Why isn't this a valid argument to the "proof" of the Axiom of Countable Choice?

I am having a little trouble identifying the problem with this argument:

Let $\{A_1, A_2, \ldots, A_n, \ldots\}$ be a sequence of sets.

Let $X:= \{n \in \mathbb{N} : $ there is an element of the set $A_n$ associated to $n \}$

(1) $A_1$ is not empty. Therefore, there exists a $x_1 \in A_1$

(2) Given an $A_n$ (and a $x_n \in A_n$), we have that $A_{n+1}$ is non-empty and, therefore, there exists a $x_{n+1} \in A_{n+1}$

So, by induction, $X=\mathbb{N}$ and the axiom of countable choice is "proven".

Where is the error?

Solution 1:

Inductions give you every finite case but not the infinite case. Induction is not "continuous" in this aspect.

Yes, $X=\Bbb N$ and from this follows the fact that every finite subcollection admits a choice function does not imply that you can actually choose from infinitely many sets at once.

Let me give you a ridiculous analogue to your argument, it is ridiculous so you can easily find the mistake there, and I hope that it will help you see your own mistake as well:

"Theorem". The set of natural numbers is finite.

"Proof". Let $X=\{n\in\mathbb N\mid\text{The set }\{k<n\mid k\in\Bbb N\}\text{ is finite}\}$; clearly $0\in X$, and if $n\in X$ then $n+1\in X$.

Therefore $X=\mathbb N$ and so $\Bbb N$ is finite as wanted. $\square\!\!\!\small?$

It is quite obvious what I did wrong here, I showed that every initial segment is finite, but not that the entire collection is finite.

Let's push this one step forward, here is a wrong argument which follows similarly, but its proof is slightly less exaggerated than the one above:

"Theorem". The power set of $\Bbb N$ is countable.

"Proof". Write $\Bbb N=\bigcup [k]$ where $[k]=\{0,\ldots,k-1\}$. Then for every $k$, $\mathcal P([k])$ is finite. Therefore $\mathcal P(\Bbb N)=\bigcup\mathcal P([k])$ is a countable union of finite sets, and so it is countable. $\square\!\!\!\small?$

In here we also proved by induction that every initial segment has a finite power set, but we made a mistake by concluding that the power set of $\Bbb N$ is this union, which if you look closely, does not even contain $\Bbb N$ itself. The mistake here is that we assume that the induction carries over to the limit stage, i.e. we falsely assume that $\sup\{2^n\mid n\in\mathbb N\}=2^{\sup\{n\mid n\in\mathbb N\}}$.

Solution 2:

As pointed out in comments and answers, the issue is that from the existence of finite sequences $(x_1,\dots,x_n)\in A_1\times \dots\times A_n$, we cannot conclude the existence of an infinite sequence $(x_1,x_2,\dots)\in A_1\times A_2\times\dots$ without some additional assumption (such as the axiom of countable choice).

This certainly seems strange when one first encounters it, so perhaps the following may indicate why there is something subtle going on:

From the work of Gödel, we know that there is no recursive consistent, complete set of axioms extending first-order Peano arithmetic $\mathsf{PA}$. What this means is that if $T$ is a set of axioms extending Peano Arithmetic, $T$ is consistent, and there is a computer program that, for each formula $\phi$, tells us whether $\phi$ is an axiom of $T$ or not, then there are statements $\psi$ such that $T$ cannot prove $\psi$, and cannot prove $\lnot\psi$ either.

(If you have issues with using Peano Arithmetic here, we can replace it with much weaker systems, such as Robinson's arithmetic $Q$; this is not an issue.)

Ok, consider now the following construction, that it is perhaps best explained as labelling the nodes of a subtree of $2^{<\mathbb N}$, the set of finite sequences of $0$s and $1$s.

Start by fixing a list $\phi_0,\phi_1,\dots$ of all sentences in the language of arithmetic. Put $\mathsf{PA}$ in the bottom node $\emptyset$. Given a finite sequence $t\in 2^{<\mathbb N}$, assume we have assigned a consistent set of axioms $T_t$ to the node $t$. This node has two successors, $t^\frown(0)$, and $t^\frown(1)$. Go through the list $\phi_0,\phi_1,\dots$, until you reach a $\phi_n$ such that $T_t$ does not prove $\phi_n$ and does not prove $\lnot\phi_n$. Assign to $t^\frown(0)$ the theory with axioms $T_t\cup\{\phi_n\}$, and to $t^\frown(1)$ the theory with axioms $T_t\cup\{\lnot\phi_n\}$.

Now, for each $n$, there is a computer program that lists the axioms of $T_s$ for all $s$ of length $n$ or less. However, this does not mean that we can find a computer program that lists the axioms through any of the branches of the tree, because any branch gives us a complete theory, and we would contradict Gödel's result. Think about this example, and note that "induction" would not help us here, as we can modify this construction slightly so that the theories at the end cannot be described in any "arithmetically definable" fashion. This is a much more generous notion than just "recursive", and it is still not enough.

The issue is the lack of uniformity: though for each $n$ there is a computer program that identifies the theories at level $n$, there is no algorithmic method for identifying the theories that are produced "at the end".

Similar issues are identified in the area of "reverse mathematics", where for example one studies the (complicated) informational content that a branch through a tree can carry, even if every node, or indeed the whole tree, is described by an easy process.

The point of countable choice is that the same issue can be carried out to the extreme. In the examples above, we have recursive information at each finite stage, and no recursive description at the end. Countable choice is needed because if we remove the requirement that the information at each finite stage is "recursive", we in fact may have no "definable" way of identifying the information at the end. So much that, in fact, we could have a whole universe of set theory where that information is not present. The way of thinking about this is that the information can be made so complicated that even having available all the definability present in a model of $\mathsf{ZF}$ set theory may not suffice to identify it.

Solution 3:

At best you show that for all $N\in\mathbb N$, there is a map $f_N\colon \{a,\ldots,N\}\to\bigcup_n A_n$ with $f_N(n)\in A_n$. That is, given $f_N$ you extend this to $f_{N+1}$ by chosing a single element $x_{n+1}\in A_{N+1}$.

What we really need, however, would be an $f_\omega$. You can't get yourself out of this trap by simply letting $f_\omega(n):=f_n(n)$ because you don't have all the $f_n$ available at once.

Solution 4:

It's true that at each $n$th stage, there is some element in $A_n$. The problem is, how do we choose which element of $A_n$ to call "$x_n$"? Countable choice lets us get around this issue, but since you're trying to prove it, we can't use it.

Your inductive argument really only shows that, for any $n$, we can find an $n$-tuple in $A_1\times\dots\times A_n.$ It does not let us "proceed to infinity", as you're attempting to do. That is, what you've proved here is the Theorem of finite choice, showing that it's fine to make arbitrary choices from finitely many non-empty sets.