# Choice function for a collection of nonempty subsets of $\{0,1\}^\omega$ [duplicate]

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

-
What's $\{0,1\}^\infty$? –  Git Gud Feb 3 '13 at 19:02
The set of all infinite sequences of 0 and 1. –  Gastón Burrull Feb 3 '13 at 19:05
So you mean $\{0,1\}^\omega$ maybe? –  Hagen von Eitzen Feb 3 '13 at 19:17
To contrast with the negative answers posted here, you may find useful the answer I just posted at math.stackexchange.com/questions/190747/… –  Andres Caicedo Feb 3 '13 at 21:37

## marked as duplicate by Asaf Karagila, Micah, Davide Giraudo, Brandon Carter, Austin MohrFeb 3 '13 at 21:06

First note that $\infty$ is not really well-defined as a set, and if you mean the set of binary sequence of index $\mathbb N$ then the proper notation is $\{0,1\}^\mathbb N$ or $\{0,1\}^\omega$ which is the same (in set theory $\omega$ denotes the set of finite ordinals which is exactly $\mathbb N$).

What you are asking, if so, is a choice function on $\mathcal P(\{0,1\}^\mathbb N)\setminus\{\varnothing\}$, but that is impossible.

Theorem: Let $X$ be a non-empty set then $X$ can be well-ordered if and only if there is a choice function on $\mathcal P(X)\setminus\{\varnothing\}$.

The proof of this theorem can be found, in a nutshell, here: How to define a well-order on $\mathbb R$? (the direction from a well-ordered to a choice function is trivial).

Theorem: It is consistent that the axiom of choice fails and $\mathbb R$ cannot be well-ordered.

This is a difficult proof which requires some deeper knowledge in set theory and consistency results, the theorem itself is due to Cohen, and this is how he proved that the failure of the axiom of choice is consistent with the other axioms of ZF.

Since $|\mathbb R|=|\mathcal P(\mathbb N)|=|\{0,1\}^\mathbb N|$, the two theorems tell us that it is impossible to construct such function without using the axiom of choice.

-
The answer is no. The axiom of choice is required to get a choice function for the reals: http://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 $(0, 1)$.