Let $(X,d)$ be a metric space and $d'$ be the [standard bounded metric][1] corresponding to $d.$ Then, $(X,d)$ is complete iff $(X,d')$ is complete

Question

Let $(X,d)$ be a metric space and $d'$ be the standard bounded metric corresponding to $d.$ Then, $(X,d)$ is complete iff $(X,d')$ is complete

From the answers it is clear that both metric spaces are not strongly equivalent. $\{x_n\}$ be a cauchy sequence with respect to $d$ implies $\{x_n\}$ be a cauchy sequence with respect to $d'$.If $(X,d)$ is complete then $\{x_n\}$ coverges to some point in $x\in X$ with respect to $d$. $\{x_n\}$ coverges to some point in $x\in X$ with respect to $d'$.How do I prove the converse?

Answer 1

I think the proof should go like this:-

Part I: If $\left( X, d \right)$ is complete, then $\left( X, d' \right)$ is complete.

To prove this, we need to start with a Cauchy sequence in $\left( X, d' \right)$. So, let $\left\lbrace x_n \right\rbrace$ be a Cauchy sequence with respect to $d'$. Then, $\forall \epsilon > 0, \exists n_0 \in \mathbb{N}$ such that $\forall n, m \geq n_0, d' \left( x_n, x_m \right) < \epsilon$. In particular, this is true for $\epsilon < 1$ and hence we shall get $\forall 0 < \epsilon < 1, \exists n_0 \in \mathbb{N}$ such that $\forall n, m \geq n_0, d' \left( x_n, x_m \right) < \epsilon < 1$. Now, $d' \left( x_n, x_m \right) = \min \left\lbrace d \left( x_n, x_m \right) \right\rbrace$. Since $d' \left( x_n, x_m \right) < 1$, we have $\min \left\lbrace d \left( x_n, x_m \right) \right\rbrace = d \left( x_n, x_m \right)$. Hence, the sequence $\left\lbrace x_n \right\rbrace$ is convergent with respect to the metric $d$. Let it converge to $x$. Now, our claim is that the sequence converges to the same limit in $\left( X, d' \right)$ as well. I think this can be easily done and encourage you to try it!

Part II: If $\left( X, d' \right)$ is complete then $\left( X, d \right)$ is complete.

To prove this, we start with a Cauchy sequence in $\left( X, d \right)$. So, let that Cauchy sequence be $\left\lbrace x_n \right\rbrace$. We have, $\forall \epsilon > 0, \exists n_0 \in \mathbb{N}$ such that $\forall n, m > n_0, d \left( x_n, x_m \right) < \epsilon$. Now, $d' \left( x_n, x_m \right) = \min \left\lbrace d \left( x_n, x_m \right), 1 \right\rbrace \leq d \left( x_n, x_m \right) < \epsilon$ for the same stage $n_0$. Thus, the sequence is Cauchy in $\left( X, d' \right)$ and hence is convergent in $\left( X, d' \right)$. Now, we need to prove that if it converges to $x \in \left( X, d' \right)$, then it converges to the same limit in $\left( X, d \right)$. Again, I encourage you to try it on your own!

Answer 2

We have $d'=\min (1,d).$ Let $\{d_1,d_2\}=\{d,d'\},$ i.e. $(d_1=d\land d_2=d') \lor (d_1=d' \land d_2=d).$

If $d_1$ is complete and $(x_n)_{n\in \Bbb N}$ is any $d_2$-Cauchy sequence then there exists $n_0\in \Bbb N$ such that $\forall m,n>n_0\;(d_2(x_m,x_n)<1).$ So $\forall m,n>n_0\;(d_1(x_m,x_n)=d_2(x_m,x_n)).$ So $\lim_{s\to \infty} \sup_{m>n>s}d_1(x_m,x_n)=\lim_{s\to \infty}\sup_{m>n>s}d_2(x_m,x_n)=0.$

So $(x_n)_{n\in \Bbb N}$ is also a $d_1$-Cauchy sequence, and since $d_1$ is complete, there exists $x$ such that $\lim_{n\to \infty}d_1(x,x_n)=0.$

Now there are only finitely many $n$ for which $d(x,x_n)\geq 1,$ so $d_2(x,x_n)=d_1(x,x_n)$ for all but finitely many $n.$ So $\lim_{n\to \infty }d_2(x,x_n)=\lim_{n\to \infty}d_1(x,x_n)=0. $