clarification on step of proof with regards to a question in completeness in metric space topology

Question

I have a question with regards to the following question:

Let $(X,d)$ be any metric space and $(X,d')$ be the standard bounded metric. The standard bounded metric is defined as: $d'(x,y)=\min(d(x,y),1).$ Show that $(X,d')$ is complete if and only if $(X,d)$ is complete.

In the if direction, I would like some clarification on a subtle point within the proof. I should mention I am not assuming nor using the fact that both metric spaces are equivalent.

Let $ \{x_n\}$ be any Cauchy sequence in $(X,d)$, we let $d'(x,y)=\min(d(x,y),1).$. Since $(X,d')$ is assumed to be both Cauchy sequence and compete, we need to show that both $(X,d)$ is also both Cauchy and complete.

It is here at this stage where I have have some confusion about a sublet point. From the definition $d'(x,y)=\min(d(x,y),1)=\min(\epsilon, 1)=\delta$ we know that since $d'(x,y)$ is Cauchy, and $\epsilon$ could be either $\leq 1$ or $\gt 1$....

Here is where I need clarification. We only considered the case where $\epsilon\leq 1$ and we don't care about $\epsilon\gt 1$. In the later case, we would have $d'(x,y)=1$, in which case, there is nothing to prove. Hence for the case $\epsilon\leq 1$, we can make use of the definition $d'(x,y)=\min(d(x,y),1).$ which would allow one to make use of the hypothesis that $(X,d')$ is Cauchy and complete. The reason I ask this is in proofs of showing how both metric spaces are equal, the case where $\epsilon\gt 1$ is considered. But for this particular question, $\epsilon \gt 1$ is never explicitly mentioned.

Thank you in advance.

Answer 1

Lemma

A sequence $\{x_n\}$ in a metric space is cauchy iff for every $\epsilon \in (0,1)$ there exists $n_0$ such that $d(x_n,x_m) <\epsilon$ for all $n, m \geq n_0$.

Proof: one way is obvious. Suppose the stated condition holds. Let $\epsilon$ be any positive number, not necessarily less than $1$. Consider two cases:

1) $\epsilon <1$

2) $\epsilon \geq 1$

In case 1) there is nothing to prove. Suppose we are in case 2). Apply the given hypothesis for $\epsilon =\frac 1 2$. We see that there exists $n_0$ such that $d(x_n,x_m) <\frac 1 2$ for all $n, m \geq n_0$. But then $d(x_n,x_m) <\epsilon$ for all $n, m \geq n_0$. We are done.

Similar result holds for convergence of a sequence.

Answer 2

Let $\{x_{n}\}$ be a Cauchy sequence in $(X,d^{\prime})$.

Let $\epsilon>0$ be arbitrary and define $\epsilon^{\prime}\equiv\min\{\epsilon,1\}$. Since $\{x_{n}\}$ is Cauchy, we can find $N$ such that $d^{\prime}(x_{n},x_{m})<\epsilon^{\prime}\leq1$ whenever $n,m\geq N$. Note, in particular, that $d^{\prime}(x_{n},x_{m})<1$ implies $d^{\prime}(x_{n},x_{m})=d(x_{n},x_{m})$, and hence we can conclude that $\{x_{n}\}$ is also Cauchy in $(X,d)$.

Now, suppose $(X,d)$ is complete. This implies that $d(x_{n},x)\rightarrow0$ for some $x$ in $X$. Therefore, $d(x_{n},x)=d^{\prime}(x_{n},x)$ for sufficiently large $n$, which in turn implies $d^{\prime}(x_{n},x)\rightarrow0$, as desired.