$n \geq N \implies |x_n-x| \leq |x_n-x_N|+|x_N-x_{n_{K_0}}|+|x_{n_{K_0}}-x|<\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon$$ Since $\epsilon>0$ is chosen arbitrarily, the sequence $(x_n)$ converges to $x$. Hence the proof.

## 1 Answer

It looks fine to me. Here's an alternative (though similar) approach, by splitting the problem into parts.

Lemma. A real Cauchy sequence is bounded.

Proof. Let $\{x_n\}_{n \geq 1} \subseteq \mathbb{R}$ be a Cauchy sequence. By hypothesis, there exists $n_0 \in \mathbb{N}$ such that if $n, m \geq n_0$ then $|x_n -x_m| < 1$. Therefore,

$$|x_n| \leq |x_n-x_{n_0}| + |x_{n_0}| \leq 1 + |x_{n_0}| \quad (\forall n \geq n_0)$$

so $|x_n| \leq 1 + \max_{1 \leq i \leq n_0}|x_i|$ for all $n$.

Lemma: let $\{x_n\}_{n \geq 1} \subseteq \mathbb{R}$ be a Cauchy sequence. If $\{x_n\}_{n \geq 1}$ has a convergent subsequence, it is convergent.

Proof. By hypothesis, there exists a convergent subsequence $x_{n_k} \to x$. Now, I claim that:

$$|x_k -x| \leq |x_k - x_{n_k}| + |x_{n_k} - x| \to 0$$

Since $x_{n_k} \to x$, we only have to note that $|x_k - x_{n_k}| \to 0$. In effect, if $\varepsilon > 0$, there exists $K \in \mathbb{N}$ such that $n,m > K$ implies $|x_n - x_m| < \varepsilon$. Now if $k \geq K$, we have that $n_k \geq k \geq K$ and thus $|x_k - x_{n_k}| < \varepsilon$. Therefore, $x_k \to x$ as claimed.

Finally, we can prove the original statement.

Theorem: $\mathbb{R}$ is complete.

Proof. Let $\{x_n\}_{n \geq 1} \subseteq \mathbb{R}$ be a Cauchy sequence. By the first lemma, the sequence is bounded. Thus, the Bolzano-Weierstraß theorem guarantees the existence of a convergent subsequence. By the second lemma, this implies that $\{x_n\}_{n \geq 1}$ itself is convergent, which completes the proof.