Show that a sequence in $(X,d)$ is Cauchy iff it is Cauchy in $(X,\sigma)$.

Question

Given metric space $(X,d)$, define a new metric on $X$ by $\sigma (x,y) =min\left \{ d(x,y),1 \right \}$. Show that a sequence in $(X,d)$ is Cauchy iff it is Cauchy in $(X,\sigma)$.

I think I have to take arbitrary Cauchy sequence $(x_n)$ in $(X,d)$ and then apply Cauchy definition. From there, show it converges to some point in $X$. Then use the $\sigma$ metric and show some Cauchy sequence $(y_n)$ in $(X,\sigma)$ converges to the same point that $(x_n)$ converges to.

Is that correct? If not, just need a tip to start.

Answer 1

If $\{x_n\}$ is Cauchy we have that$$\forall\epsilon>0\qquad,\qquad \exists N\qquad n,m>N\to d(x_n,x_m)<\epsilon\to \min\{d(x_n,x_m),1\}\le d(x_n,x_m)<\epsilon$$for proving the converse, if we have $\{x_n\}$ being Cauchy under metric $\sigma(x,y)$ we have that $$\forall\epsilon>0\qquad,\qquad \exists N\qquad n,m>N\to \sigma(x_n,x_m)<\epsilon$$Take $\epsilon<1$. Therefore $$\sigma(x_n,x_m)<\epsilon<1$$we can't have $d(x_n,x_m)>1$ because it leads to $$\sigma(x_n,x_m)=\min\{d(x_n,x_m),1\}=1>\epsilon$$which is a contradiction. Therefore $d(x_n,x_m)<1$ and so $$d(x_n,x_m)=\sigma(x_n,x_m)<\epsilon$$which completes our proof.

Answer 2

If $(x_n)$ is Cauchy in $(X,d)$, try showing it's Cauchy in $(X,\sigma)$ directly:

Let $\varepsilon > 0$. Let $\delta = \min(\varepsilon, \frac{1}{2})$. We can find $N$ such that for all $n,m \ge N$ $d(x_n, x_m) < \delta$. But as $\delta \le \frac{1}{2} < 1$ we know that $d(x_n, x_m) = \sigma(x_n, x_m)$ so...