2 Fixed up the TeX a bit

Let's say we want to define a choice function for certain particular subsets $S \subset2^{\mathbb{R}}$, i.e. we want a function $c:S \rightarrow \mathbb{R}$ such that $c(X)\in X$ for every $X\in S$. We don't want to invoke the axiom of choice. Clearly we require $\emptyset\notin S$.

For example, if $S$ is the set of non-empty open sets for the usual topology, then we can fix an enumeration of the rationals and for every open $A$ pick the first rational (in this particular enumeration) lying in $A$.

If $S$ is the set of non-empty closed sets, then for any $A\in S$ we can consider the least $n$ such that $[-n,n]\cap A$ is non empty and then pick the infimum of this non empty compact set.

The question is the following: can you define a choice function for, say, $S=F_{\sigma}\setminus {\emptyset}$ \{\emptyset\}$or$G_{\delta}\setminus{\emptyset}$G_{\delta}\setminus\{\emptyset\}$ or maybe for the higher levels of the Borel hierarchy? Is it possible to prove that such choice function exists for such $S$ without using the axiom of choice?

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# Choice function for Borel sets?

Let's say we want to define a choice function for certain particular subsets $S \subset2^{\mathbb{R}}$, i.e. we want a function $c:S \rightarrow \mathbb{R}$ such that $c(X)\in X$ for every $X\in S$. We don't want to invoke the axiom of choice. Clearly we require $\emptyset\notin S$.

For example, if $S$ is the set of non-empty open sets for the usual topology, then we can fix an enumeration of the rationals and for every open $A$ pick the first rational (in this particular enumeration) lying in $A$.

If $S$ is the set of non-empty closed sets, then for any $A\in S$ we can consider the least $n$ such that $[-n,n]\cap A$ is non empty and then pick the infimum of this non empty compact set.

The question is the following: can you define a choice function for, say, $S=F_{\sigma}\setminus {\emptyset}$ or $G_{\delta}\setminus{\emptyset}$ or maybe for the higher levels of the Borel hierarchy? Is it possible to prove that such choice function exists for such $S$ without using the axiom of choice?