Choice function for a collection of nonempty subsets of $\{0,1\}^\omega$ 
Possible Duplicate:
Finding a choice function without the choice axiom 

Is possible to construct a Choice function for a collection of nonempty subsets of $X=\{0,1\}^\omega$? (Without AC)
Attempt:
We know an explicit injection $f\colon \mathbb{N}\to X$ is so easy to construct it, I think that a possible choice function $c$ is one which $c(X)=f(1)$, $c(X\setminus\{f(\{1,\ldots,n\})\} )=f(n+1)$, but I don't know how extend it (and I don't know if is possible to do) to all collection $P(X)\setminus\{\emptyset\}$.
This exercise comes from Munkres Topology chapter 1 section 9

The answer is no.  The axiom of choice is required to get a choice function for the reals: https://mathoverflow.net/questions/45308/choice-function-on-the-powerset-of-the-reals so we just need to show that a choice function for $X$ yields a choice function for $\mathbb R$.
First note that if $Y$ has a choice function and there is a surjection $f\colon Y \to Z$ then $Z$ has a choice function: Given $U \subseteq Z$ we choose $f(x)$ where $x$ is given by applying the choice function of $Y$ to $f^{-1}(U)$.
There is a surjection $X \to [0, 1]$ defined by sending the infinite sequence $(a, b, c, \ldots) \in X$ to the real number whose base $2$ decimal expansion is $0.abc\ldots$ so $[0, 1]$ inherits a choice function from $X$.  If a choice function exists for $[0, 1]$ then it restricts to a choice function for $(0, 1)$.  Finally there are explicit bijections between $(0, 1)$ and $\mathbb R$ (for example, use the tangent function) so $\mathbb R$ inherits a choice function from $(0, 1)$.
