Bijective function

Let $f:𝕑\to 𝕑$ be a bijective function. Prove that you can find two other bijective functions $u$ , $v$ from $𝕑$ to $𝕑$ such that $f\left(x\right)=u\left(x\right)+v\left(x\right)$ , $\forall x\in 𝕑$ .


Update (23/03/2008): Solution posted! (show solution)
A solution was written in the comments (in italian), but i will still post the "official" solution because it is short and instructive. Chose an inductive ordering for $𝕑$ (for instance ). Suppose now that we have defined $u$ and $v$ on a finite set $A\in 𝕑$ such that they are injective and $f\left(x\right)=u\left(x\right)+v\left(x\right)$ , $\forall x\in A$ , and start with $A=\varnothing$ . Now we can:

Extend the domain: chose the smallest (w.r.t the ordering) $x\in 𝕑\setminus A$ , and define $u\left(x\right)$ and $v\left(x\right)$ with $f\left(x\right)=u\left(x\right)+v\left(x\right)$ and so that $u$ and $v$ are still injective .We have infinite possible choices for $u\left(x\right)$ that automatically also determine $v\left(x\right)=f\left(x\right)-u\left(x\right)$ , and they are all good but a finite number (those that make $u$ or $v$ no longer injective).

Extend the image of $u$ : let's chose the smallest $y\in 𝕑\setminus u\left(A\right)$ , and look for an $x\in 𝕑\setminus A$ such that we can define $u\left(x\right)=y$ , $v\left(x\right)=f\left(x\right)-u\left(x\right)$ keeping $u$ and $v$ injective. Again, we have infinite $x$ that we can try ( $f\left(x\right)$ are all different!), this choice also determines $v\left(x\right)$ , and only a finite number of them is bad.

The same as 2, but with $v$ in the place of $u$ .

After applying any rule the functions will stay injective. Note that if we apply the rule 1 infinite times, we cannot be sure that $u$ and $v$ are sujective, if we apply 2 infinite times, $u$ will be surjective, but possibly not defined everywhere, and similarly for 3 (w.r.t. $v$ ). But if we apply all them infinite times as 1,2,3,1,2,3,1... we can be sure that $u$ and $v$ are defined everywhere and that they are surjective, because any missing value would be definitively caught by the right rule.