If you are looking for a bijection from IRxIR β [-1; 1], I can suggest the following:
bijection from IR to] -a: a [
First, find the bijection from IR->] -1; 1 [so we just need to find a bijection from IRxIR-> IR
tan(x): ]-Pi/2;Pi/2[ -> IR arctan(x) : IR -> ]-Pi/2;Pi/2[ 1/Pi*arctan(x) + 1/2: IR -> ]0;1[ 2*arctan(x) : IR->]-Pi:Pi[
and
ln(x) : IR + -> IR exp(x): IR -> R+
Bijection from] 0.1 [x] 0.1 [->] 0.1 [
write:
(x,y) in ]0,1[ x ]0,1[ x= 0,x1x2x3x4...xn...etc where x1x2x3x4...xn represent the decimals of x in base 10 y=0,y1y2y3y4...ym...etc idem Let define z=0,x1y1x2y2xx3y3....xnyn...Oym in ]0,1[
Then, by construction, we can prove that this is an exact bijection from] 0,1 [x] 0,1 [to] 0,1 [. (I'm not sure if this is true for the zith number of infinite decimals .. but this is at least a βvery goodβ injection, tell me if I'm wrong)
denote this function: CANTOR (x, y)
then 2 * CANTOR-1 is a bijection of] 0.1 [x] 0.1 [->] -1.1 [
Then combining all of the above statements:
here you go, you get a bijection from IRxIR β] -1; 1 [...
You can combine with a bijection from IR->] 0,1 [
IRxIR -> ]-1;1[ (x,y) -> 2*CANTOR(1/Pi*arctan(x) + 1/2,1/Pi*arctan(y) + 1/2)-1
let reciproque define, we process the same:
RCANTOR: z β (x, y) (reciproque of CANTOR (x, y)
RCANTOR ((z + 1) / 2):] -1: 1 [->] 01 [x] 0.1 [
then 1/Pi*tan(RCANTOR((z+1)/2)) + 1/2 : z ->(x,y) ]-1;1[ -> IRxIR