# 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.

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The infinite decimal approach works, but there is some unpleasantness dealing with the non-uniqueness of decimal expansions, so a detailed proof turns out to be quite a bit longer than the simple "alternately use digits from one and the other." From an $f$ from $A$ to $B$, it is immediate how to construct $h$, so I imagine that's not what you are asking. –  André Nicolas Oct 4 '12 at 14:26
$h(x,y) = (f(x),f(y))$, where $(x,y) \in A^2$ –  nikita2 Oct 4 '12 at 14:31
It might be technically easier to define, once and for all, a bijection between $(0,1]$ and $\mathcal P(\mathbb N)$, and then use it forwards and backwards to wrap the easy bijection between $\mathcal P(\mathbb N)$ and $\mathcal P(\mathbb N)^2$. –  Henning Makholm Oct 4 '12 at 14:33
I think that we should use Peano curve. –  nikita2 Oct 4 '12 at 14:46
that's not a bijection, but at least continuous. –  Berci Oct 4 '12 at 14:52

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.