Construct a choice function for $Y=\mathscr{P}(X)\setminus \emptyset$, where $X=\Bbb Z,\Bbb Q,\Bbb R[x]$ and $Y=\{A_f:f\in \Bbb C[z]\}$.

Question

This is Exercise 5.12 from Goldrei's Classic Set Theory.

For each of the following families $\mathscr{F}$, construct, where possible, a choice function $h$ for $\mathscr{F}$, i.e. $h: \mathscr{F} \rightarrow \bigcup \mathscr{F}$ such that for each $A\in \mathscr{F}, h(A)\in A$.

(a)$\mathscr{F}=\mathscr{P}(\Bbb Z)\setminus \emptyset$

(b)$\mathscr{F}=\mathscr{P}(\Bbb Q)\setminus \emptyset$

(c)$\mathscr{F}=\mathscr{P}(\Bbb R[x])\setminus \emptyset$, where $\Bbb R[x]$ is the set of all polynomials with variable $x$ and real coefficients.

(d)$\mathscr{F}=\{A_f:f\in \Bbb C[z]\}\setminus \emptyset$, where $\Bbb C[z]$ is the set of all polynomials in $z$ with complex coefficients and $A_f=\{z\in \Bbb C:f(z)=0\}$.

Note that the use of the axiom of choice is not allowed.

My attempt:

(a) Given any $E\in\mathscr{P}(\Bbb Z)\setminus \emptyset$, either $E$ is bounded above or not bounded above. If $E$ is bounded above, then $E$ has a unique maximum element $x$. We may map $E$ to $x$. If $E$ is not bounded above, then $E\cap \Bbb N\neq \emptyset$. By well-ordering of $\Bbb N$, $E\cap \Bbb N$ has a least element $y$. Moreover, the least element of $E\cap \Bbb N$ is clearly unique and is in $E$. We may map $E$ to $y$.

(b) I have no idea on how to construct a choice function on this set.

(c) I have no idea on how to construct a choice function on this set.

(d) Let $E\in \{A_f:f\in \Bbb C[z]\}\setminus \emptyset$. If $E$ is the set of zeroes of the zero polynomial, then map $E$ to $0$ (clearly, $0$ is a root of the zero polynomial). Suppose $E$ isn't the set of zeroes of the zero polynomial, then $E$ is the set of zeroes of some polynomial $f\in \Bbb C[z]$ of finite degree. Then $f$ has at most finitely many zeroes. Since the real parts of the elements in $E$ are linearly ordered and $E$ has finitely many elements, we may form a subset $F$ of $E$ such that the real part of the elements of $F$ is minimal in $E$. Since $F$ is finite and the imaginary parts of the elements of $F$ are linearly ordered, $F$ has a unique element $a$ such that the imaginary part of $a$ is minimum in $F$. We may map $E$ to $a$.

Is it possible to construct choice functions for the sets in (b) and (c)? If not, how can we prove that no such choice functions exist?

Answer 1

(a) Your proof is OK, but there is a simpler one. Notice that, since $\mathbb{N}$ and $\mathbb{Z}$ are isomorphic sets (=there is a bijective map $\mathbb{N} \to \mathbb{Z}$), it is sufficient to find a choice function for the non-empty sets of natural numbers. For these we can always take the minimum. More generally, every well-ordered set provides this easy choice function. And we can well-order $\mathbb{Z}$ by defining, say,

$0 < 1 < -1 < 2 < -2 < \dotsc$

And this makes it also easy to tackle (b). Namely, $\mathbb{Q}$ is also just isomorphic to $\mathbb{N}$ as a set (see here for a collection of isomorphisms), so we are done. You can also see it that way: We can use that isomorphism to define a well-order on $\mathbb{Q}$.

Your proof for (d) is OK. You can also generalize it by noting that a choice function always exists for families of non-empty finite(!) subsets of a linearly ordered set. We always take the minimum. And the set of complex numbers is linearly ordered via the lexicographic order from the usual ordering of the real numbers (this is what you essentially did here).

Since $\mathbb{R}[x]$ and $\mathbb{R}$ are isomorphic as sets (!), the question (c) is equivalent to giving a choice function for the non-empty subsets of real numbers. I don't think we can write it down, we would need (a fragment of) the axiom of choice to find one, see this MO question for details.

You have asked for a proof that no choice function (for real numbers) exists. Well, this is not so straight forward. You cannot prove that it exists in the axiomatic system for set theory ZF, but you also cannot disprove its existence in ZF. It is an undecidable statement. If it is true or not depends on what you want to add to ZF to make it decidable. As said, in ZF + AC = ZFC we have a choice function for sure.

It would be much easier to consider $\mathbb{Q}[x]$. That set is countable and hence has a well-order and we are done.

Notice again that this only regards the underlying set of the ring of polynomials. Only the set is relevant here for the construction of choice functions. And this makes it possible to do all sorts of reductions which would not be possible if you really worked with the rings. For example, the rings $\mathbb{R}$ and $\mathbb{R}[x]$ are of course not isomorphic, but their underlying sets are. Maybe one should really write $U(\mathbb{R}[x])$ in your question to make this clear (where $U$ denotes the underlying set of a ring).