The set of all equivalence classes from Cauchy sequences is complete Let $C(x)$ be the set of all Cauchy seq. on $X$ and define for $(x_n)_{n \in \mathbb{N}}, (y_n)_{n \in \mathbb{N}}$ the following relation
$$
(x_n)_{n \in \mathbb{N}} \sim (y_n)_{n \in \mathbb{N}} \Leftrightarrow d_X(x_n, y_n) \to 0 \text{ when } n \to \infty
$$
Let 
$$
\hat{X} = C(X)/\sim, \\ 
d_{\hat{X}}([(x_n)], [(y_n)]) = \lim_{n \to \infty} d(x_n, y_n)
$$
Show that $\hat{X}$ is complete
Let $(\gamma_m)_{m \in \mathbb{N}}$ be a Cauchy seq. in $\hat{X}$ i.e $\gamma_m = ([(x_n)]_m)$. How can I show that there exists $[(x_n)] \in \hat{X}$ such that for $m> n_0 \in \mathbb{N}$ $d_{\hat{X}}([(x_n)]_m, [(x_n)]) < \epsilon, \ \forall \epsilon$?
 A: In order to avoid confusion, I will denote by $x^m$ the $m$th sequence and $x^m_n$ will be the $n$th element of the $m$th sequence: $x^m=(x^m_1,x^m_2,x^m_3,\ldots)$. We need a Cauchy sequence $y=(y_1,y_2,\ldots)$ such that $d(x^n,y)\to0$.
The idea is to take a kind of diagonal: one element from each sequence,
$y_1\in x^1$, $y_2\in x^2$, ... .
$\bullet$ Construction of the sequence $y$:
For every $n$, the sequence $x^n$ is Cauchy, so there exists some index $k_n$ such that $|x^n_i-x^n_{k_n}|<\frac1n$ for all $i\ge k_n$. Now choose $y_n=x^n_{n_k}$.
$\bullet$ Proof that $y$ is Cauchy: Take an arbitrary $\varepsilon>0$.
The sequence $x^1,x^2,\ldots$ is Cauchy in $\hat{X}$, so there is some $N$ such that $d(x^n,x^m)<\frac\varepsilon3$ for all $m,n\ge N$. Consider two arbitrary indices $m,n$ with $m,n>\max(N,\frac3\varepsilon)$. Since 
$\lim\limits_{i\to\infty}|x^n_i-x^m_i|=d(x^n,x^m)<\frac\varepsilon3$,
there is some index $I$ such that 
$|x^n_i-x^m_i|<\frac\varepsilon3$ for $i\ge I$.
By the definition of $k_n$, for all $i\ge k_n$ we have 
$|x^n_i-y_n|=|x^n_i-x^n_{k_n}|<\frac1n<\frac\varepsilon3$. 
Similarly, $|x^m_i-y_m|<\frac\varepsilon3$ holds
for all $i\ge k_m$.
Hence, with $i=\max(I,k_m,k_n)$ we have
$$
|y_n-y_m| \le |y_n-x^n_i| + |x^n_i-x^m_i| + |x^m_i-y_m| <
\frac\varepsilon3+\frac\varepsilon3+\frac\varepsilon3 =
\varepsilon.
$$
Therefore, for $m,n>\max(N,\frac3\varepsilon)$ we have 
$|y_n-y_m| < \varepsilon$.
$\bullet$ Proof that $d(x^n,y)\to0$:
Again, fix some arbitrary $\varepsilon>0$.
The sequence $y$ is Cauchy, so there is an index $I$ such that $|y_i-y_j|<\frac\varepsilon3$ for all $i,j\ge I$.
Consider an arbitrary index $n\ge\max(\frac3\varepsilon,I)$.
For $i\le \max(n,I)$ we have
$$
|x^n_i-y_i| \le 
|x^n_i-y_n| + |y_n-y_i| =
|x^n_i-x^n_{k_n}| + |y_n-y_i| < 
\frac1n + \frac\varepsilon3 < 
\frac23\varepsilon.
$$
From $i\to\infty$ we get 
$$
d(x^n,y) \le \frac23\varepsilon<\varepsilon.
$$
Therefore, for $n\ge\max(\frac3\varepsilon,I)$ we have 
$d(x^n,y) \le \frac23\varepsilon<\varepsilon$.
A: I'll assume you already proved the metric is well defined, indeed a metric, and that $X$ is dense in $\hat{X}$. Now, when you take the Cauchy sequence of points in $\hat{X}$, remember that not all the terms in it are in $X$. So choose seqeunces $(x_{nk})_{k\in \mathbb{N}}$ (of points in $X$) such that either $[(x_{nk})_{k\in \mathbb{N}}]=\hat{x}_n=$ or, if $\hat{x_n}\in X$, then choose $x_{nk}=\hat{x}_n$. Then, for a suitable choice of naturals $k_1 <k_2<\dotsc$, the sequence $(x_{nk_{n}})_{n\in\mathbb{N}}$ is Cauchy in $X$ and $(\hat{x}_n)$ will converge to its limit in $\hat{X}$.
Let me know if you'd like me to expand some details.
