This theorem comes from the book "The Real Numbers and Real Analysis" by Bloch. I am having a hard time understanding a particular part of the proof given in the book.
Prove the following theorem:
There is a unique binary operation +:ℕ×ℕ→ℕ that satisfies the following two properties for all n,m ∈ ℕ.
The following is info that can be used to prove the theorem
Definition: Let S be a set. A binary operation on S is a function S×S→S.
Axiom: There exists a set ℕ with an element 1 ∈ ℕ and a function s:ℕ→ℕ that satisfy the following three properties
1) There is no n ∈ ℕ such that s(n) = 1
2) The function s is injective (one-to-one)
3)Let G⊆ℕ be a set. Suppose 1 ∈ G, and that if g ∈ G then s(g) ∈ G. Then G = ℕ
Lemma: Let a ∈ ℕ. Suppose a ≠ 1. Then there is a unique b ∈ ℕ such that a = s(b)
Theorem (Defintion by Recursion): Let H be a set, let e ∈ H and let k: H→H be a function. Then there is a unique function f: ℕ→H such that f(1)=e, and f(s(n))=k(f(n))
The Attempt at a Solution
The following is the proof given in the book for existence of such an operation . The underlined bolded part is what I am having trouble understanding.
Suppose p ∈ ℕ. We can apply the theorem above to the set ℕ, the element s(p) ∈ ℕ and the function s: ℕ→ℕ to deduce that there is a unique function fp:ℕ→ℕ such that fp(1)=s(p) and fp(s(n)) = s(fp(n))...
Why is s(p) ∈ ℕ? I do realize the function s is defined as s:ℕ→ℕ, so the output of the function s (if it exists) must be in ℕ, but how do we know s(p) is defined from the above info or from the definition, axiom, or theorem above. The only reason I can think of is we can assume for any p that s(p) is defined, but I am not conviced because that axiom only says the function is injective, not bijective,
Any help would be appreciated,