Let $\left(X,\rho\right)$ be a complete metric space and define $\rho^*(x,y):=\min\{1,\rho(x,y)\}$ for $x,y\in X$. Show that $\left(X,\rho^*\right)$ is a complete metric space.
I know I need to take a Cauchy sequence in $X$ and show that it converges to a point in $X$ with respect to the metric $\rho^*$. That is, let $(x_n)$ be a Cauchy sequence in $X$, I need to show that there is a $x'\in X$ such that for every $\varepsilon>0$, there is an integer $N'$ for which $\rho^*(x_n,x')=\min\{1,\rho(x_n,x')\}<\varepsilon$ whenever $n\geq N'$.
So, since $(x_n)$ is Cauchy, for every $\varepsilon>0$, there is an integer $N$ for which
$\rho^*(x_n,x_m)=\min\{1,\rho(x_n,x_m)\}<\varepsilon$ whenever $n,m\geq N'$. But not sure where to go from here…
Best Answer
If $(x_n)$ is a Cauchy sequence in $(X, \rho^*)$ and $0 < \varepsilon \le 1$ then $\rho^*(x_n,x_m)=\min\{1,\rho(x_n,x_m)\}<\varepsilon$ implies $\rho(x_n,x_m) < \varepsilon$. So $(x_n)$ is a Cauchy sequence in $(X,\rho)$ and therefore convergent to some $x' \in X$.
Then $\rho^*(x_n, x') \le \rho(x_n, x')$ implies that also $x_n \to x'$ in $(X, \rho^*)$.