0

I have not studied set theory rigorously, so I will describe the problems I encountered very crudely.

I read the following two articles:

$\mathbf{(a)}$: Is induction strictly required to prove finite AC?

$\mathbf{(b)}$: The logical connotation of the axiom of choice

As mentioned in the article (a), we will not define the selection function just through a limited number of "existential instantiations", because the finite set family $A$ may be of "non-standard natural number" size, which "seems" to mean using $A$ proof of a finite number of "existence instantiations" may become a proof using an infinite number of "existence instantiations". Therefore, we need to use "mathematical induction" to ensure the existence of the selection function, because in the proof involving "mathematical induction", we only need to use "existence instantiation" twice to complete the proof.

But in the proof of $(2)\Longrightarrow (1)$ in article (b), we constructed a proposition $\forall B\in P(X)’,\exists b,b\in B$, and based on "existence instantiation", we obtained $b\in B$. In the proof at this time, are infinite times of "existence instantiation" used, or only one "existence instantiation"?

If "existential instantiation" is only used once in proof $(2)\Longrightarrow (1)$, then why does the proof of the question in article (a) not mean that "existential instantiation" is used only once, so that we can directly use it in a limited Define selection functions on families of sets, skipping mathematical induction. In this way, we don't need to worry about the proof problems caused by "non-standard natural numbers", right?

If "existence instantiation" is used an infinite number of times in proof $(2)\Longrightarrow (1)$, doesn't it mean that "introducing" "existence instantiation" an infinite number of times in the proof is allowed? But it is obvious that the explanation in article (a) is to avoid this situation, because we introduce "mathematical induction" to prove that there is a selection function on a finite family of sets, which is to ensure that our proof will only use it "a limited number of times" Go to "existence instantiation".

I don't know how to think about this...

I hope you can understand my question, thank you.

Bao Yu Bo
  • 35
  • 1
    There is no such thing as a proof using an infinite number of existential instantiations. (Or at least not in a conventional proof system for first order logic in which one would typically construct proofs from ZFC.) Proofs are finite. – spaceisdarkgreen Nov 29 '24 at 20:11
  • @spaceisdarkgreen

    Do you mean that in the proof of $(2)\Longrightarrow (1)$, for $\forall B\in P(X)’ , \exists b,b\in B$, $b\in B$ is obtained from "existential instantiation", which means that "existential instantiation" is only used once, even though there are infinite sets $B$ here? So why does the proof in the question in article (a), for $\forall i\in { 1,..,n },\exists x_{i},x_{i} \in A_{i}$, obtain $x_{i} \in A_{i}$ from "existential instantiation", does not mean that "existential instantiation" is used only once, but is used "repeatedly" many times?

     – Bao Yu Bo Nov 30 '24 at 11:56
  • 1
    Yes, the proof of "for all $B$, $S_B$ is nonempty" is 1) universally instantiate $B\in P(X)'$ 2) existentially instantiate $b\in B$. 3) $(B,b)\in S_B$, so $S_B\ne \emptyset$. 4) Universally generalize $B$ to conclude for all $B\in P(X)',$ $S_B\ne \emptyset$. This uses existential instantiation once. – spaceisdarkgreen Nov 30 '24 at 17:45
  • 1
    Now, if instead you were to claim there is a function on $P(X)'$ assigning each $B$ to an element $x_B\in S_B$ and your reasoning were "for each $B\in P(X)'$ I can existentially instantiate a $b\in B$ and take $x_B = (B,b)$", then that would nonsense that could be construed as attempting to "use existential instantiation $|B|$ times", which is analogous to the nonsense proof in (a). – spaceisdarkgreen Nov 30 '24 at 17:48

0 Answers0