If $\{x_n\}$ is a Cauchy sequence in a normed vector space, is $\frac{x_n}{\|x_n\|}$ Cauchy? Let $\{x_n\}$ a Cauchy sequence in a normed vector space $X$. Is
$$y_n = \frac{x_n}{\|x_n\|}$$
another Cauchy sequence in $D = \{x\in X : \|x\| = 1\}$?
Remark: The idea is prove that if $D$ is complete, then $X$ is also complete. Thanks so much.
 A: Let $f(x)=\frac{x}{\|x\|}$. Let $r >0$, then $f$ is uniformly continuous on $B(0,r)^c$.
To see this, suppose $\|x-y\| < \delta$. Then $|\|x\|-\|y\|| \leq \|x-y\| < \delta$ as well, and we have:
\begin{eqnarray}
\|f(x)-f(y)\|&=& \left\Vert \frac{x}{\|x\|} - \frac{y}{\|x\|} +\frac{y}{\|x\|} - \frac{y}{\|y\|} \right\Vert \\
&\le& \frac{1}{\|x\|}\|x-y \| + \left| \frac{1}{\|x\|} - \frac{1}{\|y\|} \right| \|y \| \\
&=& \frac{1}{\|x\|}\|x-y \| + \frac{1}{\|x\|} |\|x\|-\|y\| | \\
&\le& 2 \frac{\delta}{r}
\end{eqnarray}
Note that uniformly continuous functions map Cauchy sequences into Cauchy sequences. Also, if $x_n$ is Cauchy, then $\lim_n \|x_n\|$ converges, as $\mathbb{R}$ is complete. (Also, note that $\|x_n\|$ is bounded.)
Now suppose $x_n$ is Cauchy. If $l=\liminf_n \|x_n\| >0$, then for $n$ sufficiently large, $x_n \in B(0,\frac{l}{2} )^c$, and then $f(x_n)$ is Cauchy, hence $f(x_n) \to \hat{f}$ for some $\hat{f} \in D$, and $n = \lim_n \|x_n\|$ exists. Then we have
\begin{eqnarray}
\|x_n - n \hat{f}\| &=& \left\Vert \|x_n\| f(x_n) - n  \hat{f} \right\Vert \\
&=& \left\Vert \|x_n\| f(x_n) -n f(x_n) + n f(x_n)- n  \hat{f}  \right\Vert \\
&\le& \left| \|x_n\|-n \right| + n \|f(x_n)-\hat{f}\|
\end{eqnarray}
and it follows that $\lim_n x_n = n \hat{f}$.
If $l=0$, pick some element $d \in D$, let $B$ be an upper bound for $\|x_n\|$ and let $x_n'=x_n+(B+1)d$. Then $\|x_n'\| \ge 1$, and $x_n'$ is also Cauchy. By the above result, $x_n' \to \hat{y}$ for some $\hat{y}$. Hence $\lim_n x_n = \hat{y}-(B+1)d$.
It follows that $X$ is complete.
A: Consider:
$$
x_n = \frac{(-1)^n}{n}
$$
Then:
$$
y_n = (-1)^n
$$
$x_n$ is Cauchy in $\Bbb R$ as it converges to $0$. However, $y_n$ keeps alternating between $1$ and $-1$ and it's not Cauchy in $\{-1, 1\}$.
A: I can not see clear the solution at all. That's why I am trying to use the fact (hint of my professor) of if a Cauchy sequence have a convergent subsequence, then the complete Cauchy sequence converges.
Thus, in the case of $\lim\limits_{n\rightarrow +\infty}\|x_n\| \neq 0$, I considered a subsequence $\{x_{n_k}\}$ that hold $x_{n_k}\neq 0, \forall k\in\mathbb{N}$ and then the new sequence $$y_k = \frac{x_{n_k}}{\|x_{n_k}\|}$$ is good defined.
Later, $\{x_{n_k}\}$ is also a Cauchy sequence and $y_k \neq 0$ for all $k\in\mathbb{N}$. So, the next step, is prove that $y_k$ is Cauchy in $D$ and for the fact, I need to prove that exists of a $M > 0$ that $M \leq x_{n_k}$  for all $k\in\mathbb{N}$. But I don't know how I can justify the existence of $M$. Please help me. Thanks
