Prove that the metric space is Cauchy: $d(x,y)=|\frac{x}{3+x}-\frac{y}{3+y}|$ I am trying to show that metric space $([0,4],d)$ with $d(x,y)=|\frac{x}{3+x}-\frac{y}{3+y}|$ is complete.
Using the theory:

A metric space (X,d) or normed space (X,‖∙‖) is called complete if every Cauchy sequence in the space converges to some point in the space.

First i try to prove that the sequences $(n)_{n=1}^∞$ metric space is Cauchy:
Let: $\varepsilon>0$
Here: $d(n,m)=|\frac{n}{3+n}-\frac{m}{3+m}|=\frac{3|n-m|}{(n+3)(m+3)}≤ \frac{3}{9}|n-m|$
I am struggling complete this.

I have already shown that every sequences is cauchy is the interval $[0,\infty)$, from which i got $m,n≥\frac{6}{ε}$, which won't be satisfied in the interval [0,4].



 A: The key observation here is that (as you noted), for $x,y\in[0,4]$ we have
$$
d(x,y)=\frac{3\lvert x-y\rvert}{(x+3)(y+3)},
$$
and therefore
$$
\frac{3}{49}\lvert x-y\rvert\leq d(x,y)\leq\frac{1}{3}\lvert x-y\rvert.
$$
Why is this the key observation?

*

*The bound $\frac{3}{49}\lvert x-y\rvert\leq d(x,y)$ tells us that a sequence which is Cauchy with respect to $d$ is also Cauchy with respect to the usual Euclidean distance $(x,y)\mapsto\lvert x-y\rvert$. (Why?)

*The bound $d(x,y)\leq\frac{1}{3}\lvert x-y\rvert$ tells us that convergence in the usual Euclidean distance implies convergence with respect to $d$. (Why?)

So, roughly, you proceed here as follows:

*

*Take any Cauchy sequence (wrt $d$), call it $(x_n)_{n\in\mathbb{N}}$.

*By (1) above, this sequence is also Cauchy with respect to the Euclidean distance on $[0,4]$.

*But $[0,4]$ equipped with Euclidean distance is complete, and therefore $x_n\to x$ in Euclidean distance as $n\to\infty$ for some $x\in[0,4]$.

*Then (2) above implies that $(x_n)\to x$ with respect to $d$ as $n\to\infty$ as well.


Note that the pair of inequalities above is a specific example of what are called equivalent metrics.  Specifically, two metrics $d_1,d_2$ on the same space $X$ are called equivalent if you can find positive constants $c,C$ such that for all $x,y\in X$,
$$
c d_1(x,y)\leq d_2(x,y)\leq Cd_1(x,y).
$$
The sketch of the proof above shows why this is a useful concept -- convergence with respect to $d_1$ and $d_2$ is actually equivalent.
