I was recently encouraged to develop a deeper understanding of the Principle of Infinite Descent so that I could become more comfortable with its usage in proofs. I have a few questions that I wanted to run by the community in order to confirm that I have sufficiently absorbed the concept.
The below excerpt, taken from http://elib.mi.sanu.ac.rs/files/journals/tm/31/tm1622.pdf, provides this statement:
Given what we know about the Well-Ordering Principle for Natural Numbers, this statement is logically sound.
Equipped with the above logical statement, the aforementioned math website goes on to prove the following proposition:
There is no infinite strictly decreasing sequence of natural numbers
This proposition is proven using the following argument:
Firstly, I wanted to confirm that the reason "we know" $1 \notin A$ (where $1$ here is functioning as the $0$ that I am more familiar with) is that by Peano's Axioms and the Well-Ordering Principle, there is no element smaller than $1$.
Therefore, if $1$ WAS a member of set $A$, there could be no element less than $1$ to continue the infinite sequence...which would necessarily make $A$ finite (contradiction). So, to avoid contradiction, the author moves on with the claim $1 \notin A$.
Secondly, why is the initial assumption that "$\exists$ an infinite set $A$" an acceptable assumption? Is it purely because we know that there are an infinite number of natural numbers and therefore, by defining members of $A$ as "$a_n \in \mathbb N$ and $n \in \mathbb N$" we recognize that the natural number indexing of the elements permits an infinite list?
Finally, I recognize the contradiction. Specifically, we arrive at the simultaneous conclusion that the statement $p(n)$ is true for $\forall n \in \mathbb N$ (i.e. no elements of $n$ are in $A$) while claiming the existence of an infinite set that contains members of $\mathbb N$ (and therefore provide infinitely many instances of $n$'s where $p(n)$ is false). My question is thus, "What is the negation that is performed on the initial assumption"? Is it simply that The set $A$ does not exist? Consequently, given the properties attached to $A$, the conclusion is:
No infinitely sized set can be constructed from strictly decreasing elements selected from the $\mathbb N$.
Is this all correct? Cheers~
Regarding your first question, yes that is precisely the reason we know $1 ∉ A$.
Regarding your second question, the existence of an infinite set is typically an axiom in whatever system you are working in. Since I see that the $∈$ relation is being used in the text that you cite, I am assuming that the system is something at least adjacent to ZFC, for which you can find a list of its axioms (including the axiom of infinity). What you state is more or less correct. $ℕ$ (but actually $ω$) is constructed with the axiom of infinity as the intersection of all inductive sets, and from there we can define functions with the whole of (or only a part of) $ℕ$ (but really $ω$) as the domain. So when we write $\{a_{1},a_{2},\dots \}$ we are in some sense saying that there is a function with $\{1,2,\dots \}$ as the domain and $\{a_{1},a_{2},\dots \}$ as the range.
Your last comment is correct in that the negation of the initial assumption is that the set $A$ does not exist. Your observation of the consequence is also correct in the system I assume we are working in.