Bloch Analysis proof of Theorem 2.5.5 (Definition by recursion)

[BugKingpin]

TL;DR Summary

Not sure why ##N## X H ##\epsilon## C in the existence proof of theorem 2.5.5 in Bloch.

Want to understand how set C contains ##N## x H. H is only defined to be a set with element e and as the domain/range of function k. Is this enough information to conclude that the second set in the cartesian product W is H and not a subset of H?

My thinking is to show that ##N## and H satisfy the definition of sets A and B in an element W = A x B ##\epsilon## C. Then ##N## x H ##\epsilon##. Since 1 ##\epsilon## A so A ##\subseteq N##. If n ##\epsilon## A then s(n) ##\epsilon## A then by Peano Axioms/induction A = ##N##. Then we have to show that B = H. This is the part I am confused by.

Let e ##\epsilon## B. By the set builder rule of C, if (n,y) ##\epsilon## W then (s(n),k(y)) ##\epsilon## W which means if y ##\epsilon## B then k(y) ##\epsilon## B. This means that k(1) ##\epsilon## B and k^{n}(1) ##\epsilon## B. But according to this definition, B is infinite as there are infinite n ##\epsilon## N.

To show that B = H, I either need to use some theorem or show that H and B are subsets of each other. But if I don't know what's in H how do I show that every element of B is also in H and vice versa?

 

[Office_Shredder]

I don't fully understand your post but I think you've missed the point because it's trivial. ##\mathbb{N}\times H## just simply satisfy that if ##(x,y) \in \mathbb{N}\times H## then ##(s(x),k(y))\in \mathbb{N}\times H##. But s and k by definition return elements of ##\mathbb{N}## and ##H## so of course it's true.

Also we need ##(1,e)\in \mathbb{N}\times H## which is similarly trivial, since we already know they belong to these sets. Hence ##\mathbb{N}\times H## is an element of ##C##

 

[BugKingpin]

I don't fully understand your post but I think you've missed the point because it's trivial. ##\mathbb{N}\times H## just simply satisfy that if ##(x,y) \in \mathbb{N}\times H## then ##(s(x),k(y))\in \mathbb{N}\times H##. But s and k by definition return elements of ##\mathbb{N}## and ##H## so of course it's true.

Also we need ##(1,e)\in \mathbb{N}\times H## which is similarly trivial, since we already know they belong to these sets. Hence ##\mathbb{N}\times H## is an element of ##C##

I really overthought this. So that condition is really just the definition of a function H->H; H contains every element in the domain and range of k and there must be a k(y) in H for every element y in H. So that definition starting with the element e completely populates H as I mentioned above?

Great signature btw. Combinatorics is indeed very subtle.