Let $(X,p) $ , $ p(x,y)=\sup\limits_k\dfrac{|x_k-y_k|}{1+|x_k-y_k|} $. Show that $(X,p)$ is complete. Let $(X,p) $ be the metric space of the real bounded sequences with the metric  $$ p(x,y)=\sup\limits_k\dfrac{|x_k-y_k|}{1+|x_k-y_k|} $$
Show that $(X,p)$ is complete.
I know the metric space of the real bounded sequence with the metric $d=\sup\limits_k|x_k-y_k| $ is complete, but I don´t know how to prove it by definition or some property.
 A: The real map $f: t \to \frac{t}{1+t}$ is continuous, strictly increasing on $[0,+\infty)$, $f([0,+\infty))=[0,1)$, $f(1)=\frac{1}{2}$ and $$f^\prime(t)=\frac{1}{(1+t)^2}.$$
Hence for $u \in [0,\frac{1}{2}]$, you have using the mean value theorem: $$u \ge f(u)= f^\prime(\xi) u \ge f^\prime (\frac{1}{2}) u =\frac{4}{9} u$$ where $0 \le \xi \le u \le \frac{1}{2}$. Consequently for $x,y \in X$ with $p(x,y) \le \frac{1}{2}$, you have $$\frac{4}{9} d(x,y) \le p(x,y) \le d(x,y).$$
Based on those inequalities, you can prove that a Cauchy sequence for $(X,p)$ is a Cauchy sequence for $(X,d)$ hence converges in $(X,d)$ according to the result you mentionned. And also converges (to the same limit) for $(X,p)$ again according to above inequalities again.
A: To prove that a given metric space is complete, you need to prove that every Cauchy Sequence has a limit.
So take $x_n$  a sequence of $x_n\in X $  which is also a Cauchy Sequence, by the given metric.
From the Cauchy quality what do you know?
How does this help you with the Cauchy interpretation of Limit*?


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*Cauchy said that a sequence has a limit iff for every $0<\varepsilon  $  there exists $ N  $, an index such that:  $  N<m,n \:then \:p(x_m,x_n)<\varepsilon$.

