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I am aware of this and this topic, but I would like to receive a clarification concerning the foregoing naive attempt to prove Countable Choice, to see if I understood properly the question.

Assume that $A$ is a countable set, that is to say, assume that we have a bijection $I\subseteq \mathbb{N} \times A$. For the sake of simplicity, we may assume that $A = \left\{X_n\mid n \in \mathbb{N}\right\}$ is a set os sets.

For every $n$ in $\mathbb{N}$, I set $$A_n := \left\{x\mid x\in A \text{ and }(m,x)\in I \text{ for }m\leq n\right\} = \left\{X_m\mid 0\leq m\leq n\right\}.$$ This is a finite set, whence I know, by ordinary induction, that there exists a choice function $$f_n:A_n\setminus \emptyset \to \bigcup A_n \qquad \text{s.t.} \qquad f_n(x) \in x \text{ for all }x\in A_n\setminus\emptyset.$$ In the easier setting, $$f_n:A_n\setminus \emptyset \to \bigcup_{m=0}^nX_m \qquad \text{s.t.} \qquad f_n(X_m) \in X_m \text{ for all } 0\leq m \leq n.$$

As a consequence, I would like to consider the function $$F := \left\{y \mid y \in \mathbb{N}\times \left(\bigcup A\right) ^{A\setminus\emptyset} \text{ and }y=(n,f_n)\right\}$$ i.e. $$F : \mathbb{N} \to \mathsf{Fun}\left(A\setminus \emptyset,\bigcup A\right), \qquad n \mapsto f_n.$$

Now, I would like to define a choice function on $A$ as follows. Since for every $x$ in $A\setminus \emptyset$ there exists a unique $n$ in $\mathbb{N}$ such that $(n,x) \in I$, I would set $$\mathcal{F} := \left\{z \mid z \in (A\setminus \emptyset)\times \bigcup A \ \text{ and } \ z=(x,f_n(x)) \ \text{ for } \ (n,f_n) \in F\right\} \subseteq (A\setminus \emptyset)\times \bigcup A.$$

Question: Does the problem lie in the fact that $F$ is not necessarily well-defined, because I am impliclty performing an infinite choice (since $f_n$ is not uniquely determined)?

Is this similar to the argument concerning the fact that "being $n$ finite for every $n \in \mathbb{N}$ does not imply that $\mathbb{N}$ is finite?"

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Ender Wiggins
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1 Answers1

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The problem with the usual inductive arguments is that if $A_n$ has more than one element, and you are unable to specify exactly a unique element of $A_n$ being chosen at each step, then you have a "splitting" in your recursion which requires you making some arbitrary choice.

As long as you're going along a finite path, making a choice is fine, because you've only had to make finitely many of theses choices. But if you want to amalgamate them, well, then you have to already have had a set of coherent choices, that is to say, an infinite path that you could follow in your recursive definition.

The problem is when you have splittings and you need to make arbitrary choices every so often, you cannot guarantee to have had this path, unless you assume countable choice. And thus cause a circularity. This is exactly the same failure with "every $n\in\Bbb N$ is finite, therefore $\Bbb N$ is finite". The only way to have proved that $\Bbb N$ is finite, is to have assumed it was finite to begin with.

Compare this with the actual proof of the recursion theorem, well there you do have a well-defined singular choice. You are given a function $f\colon A\to A$, and you are using it to define your sequence $F(0)=f(a)$, for some fixed $a\in A$, and as a function spits a single element back at you, there is no arbitrary choices when you want to define $F(n+1)$, it is simply $f(F(n))$.

Asaf Karagila
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