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I'm going to explain the difference between "existential instantiation" and "the axiom of choice". I'm not sure if I understand it correctly, so I need others to help me point out my mistakes.

First, I read this article:

Confusion about Axiom of Choice

Confused about Axiom of Choice

According to Asaf Karagila, the "axiom of choice" is a stronger statement than "existential instantiation". When the number of sets is finite, we can use "existential instantiation" and "mathematical induction" to prove "finite choices", but when the number of sets is infinite , we cannot just Prove "infinite choices" through the above method.

I want to show that $\mathbf{Proposition(1)}$: Given the family of sets $\mathscr{A}=\{ X_{i} \}_{i\in I}$ and $\forall i\in I,X_{i}\neq\emptyset$, where $I=\{ 1,2,...,n \},n\in \mathbb{N}$. Then there is a choice function on the set family $\mathscr{A}$.

We use "existential instantiation" and "mathematical induction" to prove.

$\mathbf{[Proof]}$: When $n=0$, the empty function $f=\emptyset$ is a choice function. Let's start with $n=1$, since $X_{1}\neq\emptyset$, which represents $\exists a(a\in X_{1})$, gets $a\in X_{1}$ through existential instantiation. Therefore we define the choice function:$f:\{ 1 \}\rightarrow X_{1} ,1\mapsto a$.

Now assume inductively $\forall 1\leq i\leq (n-1)$, there is a choice function in set family $\mathscr{B}=\{ X_{i} \}_{i\in \{1,2,..,(n-1) \} }$. We need to prove that there is a choice function in set family $\mathscr{A}$. For the set $X_{n}\neq\emptyset$, this represents $\exists b(b\in X_{n})$, and by "existential instantiation", we get $b\in X_{n}$. By assumption, we know that there is a choice function $f:\{ 1,2,..,(n-1) \} \rightarrow \bigcup_{i\in \{ 1,2,..,(n-1) \} } X_{i}$,satisfies $\forall i\in \{ 1,2,..,(n-1) \},f(i)\in X_{i}$. Therefore we can define a choice function on the family of sets $\mathscr{A}$:

  • $f’:I\rightarrow \bigcup_{i\in I}X_{i}$

$\forall i\in \{ 1,2,..,(n-1) \} ,f’(i):=f(i)\in X_{i}.$

If $i=n$, then $f’(i):=b\in X_{n}. \,\,\,\, \mathbf{(Q.E.D)}$

However, when given the set family $\mathscr{C}=\{ X_{i} \}_{i\in I}$ and $\forall i\in I,X_{i}\neq\emptyset$, where $I$ is an infinite set. We cannot prove that there is a choice function for set family $\mathscr{C}$ through the above means. Because this is beyond the proof capability of mathematical induction.

Although $\forall i\in I,X_{i}\neq\emptyset$ represents $\forall i\in I,\exists x(x\in X_{i})$, at this time $x\in X_{i}$ can be obtained by "existential instantiation", we cannot recursively define the choice function $f:I\rightarrow \bigcup_{i\in I}X_{i},\forall i\in I,f(i):=x\in X_{i}$, because $I$ is an infinite set. Therefore, we need the “axiom of choice” to ensure that there is a choice function on the set family $\mathscr{C}$.

  • $\mathbf{Summary}$: When the number of sets is limited, we use "existential instantiation" and "mathematical induction" to help us "select" elements from each non-empty set. But when the number of sets is infinite , "mathematical induction" will no longer apply. Without introducing the axiom of choice, we cannot "select" elements from every non-empty set. More strictly speaking, , we cannot define a selection function.

Is my understanding correct? If it is incorrect, please specify what is incorrect, thank you.

