I am having trouble with a specific direction in proving the equivalence of certain conditions for being Noetherian.
Let $A$ be a commutative ring with unity. Let $I(A)$ denote the set of ideals of $A$. The claim I want to prove is this: If every ascending chain of ideals of $A$ is stationary, then every non-empty set $S \subset I(A)$ has a maximal element.
I came up with an argument that I think makes 'intuitive' sense. However I cannot figure out how to make it formal. Here is my argument:
Suppose for every ascending chain, the condition holds. Now suppose that $S \in I(A)$. We will prove the result via contradiction. Suppose that $S$ does not contain a maximal element. This is equivalent to saying $\forall J \in S \exists I \in S (J \subset I \land J \neq I)$. Now suppose $J_1 \in S$. By universal instantiation, we have that $\exists I \in S (J_1 \subset I \land J \neq I)$. Let us use existential instantiation and denote $I = J_2$. By continuing this process, we form an ascending chain of ideals that is not stationary, contradicting the stationary property of all ascending chains of ideals. Hence $S$ must contain a maximal element.
The bold statement above is where I am really struggling to see if my argument is actually correct. My issue is this: An ascending chain of ideals may be written formally as a map $f: \mathbb{N} \rightarrow I(A)$, in other words it is a map defined over all the natural numbers. I cannot figure out how to formalise my bolded statement in such a way that I construct a new function $\bar{f}: \mathbb{N} \rightarrow S$, such that it fails to satisfy the stationary condition.
The reason I have doubts is that if I want to (really) formalise this proof, I must be able to start from the axioms, and in a finite number of steps, arrive at the conclusion. But I don't see how this is possible: At each stage in my argument, I first use universal instantiation on some specific ideal $J_k$, and then I use existential instantiation and introduce a new ideal $J_{k+1}$. But at any point in my argument, I am never able to construct a function $\bar{f}$ whose domain is all of $\mathbb{N}$; the domain is only ever a finite subset, and as such, I cannot call $\bar{f}$ an ascending chain.
Can someone please advise me on whether my concern is valid? If so, is there any rule of inference that would allow me to perform these 'infinite steps' in one go?
Some additional (not necessary) thoughts:
I realise that the most likely answer to the situation would be to use some sort of inductive argument. But the problem I face is (1) I cannot think of a good statement on $n$ to use, and (2) I understand that induction is usually used to conclude the truth of statements; whereas here I am trying to construct some function/mathematical object. I'm not sure if this means induction won't work.
I also thought about simply saying $\bar{f}$ is defined to be a function where I simply assert that for every $n$, $f(n) \subsetneq f(n+1)$. I thought my old question would be relevant here Why is AC needed to assert the existence of some infinite sets but not others?, because the answer explains how when we define sets we do not necessarily 'construct' every element, step by step. I thought that would be good news in this case, however, in order to justify the that $f(n) \subsetneq f(n+1)$ is indeed true for any $n$, I would surely have to perform one UI and then one EI to ensure that the condition makes sense. So I don't understand how to 'sidestep' the infinite application of UI/EI.
