How can you define a function with domain the whole of RR and range the whole of CC?

(It is possible to define a bijection constructively)

1 Answer
Dec 31, 2016

See explanation...

Explanation:

Define h(x): (0, 1) -> (-oo, oo):

h(x) = (1-2x)/(x(x-1))

graph{(sqrt(1/4-(x-1/2)^2))/(sqrt(1/4-(x-1/2)^2))(1-2x)/(x(x-1)) [-1, 2, -10, 10]}

Then:

h^(-1)(y) = ((y-2)+sqrt(y^2+4))/(2y)

So h(x) is a bijection between (0, 1) and RR, with inverse h^(-1)(y)

Let S = { a+bi : a, b in (0, 1) } i.e. the open unit square in Q1.

We can use h(x) on the Real and imaginary parts of a Complex number to define a bijection k(x) between S and CC:

k(a+bi) = h(a)+h(b)i

k^(-1)(a+bi) = h^(-1)(a) + h^(-1)(b)i

Having found these bijections, next consider decimal representations of numbers in (0, 1).

Some numbers have two possible decimal representations; one with a tail of repeating 0's and the other with a tail of repeating 9's. For our purposes, we disallow representations with a tail of repeating 0's, choosing to use the representation with repeating 9's.

Then there is a bijection between (0, 1) and representations of the form 0.a_1 a_2 a_3 a_4 ..., where a_i in { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 } are decimal digits and no sequence has a tail of repeating 0's.

Given a number 0.a_1 a_2 a_3 a_4 ..., split the sequence into a sequence of subsequences, each of which is as short as possible, but ends with a non-zero digit.

For example:

0.0230040291...

splits as:

0." "02" "3" "004" "02" "9" "1" "...

Then recombine alternate subsequences to form two Real numbers in (0, 1):

0.020049...

0.3021...

These are the Real and imaginary parts of a corresponding Complex number in S

This process defines a function:

s(x): (0, 1) -> S

Conversely, given a Complex number in S, disassemble the Real and imaginary parts, then interlace them to form a Real number in (0, 1):

So given:

0.020049... + 0.3021... i

Split into subsequences:

0." "02" "004" "9" "...

0." "3" "02" "1" "...

Interlace:

0.0230040291...

This defines the inverse function s^(-1)(y): S -> (0, 1)

Hence we can define a function with domain RR and range CC by:

f(x) = k^(-1)(s(h(x)))

This is a bijection, with inverse:

f^(-1)(y) = h^(-1)(s^(-1)(k(y)))