Bijection from $(0,1]$ to $[0, \infty)^2$ Define a bijection from $(0,1]$ to $[0, \infty)^2$
Route to follow,
A-) First define a bijection from $(0,1]$ to $(0,1]^2$
B-) Since there is a bijection from $(0,1]$ to $[0, \infty)$, namely $f(x) = (1/x) -1$, there is a bijection from $(0,1]^2$ to $[0, \infty)^2$
B says, if $f:A \rightarrow B$ is a bijection then there is a bijection $h:A^2 \rightarrow B^2$
Can anyone define me a function that satisfies A, and a function h for proof of B. 
Rigor at elementary - intermediate analysis level will be appericiated. 
Note: If possible I wonder the validity of infinite decimal approach for defining a function for part A. 
 A: For B): $h:=\langle a_1,a_2\rangle \mapsto \langle f(a_1),f(a_2)\rangle$, easy to see that it is a bijection, if $f$ was.
A) I think, can be valid, but should take care: first of all, I would apply $x\mapsto (1-x)$ because prefer $[0,1)$. Then, each $z\in [0,1)$ can be written in infinite decimal form, at most in two ways (since $0.1=0.99999\dots$), and choose the simpler one, so assume the digits are:
$$z=0.a_1b_1a_2b_2a_3b_3\dots$$
Then $z\mapsto \langle 0.a_1a_2a_3\dots,\ 0.b_1b_2b_3\dots\rangle$ will almost be a bijection, but problems may occur only because of the above phenomenon, some pairs of numbers like
($0.0109090909\dots$ and $0.02$) will violate injectivity.
So, probably best is to consider first a bijection $[0,1)\to\{0..9\}^{\mathbb N} $, and then by this comb procedure conclude a bijection   $\left(\{0..9\}^{\mathbb N}\right)^2 \to \{0..9\}^{\mathbb N}$
in an exact way.
The above map of decimal digits $g:[0,1)\to\{0..9\}^\mathbb N$ (which omits the sequences ending full of $9$'s from the range) is injective, and the number of omitted sequences is countable..
Note: Of course, we could also have used binary fractions and $\{0,1\}$ instead of $\{0..9\}$, but doesn't really matter.
