Proof. This proof imitates that of Theorem.
Let $\left(x_{n}\right)$ be a Cauchy sequence in $\mathbb{R}$. Let $\delta>0$ be arbitrary. There exists a positive integer $N=N(\delta)$ such that for all $m \geq N$ and $n \geq N$, we have $\left|x_{n}-x_{m}\right|<\delta / 2$. In particular we have $\left|x_{n}-x_{N}\right|<\delta / 2$. Or, equivalently,
$$
x_{n} \in\left(x_{N}-\delta / 2, x_{N}+\delta / 2\right) \quad \text { for all } n \geq N .
$$
From this we make the following observations:
(i) For all $n \geq N$, we have $x_{n}>x_{N}-\delta / 2$.
(ii) If $x_{n} \geq x_{N}+\delta / 2$, then $n \in{1,2, \ldots, N-1}$. Thus the set of $n$ such that $x_{n} \geq x_{N}+\delta / 2$ is finite.
We shall apply these two observations below for $\delta=1$ and $\delta=\varepsilon$.
We claim that $S$ is nonempty, bounded above and that $\sup S$ is the limit of the given sequence.
From (i), we see that $x_{N}-1 \in S$. Hence $S$ is nonempty.
5CCM226A COURSE NOTES ：
$$
\left|x_{N}-\ell\right| \leq \varepsilon / 2
$$
For $n \geq N$ we have
$$
\begin{aligned}
\left|x_{n}-\ell\right| & \leq\left|x_{n}-x_{N}\right|+\left|x_{N}-\ell\right| \
&<\varepsilon / 2+\varepsilon / 2=\varepsilon
\end{aligned}
$$
We have thus shown that $\lim {n \rightarrow \infty} x{n}=\ell$.