Let $(X,d)$ be a complete metric space.Prove $(X,d')$ is a complete metric space $d'(x,y)=\begin{cases}\max\{1,d(x, y)\}&x\neq y\\0&x=y\end{cases}$

Question

Let $(X,d)$ be a complete metric space.

Prove $(X,d')$ is a complete metric space such that $d'(x,y)=\begin{cases} \max\{1, d(x, y) \}&x\neq y\\0&x=y\end{cases}$

Can't figure out why this statement is correct.

Let $\{X_n\}$ be a Cauchy-sequence.

By definition of Cauchy-sequence $\exists N\geq1$ such that $d(x_n,x_m)<1,\forall n\neq m>N.$

It follows that $d'(x_n,x_m)=1 \forall n,m>N$.

In another words , $\forall \epsilon$ where $0<\epsilon<1 ,d'(x_n,x_m)\ge 1$, this is not cauchy-sequence..

Where am I wrong ?

Appreciate any help.

Answer 1

Note that $d'$ is the metric $\max\{d_{\text{dis}}, d\}$, where $d_\text{dis}$ is the discrete metric on $X$.

Let $(X,d)$ be any metric space. Then the metric space $(X, d'=\max\{d_{\text{dis}}, d\}) $ is complete.

Proof :

Let $(x_n)\subset X$ be $d'$-Cauchy.

Since $d'=\max\{d_{\text{dis}}, d\}\ge d_{\text{dis}}$, any $d'$-Cauchy sequence $(x_n) \subset X$ is $d_{\text{dis}}$-Cauchy, because $d_{\text{dis}}(x_m, x_n) \le d'(x_m, x_n) \to 0$ as $m,n\to \infty$

Now $(x_n)\subset X $ is $d_{\text{dis}}$-Cauchy implies $(x_n) $ is eventually constant implies $(x_n)$ is convergent in $(X, d').$

Note: We don't need the completeness of $(X, d) $.

Answer 2

Let $(a_n)_{n \in \mathbb{N}}$ be a $d'$-Cauchy sequence.

Then for any $\varepsilon \in \mathbb{R}^+$, there exists some $N \in \mathbb{N}$ such that for all integers $m,n > N$, we have $d'( a_m, a_n) < \varepsilon$.

If we take $\varepsilon = 1$, then, taking $N$, we can see that for all $m, n > N$, we have $d'( a_m, a_n) < 1$.

This can only be the case if $d'( a_m, a_n) = 0$, and so $a_m = a_n$, by the definition of the metric $d'$.

Thus, the Cauchy sequences in $(X, d')$ are precisely the eventually-constant sequences, and any eventually-constant sequence trivially converges to the eventual constant, which is in $X$ by definition.

Thus $(X, d')$ is Cauchy-complete.