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