I want to prove $[0, \infty)$ with $d(x, y) = |\sqrt x - \sqrt y|$ is not complete. 
prove that $[0, \infty)$ with $d(x, y) = |\sqrt x - \sqrt y|$ is not complete.

I want to find Cauchy sequence but not convergent with $d$.
It seems too hard for me.
Please tell me the hint.
 A: Actually the problem has nothing to do with square roots except for the property of $x\mapsto\sqrt x$ to be a bijection. If $f:X\to(Y,d)$ is any injective function, where $X$ is a set and $Y$ is a space with metric $d$, and if we equip $X$ with the metric $d_f(x,x')=d(f(x),f(x'))$, then $f$ becomes an isometry $(X,d_f)\to(Y,d)$, i.e. a map which preserves distances. You can then use the fact that for any bijective isometry between two metric spaces, one space is complete if and only if the other is complete.
A: The space $A=[0,\infty)$ with the norm $d(x,y)=|\sqrt{x}-\sqrt{y}|$ is complete. Here's a more elementary proof.
Let $(a_n)$ be a Cauchy sequence in $(A,d)$, and $\varepsilon$ a positive real number. Then, there is a positive integer $N_\varepsilon$ such that
$$
d(a_n,a_m)\le \varepsilon \quad \forall m,n\ge N_\varepsilon.
$$
For $\varepsilon=1$, we get
$$
d(0,a_n)\le d(0,a_{N_1})+d(a_{N_1},a_n) \le  d(0,a_{N_1})+\max\{1,\max_{1\le i,j\le N_1}d(a_i,a_j)\} \quad \forall n,
$$
i.e.
$$
0\le a_n\le M \quad \forall n, 
$$
where
$$
M^2\le d(0,a_{N_1})+\max\{1,\max_{1\le i,j\le N_1}d(a_i,a_j)\}.
$$
Now, for every $m,n\ge N_\varepsilon$ we have
$$
|a_n-a_m|=(a_n+a_m)d(a_n,a_m) \le 2M\varepsilon,
$$
i.e. $(a_n)$ is a Cauchy sequence in $(\mathbb{R},|\cdot|)$, and therefore is convergent because $(\mathbb{R},|\cdot|)$ is complete. Since $A$ is closed in $\mathbb{R}$, we have $\lim_na_n\in A$, and therefore $(A,d)$ is complete.
