[proofplan]
We prove directly that every [Cauchy sequence](/page/Cauchy%20Sequence) of [real numbers](/page/Real%20Numbers) converges. The Cauchy condition first implies boundedness, so every tail has a real infimum and supremum by the least-upper-bound property of $\mathbb{R}$. The lower tail bounds and upper tail bounds squeeze together because the sequence is Cauchy, and the common squeezed real number is the limit of the original sequence.
[/proofplan]
[step:Bound an arbitrary Cauchy sequence in $\mathbb{R}$]
Let $(x_n)_{n\in\mathbb{N}}:\mathbb{N}\to\mathbb{R}$ be a Cauchy sequence in the metric $d(x,y)=|x-y|$. By the Cauchy property with tolerance $1$, there exists $N_0\in\mathbb{N}$ such that, for all $m,n\geq N_0$,
\begin{align*}
|x_n-x_m|<1.
\end{align*}
Define the real number
\begin{align*}
M:=1+\max\{|x_1|,\dots,|x_{N_0}|\}.
\end{align*}
If $n<N_0$, then $|x_n|\leq M$. If $n\geq N_0$, then the triangle inequality gives
\begin{align*}
|x_n|\leq |x_{N_0}|+|x_n-x_{N_0}|<|x_{N_0}|+1\leq M.
\end{align*}
Thus $(x_n)_{n\in\mathbb{N}}$ is bounded in $\mathbb{R}$.
[/step]
[step:Define lower and upper bounds for every tail]
For each $n\in\mathbb{N}$, define the tail set
\begin{align*}
A_n:=\{x_m\in\mathbb{R}:m\in\mathbb{N}\text{ and }m\geq n\}.
\end{align*}
Each $A_n$ is nonempty because $x_n\in A_n$, and each $A_n$ is bounded because the sequence is bounded. By the least-upper-bound property of $\mathbb{R}$, define
\begin{align*}
\alpha_n:=\inf A_n\in\mathbb{R},\qquad \beta_n:=\sup A_n\in\mathbb{R}.
\end{align*}
Since $A_{n+1}\subset A_n$, the lower bounds increase and the upper bounds decrease:
\begin{align*}
\alpha_n\leq \alpha_{n+1}\leq \beta_{n+1}\leq \beta_n.
\end{align*}
Moreover, because $x_n\in A_n$, we have
\begin{align*}
\alpha_n\leq x_n\leq \beta_n
\end{align*}
for every $n\in\mathbb{N}$.
[/step]
[step:Use the Cauchy property to make the tail bounds squeeze together]
Fix $\varepsilon>0$. Since $(x_n)_{n\in\mathbb{N}}$ is Cauchy, there exists $N_\varepsilon\in\mathbb{N}$ such that, for all $m,k\geq N_\varepsilon$,
\begin{align*}
|x_m-x_k|<\varepsilon.
\end{align*}
Hence every two elements of $A_{N_\varepsilon}$ differ by less than $\varepsilon$. In particular, $\beta_{N_\varepsilon}$ is an upper bound for $A_{N_\varepsilon}$ and $\alpha_{N_\varepsilon}+\varepsilon$ is also an upper bound for $A_{N_\varepsilon}$: indeed, if $m\geq N_\varepsilon$, then $x_m\geq \alpha_{N_\varepsilon}$ and the Cauchy estimate applied to $x_m$ and any element of the tail gives that the diameter of the tail is at most $\varepsilon$. Therefore
\begin{align*}
0\leq \beta_{N_\varepsilon}-\alpha_{N_\varepsilon}\leq \varepsilon.
\end{align*}
Since $A_n\subset A_{N_\varepsilon}$ for every $n\geq N_\varepsilon$, we also have
\begin{align*}
0\leq \beta_n-\alpha_n\leq \beta_{N_\varepsilon}-\alpha_{N_\varepsilon}\leq \varepsilon
\end{align*}
for every $n\geq N_\varepsilon$.
[guided]
Fix $\varepsilon>0$. The Cauchy condition says that, beyond some index, all terms of the sequence lie within $\varepsilon$ of one another. Formally, there exists $N_\varepsilon\in\mathbb{N}$ such that, for all $m,k\geq N_\varepsilon$,
\begin{align*}
|x_m-x_k|<\varepsilon.
