I know that this question has been asked a lot here, some of them are duplicate of each other. I’ve read every single of them, but my problem has not been resolved. I can just memorize that Axiom of Choice (AC) is needed, but I want to be clear of this, logically.
Note : “Countable” here means infinitely countable.
Recall the proof, it goes like this:
First, I have countably many of countable sets. Let me enumerate them to $\langle A_i\rangle_{i\in\mathbb{N}}$. For each $i$, $A_i$ is countable. Since $A_i$ is countable for each $i$, there exists a bijection $h_i:\mathbb{N}\rightarrow A_i$ for each $i$. Then this is where Axiom of Choice (AC) comes into play. There are countably many $A_i$, so I have to choose ‘countably many’ times. Then I’ve got $h_i:\mathbb{N}\rightarrow A_i$ for each $i$ I want.
But why is AC needed?
What I understand about AC is, if I have a collection of non-empty sets, and I want to construct a new set by picking an element from each set in the collection, then AC allows me to pick them to ‘construct’ my new set. This is explicitly stated in its logical formula. For example, if I have a collection $\mathcal{A}=\lbrace A_i\rbrace_{i\in\mathbb{N}}$ where $A_i$ is non-empty for each $i$ and I want to construct a new set $\lbrace a_i\rbrace_{i\in\mathbb{N}}$ such that $a_i\in A_i$ for each $i$, then I need AC to guarantee that such set can be made. However, if my collection $\mathcal{A}=\lbrace A_i\rbrace_{i=0}^n$ is finite, then I can construct the new set without AC, i.e., it can be proved that such $\lbrace a_i\rbrace$ can be constructed without the need of AC.
But what I don’t understand is, why am I not even permitted to just pick an element and manipulate them? Return to the same example, if I have a collection $\mathcal{A}=\lbrace A_i\rbrace_{i\in\mathbb{N}}$ where $A_i$ is non-empty for each $i$, then I know that there exists $a_i\in A_i$ for each $i$. Am I permitted to manipulate $a_i$ for each $i$?, like constructing $\lbrace a_i\rbrace$ by the Axiom of Pair? Can I construct a set $A_i’:=A_i-\lbrace a_i\rbrace$ for each $i$? Each $a_i$ exists logically, and I want to put bracket around them like this $\lbrace a_i\rbrace$. This is allowed by the Axiom of Pair. Of course, I cannot make $\lbrace a_i\rbrace_{i\in\mathbb{N}}$ since this is not implied by any axiom (even by the Axiom of Union or Axiom of Infinity) except AC.
Most answer I found here is, I cannot even 'pick' an element from each set infinitely many times without AC (not to mention constructing a new set from them). Some people even said that each $A_i$ in the proof being countable does not mean that its enumeration is given. Well, in that case, can't I just say that since it does not have enumeration, then it's not countable and hence a contradiction to the assumption? An enumeration must exists. Some might say that it exists but is not given. Then what does 'given' mean in logic? Even in the proof that utilizes AC, we claim that a choice function exists, but not clearly stated 'which' choice function. We only know that it exists and use it to finish the proof. Why is this situation is different from knowing that each bijection $h_i$ exists for each $i$. Why are we permitted to manipulate existing choice function, but not allowed to manipulate existing bijection? Is this related to the fact that it is formulated in First-order logic?
Thank you!
Best Answer
A weak form of AC is used, for while each countable set $A_i$ has a bijection $f_i\colon A_i\rightarrow \mathbb{N}$ this bijection is not unique and AC is required to chose the sequence $\langle f_i:i\in\mathbb{N}\rangle$.