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I've been learning Set Theory from both Danel Cunningham and Herbert Enderton's books and I can't wrap my head around how they prove the Recursion theorem on $\omega$:

Let $A$ be a set and $a \in$ A. Suppose that $f : A → A$ is a function. Then there exists a unique function $h : ω → A$ such that

(1) $h(0) = a$,

(2) $h(n + ) = f (h(n))$, for all n $\in$ ω.

Both of them starts off the proof by constructing a set that is contains all the acceptable functions i.e. satisfy the above $2$ conditions.

My question is why is it even necessary to have a set of functions in the first place? Why don't just assume a relationship $R \subset \omega \times A $ that satisfies the $2$ condition (using Subset Axiom) and work from there?

I do understand the basic ZF axioms (Extensionality, Empty Set, Pairing, Union, Power Set, Subset) that I deem handy when solving some other problems but not this set of functions thingy. Is this supposed to be a proving trick or some logical connection that I missed out on? Or am I not understanding the basic ZF axioms fully?

From that point onwards, I could understand the proofs fully, but it leaves me with much insecurity and anxiety that I don't have good enough intuition or the smarts to flow along with a textbook because I do believe there must be some kind of intuition that this approach is built on top of.

UNAN
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Wei Minn
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1 Answers1

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Why don't just assume a relationship $R\subseteq\omega\times A$ that satisfies the $2$ condition (using Subset Axiom).

Note that the difficulty of the proof is to find the existence of the relation. You can only assume that $R$ exists (and is a function with the $2$ properties) once you are able to show this...

Maybe you want to use something else to show the existence: What do you mean be the "Subset axiom"?

Maybe you mean the subset schema of specification/comprehension?

But for that you need to describe a property $F$ such that $(n,a)\in R\iff F(n,a)$. Note that the description of the property $F$ cannot refer to $R$, because you want to use this property in the comprehension schema to define $R$ in the first place...

Martin Väth
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  • I think I get why Relation $R$ because it's not a definable subclass of the Cartesian Product of $\omega \times A$ using the 2 properties and the Axiom Schema of Specification. Because it would be $R = { (n,x) \in \omega \times A : (n,x)=(0,a) \ \vee \ [ n = k^+ \wedge \ \exists(k, y) \in R (f(y)=x) ] } $ where $(k, y) \in R$ is not valid cos $R$ is not defined yet! So, the only possible way is to find all combinations of subsets of $\omega \times A$ which is the set of all $R \subset \omega \times A$ and work from there. – Wei Minn Jul 09 '21 at 03:29
  • @WeiMinn - Correct; as per answer above, you cannot use $R$ in the definition of $R$ itself... But, proceeding step-by-step, you can use the "portion" of $R$ already defined to define the "next step" (and thus you can see why Recursion is simply Induction in "set-clothes"). – Mauro ALLEGRANZA Jul 09 '21 at 12:43
  • @MauroALLEGRANZA - But doesn't having the Set $S$ of acceptable functions mean that its members should be properly defined in the first place? Because for each $R \in S$, $R$ needs to defined with the 2 properties as well! Doesn't it make the defining of the set itself invalid? – Wei Minn Jul 09 '21 at 14:21
  • @WeiMinn - see G. Tourlakis, Logic and Set Theory. Vol II: Recursive Definitions over ω, page 244. – Mauro ALLEGRANZA Jul 09 '21 at 14:27
  • @WeiMinn, This is actually avoided in Halmos's book naive set theory, where the candidate relation $R$ is taken to be the intersection of all those subsets $S$ of $\omega\times A$ for which $(0,a)\in S$ and for which $(n^+,f(x))\in S$ whenever $(n,x)\in S$. Also such subsets exists since $\omega\times A$ is one (so that the intersection is not meaningless). Also on this site, see https://math.stackexchange.com/q/1421626/. But of course, I don't think I understand fully of the nuances. – davyjones May 01 '22 at 10:54