\end{align*}
We want to translate this statement about pairwise distances into a statement about the infimum and supremum of the tail. The tail set is
\begin{align*}
A_{N_\varepsilon}=\{x_m\in\mathbb{R}:m\geq N_\varepsilon\}.
\end{align*}
The preceding inequality says precisely that the diameter of this set is at most $\varepsilon$: no two points in the tail can be separated by $\varepsilon$ or more. Since $\alpha_{N_\varepsilon}=\inf A_{N_\varepsilon}$ and $\beta_{N_\varepsilon}=\sup A_{N_\varepsilon}$, this forces
\begin{align*}
0\leq \beta_{N_\varepsilon}-\alpha_{N_\varepsilon}\leq \varepsilon.
\end{align*}
To see the last inequality directly, suppose instead that $\beta_{N_\varepsilon}-\alpha_{N_\varepsilon}>\varepsilon$. By the defining property of [supremum and infimum](/page/Supremum%20and%20Infimum), there exist indices $m,k\geq N_\varepsilon$ such that
\begin{align*}
x_m>\beta_{N_\varepsilon}-\frac{\beta_{N_\varepsilon}-\alpha_{N_\varepsilon}-\varepsilon}{2}
\end{align*}
and
\begin{align*}
x_k<\alpha_{N_\varepsilon}+\frac{\beta_{N_\varepsilon}-\alpha_{N_\varepsilon}-\varepsilon}{2}.
\end{align*}
Subtracting these inequalities gives $x_m-x_k>\varepsilon$, contradicting $|x_m-x_k|<\varepsilon$. Hence the tail width satisfies $\beta_{N_\varepsilon}-\alpha_{N_\varepsilon}\leq\varepsilon$.
Finally, if $n\geq N_\varepsilon$, then $A_n\subset A_{N_\varepsilon}$. Passing to a smaller set can only increase the infimum and decrease the supremum, so
\begin{align*}
0\leq \beta_n-\alpha_n\leq \beta_{N_\varepsilon}-\alpha_{N_\varepsilon}\leq \varepsilon.
\end{align*}
This is the squeezing mechanism that will produce the limit.
[/guided]
[/step]
[step:Extract the limit from the lower tail bounds]
The sequence $(\alpha_n)_{n\in\mathbb{N}}$ is increasing and bounded above by $\beta_1$, so the least-upper-bound property of $\mathbb{R}$ gives a real number
\begin{align*}
x:=\sup\{\alpha_n:n\in\mathbb{N}\}\in\mathbb{R}.
\end{align*}
For every $n\in\mathbb{N}$, since $\alpha_n\leq x$ and $x\leq\beta_n$, we have
\begin{align*}
\alpha_n\leq x\leq \beta_n.
\end{align*}
The inequality $x\leq\beta_n$ follows because $\beta_n$ is an upper bound for $\{\alpha_k:k\in\mathbb{N}\}$: if $k\leq n$, then $\alpha_k\leq\alpha_n\leq\beta_n$, while if $k\geq n$, then $\alpha_k\leq\beta_k\leq\beta_n$.
[/step]
[step:Show the original sequence converges to the squeezed real number]
Let $\varepsilon>0$. Choose $N_\varepsilon\in\mathbb{N}$ as in the squeezing step, so that
\begin{align*}
0\leq \beta_n-\alpha_n\leq \varepsilon
\end{align*}
for every $n\geq N_\varepsilon$. For such $n$, the inequalities $\alpha_n\leq x_n\leq\beta_n$ and $\alpha_n\leq x\leq\beta_n$ imply
\begin{align*}
|x_n-x|\leq \beta_n-\alpha_n\leq \varepsilon.
\end{align*}
Thus $x_n\to x$ in the metric $d$, because $d(x_n,x)=|x_n-x|$. Since the Cauchy sequence $(x_n)_{n\in\mathbb{N}}$ was arbitrary, every Cauchy sequence in $(\mathbb{R},d)$ converges to a point of $\mathbb{R}$. Therefore $(\mathbb{R},d)$ is complete.
[/step]
