finding $f$ function For which of the sets $\mathbb{X}:= \mathbb{N},\mathbb{Z},\mathbb{Q},\mathbb{R},\mathbb{C}$, there exist a function $f:\mathcal{P}(\mathbb{X})\rightarrow\mathbb{X}$ that for all $A\in\mathcal{P}(\mathbb{X})\setminus \lbrace\phi\rbrace$, $f(A)\in A?$ I want to find $f$ function? please help me.
 A: A function $f:\mathcal{P}(A)-\{\varnothing\} \to A$ satisfying the condition $f(A)\in A$ is called a choice function on $A$. 
In the case of $\Bbb{N}$, $\Bbb{Z}$, $\Bbb{Q}$, you can construct the choice function explicitly. For example, you can define $f:\mathcal{P}(\Bbb{N})\to\Bbb{N}$ as like:
$$
f(A):=\min A 
$$
In the case of $\Bbb{Z}$ you can construct the bijection $g:\Bbb{Z}\to\Bbb{N}$. If we define $h(A)=g^{-1}(f(g_*(A)))$ (where $g_*(A)$ is image of $A$ under $g$) then $h$ satisfies $f(A)\in A$ for all nonempty $A$. Similarly, you can construct the bijection between $\Bbb{Q}$ and $\Bbb{N}$ and you can also define the function satisfy the given condition.
In the case $\Bbb{R}$ and $\Bbb{C}$, you can prove the existence of these functions. However, there is no explicit construction for the function satisfy the given condition. In fact, the existence of these function requires the axiom of choice and it is necessary to prove the existence of choice function on $\Bbb{R}$. It is known that '$\Bbb{R}$ has no choice function' consistent with ZF. 
