The Axiom of Choice and a definition of addition in a quotient space of a vector space

Question

I am thinking about the Axiom of Choice and I am trying to understand the Axiom with some but a little progress. Some time ago I could not understand why the obvious "proof" of the Axiom of Choice is not correct. Namely, let $A_i$, $i\in I$, be a collection of nonempty sets; in what follows we assume that the index set $I$ is infinite. We construct a choice function for this collection as follows. Consider arbitrary $i\in I$. Since the set $A_i$ is nonempty, $\exists x(x\in A_i)$. By Existential Instantiation we get $a\in A_i$. Set $f(i):=a$. It is clear that $f$ is a needed choice function. Certainly the argument above is not correct, otherwise the argument would prove the Axiom of Choice so it would not be an axiom. But why the argument is not correct? Now my imperfect understanding is the following. We can indeed consider arbitrary $i\in I$. We can indeed apply Existential Instantiation to get $a\in A_i$. We can indeed write $f(i):=a$. But this defines $f$ only at one point $i$. At arbitrary point, but only at one. So currently $f$ is not a choice function for the whole given collection of sets. We can choose another index, say $j\neq i$, apply Existential Instantiation to get $b\in A_j$, and set $f(j):=b$. But now $f$ is defined only at two points $i$ and $j$. At arbitrary two points, but only at two. So currently $f$ is not a choice function for the whole given collection of sets. Continuing this line of reasoning, we can define $f$ at, say, 100 arbitrary points of $I$. At arbitrary 100 points, but only at 100. So currently $f$ is not a choice function for the whole given collection of sets. Finally, we cannot continue this process indefinitely. Why? Because in mathematics a proof must be finite! Indeed, a proof is a sequence of (number of statement, statement, justification), and this sequence must be finite; the last statement is exactly what we wanted to prove. In particular, in a proof we can use only a finite number of Existential Instantiations (as justifications). So, we must stop at some step. At this step $f$ is defined only at a finite number of points of $I$. So $f$ is not a choice function for the whole given collection of sets! Thus by the obvious argument above we do not get a choice function.

Question 1: is my understanding of the failure of the obvious "proof" of the Axiom of Choice right and complete, or perhaps I am wrong, or perhaps I am missing something? I am also very very interested in opinions of other people on the failure of the obvious "proof" of the Axiom of Choice. Please do not hesitate to write me your opinion, here or on my e-mail ivanmath007@gmail.com.

Let my move to my second question. It is about a definition of addition in a quotient space of a vector space. So let $V$ be a vector space and $U$ be a subspace of $V$. Define the equivalence relation $\sim$ on $V$ as follows: $x\sim y$ if $x-y\in U$. It is easy to check that $\sim$ is indeed an equivalence relation on $V$. Denote by $V/U$ the set of equivalence classes. It is indeed a set, it follows from ZF axioms. We want to define a vector space structure on the set $V/U$. In particular, we want to define addition on $V/U$, i.e., we want to define a function $+:V/U\times V/U\to V/U$ which satisfies certain known axioms. I will write $\varphi$ instead of $+$, and so we want to define a function $\varphi:V/U\times V/U\to V/U$. The standard way to do this is the following. Consider arbitrary $\alpha,\beta\in V/U$. Choose $x\in\alpha$ (by Existential Instantiation, because we have no canonical representative of $\alpha$) and $y\in\beta$ (also by Existential Instantiation). Now define $\varphi(\alpha,\beta)$ to be the equivalence class of $x+y$. Of course, it is easy to check that this equivalence class will be the same for any other representatives of $\alpha$ and $\beta$, but we need to choose $x$ and $y$ to define it. Thus we defined a function $\varphi$. But now analyze this definition of the function $\varphi$. It is essentially the same as the construction of a choice function from the obvious "proof" of the Axiom of Choice! Thus this definition of $\varphi$ is not correct, it does not define a function $\varphi:V/U\times V/U\to V/U$! This makes me crazy. I am used to the above definition of addition in a quotient space and I "believed" in it for the last 15 years. So it is hard for me, but I will ask:

Question 2: am I right that the above definition of the function $\varphi$ is not correct and it does not define a function $\varphi:V/U\times V/U\to V/U$?

Finally I should note that we can save the day --- define $\varphi(\alpha,\beta):=\{x+y\,|\,x\in\alpha,y\in\beta\}$ for $\alpha,\beta\in V/U$; of course one must check that for arbitrary $\alpha,\beta\in V/U$ the set $\{x+y\,|\,x\in\alpha,y\in\beta\}$ is an equivalence class. In spite of this, Question 2 remains very very important for me.

Despite the fact that this question turned out to be very long I hope it will be useful. Once more, please do not hesitate to write me.

Answer 1

Actually, the reason the definition you give isn't quite proper has nothing to do with the axiom of choice. Rather, it's because a definition is not a process: if we want to be extremely precise, language like "Pick an $x$" has no place in a definition regardless of what we assume about choice, the issue being the word "Pick."

Here is a grammatically-valid definition:

$\alpha\boxplus_\exists\beta=\gamma$ iff there are $x\in\alpha,y\in\beta$ such that $x+y\in\gamma$.

(I'm using the new "$\boxplus$" symbol for clarity.) And here's another one:

$\alpha\boxplus_\forall\beta=\gamma$ iff for all $x\in\alpha,y\in\beta$ we have $x+y\in\gamma$.

Note that their equivalence is not logically trivial: in particular, without knowing anything about vector spaces I might worry that $\boxplus_\exists$ is actually multi-valued and $\boxplus_{\forall}$ is actually non-total. But this still has nothing to do with choice! If $\boxplus_\exists$ were multivalued then we might want to "refine" it to a single-valued function, and at this point choice and the question of its necessity could conceivably enter the picture. However, we can prove - without using choice at all - that this is not the case, so the issue doesn't arise in the first place.

(Note that if we were really concerned about these issues we'd actually go one step further and avoid using function notation at all - we'd first consider e.g. the formula $\sigma_\exists(\alpha,\beta,\gamma)\equiv\exists x\in\alpha,y\in\beta(x+y\in\gamma)$, and then show that this formula defines a function in the sense that $\forall \alpha,\beta\exists!\gamma(\sigma_\exists(\alpha,\beta,\gamma)$ - but this doesn't really add to the discussion above.)

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What about your original proposed "proof" for $\mathsf{AC}$? Well, again, natural language issues - in particular the intrusion of "imperative" language - are going to obscure the underlying formalism. I strongly recommend learning some particular formal proof system (e.g. sequent calculus) so that you can write and compare correct arguments with superficially-similar incorrect arguments.

(Meanwhile, one of the best sells for the axiom of choice in my opinion is that it makes imperative reasoning safer.)