Bao Yu Bo
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    You can't even do countable choice. Induction shows that you can make $n$ choices for any finite $n$. It does not prove that you can make a simultaneous choice for all of $\omega$. This is analogous to showing that there are natural numbers whose decimal representation has length $n$ for each and every $n$; this does not prove that there is a natural number whose decimal representation has length $\omega$. – Arturo Magidin Nov 18 '24 at 19:13
  • So how can I prove that selection functions also exist on finite set families? The proof in my question is based on the proof of "finite choices" in Chapter 3 of the book Real Analysis written by Professor Terence Tao. However, the book does not mention how to construct a choice function, but only uses "existential instantiation". ” This completes the proof. – Bao Yu Bo Nov 18 '24 at 19:20
  • You can use induction to prove choice functions exist on finite families. But you seem to be claiming that induction will prove it for countably infinite families, and that is not true. Your argument shows that for all $n\in\omega$, if ${X_i}{i\in n}$ is a family with $n$ nonempty sets, then there is a choice function $f\colon n\to\cup X_i$ with $f(j)\in X_j$ for each $j\in n$. It does not prove that if ${X_n}{n\in\omega}$ is a countably infinite family of nonempty sets, then there exists a function $f\colon\omega\cup X_n$ such that $f(n)\in X_n$ for every $n\in\omega$. – Arturo Magidin Nov 18 '24 at 19:23
  • My proof is to use mathematical induction to prove that there is a selection function on the "finite set family". I do not claim that mathematical induction can prove that selection functions also exist on "infinite sets". – Bao Yu Bo Nov 18 '24 at 19:25
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    You keep bringing up "uncountable" in your text, as if the distinction is between countable and uncountable. That is not the correct distinction. The correct distinction is between finite and infinite. Countability is irrelevant, so why do you keep mentioning "uncountable"? "Uncountable" is not a synonym for "infinite". Uncountable means "cannot be put in bijective correspondence with either $\omega$ or with any $n\in \omega$." In this context "Countable choice" refers to finite and countably infinite choice. – Arturo Magidin Nov 18 '24 at 19:26
  • The set of real numbers is famously uncountable. If you don't know what the word means, then don't use it. – Arturo Magidin Nov 18 '24 at 19:32
  • I do know that the set of real numbers is uncountable. Uncountable sets mean that bijective functions cannot be established with the set of natural numbers. I also know that the so-called "countable set" has two meanings. The first one means "the number of elements of the set". The number is finite", the second means "it can establish a bijective function with the set of natural numbers". There was a typo in my comment, so I deleted it. What I just want to confirm is whether my $\mathbf{Proposition (1)}$ certificate is correct? – Bao Yu Bo Nov 18 '24 at 19:38
  • If you know, why do you say "the 'countable set' pointed out in my article is, for example, the set of real numbers, or the power set of real numbers" in the comment you deleted (to hide the error?) ? The argument in proposition 1 is essentially correct, though it would be better if you described the function as a function (set of ordered pairs); $f' = f\cup{(n,b)}$. Every time you say "uncountable", though, you are conflating "uncountable" with "infinite", which, for the $n$th time, is wrong and which you apparently just don't care about. – Arturo Magidin Nov 18 '24 at 19:42
  • Sorry, there was a typo in my comment, I know the set of real numbers is uncountable, I made a typo. I want to explain why I use the word "uncountable", because mathematical induction is true for the entire set of natural numbers, so when the index set is the set of natural numbers (countable), I think we also don't need to choose the axiom. The selection function can be found. – Bao Yu Bo Nov 18 '24 at 19:46
  • And what I've been telling you since the first comment is that that is false. Induction proves it for each $n\in\omega$, it does not prove it for $\omega$ itself. You cannot use induction to prove that there is a choice function for a countably infinite family of nonempty sets. – Arturo Magidin Nov 18 '24 at 19:52
  • Consider the statement "the set is finite". By induction, we can prove that $\varnothing$ is finite, and that if $n$ is finite, then $s(n)=n\cup{n}$ is finite. Does that mean that $\mathbb{N}$ itself is finite? No. Because induction lets you state that "is finite" holds for each $n\in\mathbb{N}$, but it does not let you conclude that it holds for $\mathbb{N}$ itself. You cannot use induction to prove countable choice. – Arturo Magidin Nov 18 '24 at 19:54
  • Thank you for pointing out the error in the article, which I have corrected. In addition, I would like to ask, are there any other errors in my article? – Bao Yu Bo Nov 18 '24 at 20:01
  • "Is this wall of text correct" is off-topic for the site. – Arturo Magidin Nov 18 '24 at 20:02
  • Okay, if my article no longer contains any errors, then we can end this discussion, thank you. – Bao Yu Bo Nov 18 '24 at 20:11
  • Possibly related: https://math.stackexchange.com/a/3354361/279914 – Karl Nov 18 '24 at 20:20

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I think your terminology is off here. Existential instantiation is a syntactic rule in a formal proof which states that an arbitrary constant can temporarily stand in place of an existentially quantified variable, so long as it is discharged from the result.

The axiom of choice states that there is a function for all nonempty sets that takes an element from each of those sets and maps it to a new set, and that that resulting set is a member of the union of all of those nonempty sets.

You use existential instantiation in formal proofs involving the axiom of choice, because of its first-order formulation (see here (A7)).

Existential instantiation works whether the domain of discourse or a set in it is infinite, finite, or empty. The axiom of choice only applies to nonempty sets.

Joshua Harwood
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  • Regarding the logical inference rule "existence instantiation", the book SYMBOLIC LOGIC written by Professor I.M. Copi explains it this way. $\mathbf{(EI)}$,$(\exists x)(\phi x)/ \therefore (\phi y)$, where y is an individual constant, other than “y”, that has no prior occurrence in the context. The book Basic Logic written by Professor Peng Mengyao of National Taiwan University gives a different explanation. $\mathbf{(EI)}$, $(\exists x)(\phi x)/ \therefore (\phi y)$, among them, y cannot be an individual constant, and y can only be an individual variable. – Bao Yu Bo Nov 19 '24 at 10:15
  • Professor Peng Mengyao believes that treating y as an individual constant that has never appeared before will not cause any technical problems. But philosophically speaking, this confuses the difference between "individual variables" and "individual constants". Except that the "Existential instantiation" rule is somewhat controversial. I think your explanation of the difference between "Axiom of Choice" and “Existential instantiation" is correct. – Bao Yu Bo Nov 19 '24 at 10:21
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    From a proof-theoretic standpoint, you'll have a bit of a hurdle creating some system for dealing with free variables if instantiation from variables to other variables is allowed, but that won't cause technical problems if they're handled properly. – Joshua Harwood Nov 20 '24 at 01:34