Prove that a metric space is complete.

Question

Let $a,b \in [0,\infty)$ with $a \leq b$.

$D(x,y)=\mid \frac{x}{2+x}-\frac{y}{2+y} \mid$.

There exist constants $c_{1}, c_{2} \in [0,\infty)$ such that $$c_{1}\mid x - y \mid \leq D(x,y) \leq c_{2}\mid x - y \mid \forall x,y\in [a,b]$$

Show that $([a,b],D)$ is a complete.

I am very confused about how to prove that a metric space is complete. There are multiple theorems involving Cauchy sequences and closed subsets. I have seen solutions involving the standard metric but I am not sure how to use it in this proof. Any help would be appreciated.

Answer 1

Let $ (u_n) $ be a Cauchy sequence in $ ([a,b],D)$.

Given $ \epsilon >0$.

$$\exists N \;: \forall p,q\ge N \; D(u_p,u_q)<\epsilon.c_1$$

$$\implies c_1|u_p-u_q|<c_1D(u_p,u_q)<\epsilon$$

$$\implies (u_n) \text{ is Cauchy in } ([a,b],| |) $$

$$\implies (u_n) \text{ converges in } \Bbb R$$

but $ [a,b] $ is closed, so $$\lim_{n\to +\infty}u_n\in [a,b] $$

Answer 2

The key is: $$c_1 d(x,y) \leq D(x,y) \leq c_2 d(x,y)$$ If $(x_n)_n$ is Cauchy wrt to $D$ use the first part of the inequality to show that it is Cauchy wrt to the standard metric (call it $d$), so it converges to $x \in [a,b]$ wrt to $d$ since $([a,b],d)$ is complete. Use the second part of the inequality to show that $(x_n)_n$ converges to $x$ also wrt $D$.