The logical connotation of the axiom of choice

Question

My native language is Chinese, so I use a translation machine.

I don't know how to rigorously prove the equivalence of the following two propositions:

(1)$\textbf{Axiom of choice}$:$Let\,P(X)’=P(X)/ \{ \emptyset \}$.$\forall X,\exists f:P(X)’\rightarrow X,\forall S\in P(X)’,f(S)\in S.$

(2)$\textbf{Proposition}$:$\{ S_{i} \}_{i\in I}$ is a family of sets, where $\forall i\in I,S_{i}\neq\emptyset$ and $\forall i,j\in I,i\neq j\iff S_{i}\cap S_{j}=\emptyset$,then $\exists C,C\cap S_{i}\neq\emptyset$ and $card(C\cap S_{i})=1$.

I have proved $(1)\Longrightarrow (2)$. However, I don’t know if the proof of $(2)\Longrightarrow (1)$ is suspected of being a circular argument. Its proof is as follows: Given $X\neq\emptyset$,$\forall B\in P(X)’$, let $S_{B}=\{ B \} \times B=\{(B,b)|B\in\{ B \},b\in B \}$, we get the set family $\{ S_{B} \}_{B\in P(X)’}$,then $\forall B\in P(X)’,S_{B}\neq\emptyset$ and $\forall i,j\in P(X)’,i\neq j\iff S_{i}\cap S_{j}=\emptyset….$

“$\forall B\in P(X)’,\exists b ,b\in B$” is used in the proof. Does this imply the use of the axiom of choice?

Dr. Naïm Favier told me that $b\in B$ here does not actually make any selection, it is derived based on $B\neq\emptyset$.

Dr. Naïm Favier said: “ You can get some $b\in B$ from $B\neq\emptyset$ using excluded middle, but that $b$ is not unique, so there's no choice being made here. The fact that something exists does not give you a specified witness of that thing.” In logical terms, this means that element $b$ is an individual variable.

And I read the following: Do We Need the Axiom of Choice for Finite Sets?

Professor Carl Mummert uses "existential instantiation" to explain this problem. When we have a non-empty set, through existential instantiation, I can get $b\in B$, where $b$ is an individual variable, so no selection is actually made.

And I have a question, for “$…\forall S\in P(X)’,f(S)\in S$” in the axiom of choice, is $f(S)$ an individual variable or an individual constant?

(1)If $f(S)$ is an individual variable, does $\forall B\in P(X)’,\exists b,b\in B$ already assume the axiom of choice, leading to a circular argument?

(2) If $f(S)$ is an individual constant, then according to Professor Carl Mummert, given a finite number of non-empty sets $S$, $s\in S$ obtained through existential instantiation does not represent a real choice. This is the view of classical logic on the symbol "$\exists$” non-constructive results. In other words, given a finite number of non-empty sets, we don't have to make an actual choice. But why do we use the axiom of choice "$…f(S)\in S$" to actually select a certain value $f(S)$ from $S$ when given infinitely many non-empty sets?

In addition, I read Professor Terence Tao analysis textbook, and there is a theorem in Chapter 3, that is, given a finite number of non-empty sets, we may select an element from the non-empty set. However, judging from the context, we can know that the element selected in the proof of the theorem is actually an individual variable, rather than a certain value (individual constant).

I hope you can understand my question, thank you.

• Clarification (a): My question is not, “in the entire proof of $(2)\Longrightarrow (1)$, is $b$ a variable or a constant?” I know that $b$ in the set $S_{B}$ is a variable at the beginning, and then we can use $\textbf{Proposition}(2)$ to get that $b$ is a specific value.

• Clarification (b): My question is: Taking “ Clarification (a)” position, why do we need the axiom of choice to ensure that one element can be specifically selected from each non-empty set when there are infinitely many non-empty sets? Because according to Professor Carl Mummert, when the number of non-empty sets is finite, we do not need to use the axiom of choice to accurately select an element from the non-empty set, but the non-constructive nature of the existential quantifier guarantees this. I mean, when the number of sets is limited, why do we not need to specifically select a value from each non-empty set?

Answer 1

You have mentioned an interpretation of "existential instantiation" as saying "no selection is actually made". And you have also mentioned it as saying "we don't have to make a choice". But I disagree with those interpretations.

I think of existential instantiation as simply saying: we can select (or choose, or take) one element $b$ out of one nonempty set $B$. Here we are forced to use a symbol for that element we took, and the symbol we are using in this discussion is $b$. While I am unsure what you mean by an "individual variable", whatever it might mean I think that it is not an appropriate description of $b$ after existential instantiation. The effect of existential instantiation is that the existential quantifer binds the symbol $b$, making it represent a well-defined element of the set $B$. So the symbol $b$ becomes a constant, representing an actual element of $B$.

And then, to address your paragraph referring to Tao's book regarding choice for a finite collection of nonempty sets, it's already proved for one set, and that's the basis step for the induction argument. To complete the argument I'll state the induction step: suppose that we can select (or choose, or take) $k$ elements $b_1,\ldots,b_k$ out of $k$ nonempty sets $B_1,\ldots,B_k$, respectively. Consider another nonempty set $B_{k+1}$. Using existential instantiation, we can select $b_{k+1} \in B_{k+1}$. Putting our choices together, namely the $k$ elements $b_1,\ldots,b_k$ and the element $b_{k+1}$, we have chosen $k+1$ elements $b_1,\ldots,b_{k+1}$ out of the sets $B_1,\ldots,B_{k+1}$, respectively.

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Finally, except for a minor notation problem your proof of (2)$\implies$(1) starts out just fine. The notation problem is simple: instead of writing $\{B\} \times B$ you should write $\{\{B\} \times B \mid B \in P(X)'\} = \{(B,b) \mid B \in P(X)', b \in B\}$; and presumably also you intended to write $S_B = \{B\} \times B$.

So far there is no "circularity", because you haven't used (1).

Your next step does not use (1) either, although it does use existential instantiation. You have to prove the statement If $B \in P(X)'$ then $S_B = \{B\} \times B$ is nonempty. You start the proof by letting $B$ be any element of $P(X)'$. It follows that $B$ is not empty. So we may apply existential instantiation to the set $B$: we may choose $b \in B$. Therefore we have found an element of the set $\{B\} \times B$, namely $(B,b)$. This proves that the set $S_B = \{B\} \times B$ is nonempty.

Notice: All we have done is to combine nonemptyness of one set, namely $B$, together with existential instantiation, to prove nonemptiness of another set, namely $S_B$. We have not used the axiom of choice here, in any version --- except the version which goes under the title of existential instantion and which allows us to choose one element out of any one nonempty set.

After that your proof dwindles to an uncertain finish, but I suspect that you can now continue onward to complete the proof.