Simple question on cauchy sequences and convergence Let $K$  be a field with metric defined on it.
$(x_n)$ be a cauchy sequence in  $K$ which does not converge to $0$.
Question is to see if we can  get a another cauchy sequence $(y_n)$ in $K$  such that $(x_ny_n)\rightarrow 1$.
I believe this happens but could not prove. Any hints are welcome. 
We do not assume any topological completion. Everything we do has to be inside $K$. 
 A: Theorem: If $(x_n)_{n \ge 1}$ is Cauchy in a normed field $(K, | \cdot |)$ and $x_n \not\to 0$, then there exist $N \in \Bbb N$ such that $\left( \frac 1 {x_n} \right)_{n \ge N}$ is Cauchy in $K$.
Proof: Since $x_n \not\to 0$ but $(x_n)$ is Cauchy, then there exist $r>0$ and $M \in \Bbb N$ such that $|x_n| \ge r$ for $n \ge M$ (otherwise, there would exist a subsequence tending to $0$ and, since the sequence is Cauchy, this would imply that the sequence itself tends to $0$, which would be a contradiction).
Let $\epsilon > 0$. Since $(x_n)$ is Cauchy let $M' \in \Bbb N$ be such that $|x_n - x_m| < \epsilon r^2$ for $m,n \ge M'$. Let $N = \max (M, M')$. Then
$$\left| \frac 1 {x_m} - \frac 1 {x_m} \right| = \frac {| x_n - x_m |} {|x_n| |x_m|} \le \frac {\epsilon r^2} {r^2} = \epsilon$$
whih shows that, indeed, $\left( \frac 1 {x_n} \right)_{n \ge N}$ is Cauchy in $K$.
Define now $y_n = \begin{cases} 0, & n < N \\ \frac 1 {x_n}, & n \ge N \end{cases}$. It follows that $(y_n)$ is Cauchy and $x_n y_n = 1$ for $n \ge N$, therefore, in particular, $x_n y_n \to 1$.
