Prove that there is a bijection $f: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$
$f(x,y) = h(g(x) + r(y))$
As far as I understand, one need to prove that there exist such $g(x)$ and $r(y)$ so for any $(x_1,y_1), (x_2,y_2) \Rightarrow g(x_1) + r(y_1) \neq g(x_2) + r(y_2)$.
Is there any way to prove that such functions exist or is there another convenient way to solve the problem?
Best Answer
Hint:
You may create a number alternating the digits of $x$ and $y$.
(This is not the full answer and it has some problems, but it might help you get started.)
Edit:
Since the answer is already accepted, I will post an example for $h$, $g$ and $r$.
Initially, I though about alternating digits, but negative numbers might be a problem. So, using the exponential function (that is injective) will allow to get positive numbers. Afterwards, we could alternate the digits between the transformed numbers.
Let $a_k \in \{0,1,\ldots,9\}$ be the decimal representation of $e^x$, i. e.: $$ e^x = \sum_{k=-\infty}^{+\infty} a_k 10^k $$ with $a_k$ chosen such that for any $J$ there exists at least one $j > J$ such that $a_j \neq 9$. (Thanks due to fleablood, this way we prevent two sequences giving the same number.)
The function $g(x)$ will be defined as: $$ g(x) = \sum_{k=-\infty}^{+\infty} a_k 10^{2k} $$ Note the exponent $2k$.
Similarly, let $b_k \in \{0,1,\ldots,9\}$ be the decimal representation of $e^y$, i. e.: $$ e^y = \sum_{k=-\infty}^{+\infty} b_k 10^k $$ with $b_k$ chosen such that for any $J$ there exists at least one $j > J$ such that $b_j \neq 9$.
The function $r(x)$ will be defined as: $$ r(x) = \sum_{k=-\infty}^{+\infty} b_k 10^{2k+1} $$ Note the exponent $2k+1$.
And the function $h$ can be simply: $$ h(z) = \log z $$
This way, the sum between $g$ and $r$ will give a positive number with alternating digits of $e^x$ and $e^y$. Since $h$ is the logarithm function, the outcome can be either positive or negative or zero.