# Generalising the recursion theorem to well-founded posets

The recursion theorem states the following.

Suppose $$(A,\preccurlyeq)$$ is a chain with order type $$\omega$$. Let $$X$$ be any set, let $$x\in X$$ and let $$g\colon X\times A \to X$$ be a function. Then there exists a unique function $$f\colon A\to X$$ such that $$f(a) = \begin{cases} \hfil x\hfil & \text{if a=\min A}\\ g(f(t),a) & \text{if a=\operatorname{succ}(t)}. \end{cases}$$

(E.g., for the factorial function, we put $$A=\mathbb N=X$$, $$x=1$$ and $$g(m,n)=mn$$.)

Now if we instead have a poset $$(A,\preccurlyeq)$$ which is well-founded, we can partition it so that the classes of the partition form a chain which has order type $$\omega$$. Indeed, define the idea of a "minimal set" as follows.

Let $$(A,\preccurlyeq)$$ be a poset. The minimal set of a subset $$B\subseteq A$$, denoted $$M_\preccurlyeq(B)$$ or just $$M(B)$$, is the set defined by $$M(B):= \{b\in B : \text{there is no m\in B such that m\preccurlyeq b}\}.$$

Then the desired partition of $$A$$ is \begin{align*} A_1 &= M(A)\\ A_2 &= M(A\smallsetminus A_1) \\ A_3 &= M(A\smallsetminus (A_1\cup A_2)) \\ &\kern6pt\vdots\\ A_{n+1} &= M(A\smallsetminus(A_1\cup \cdots\cup A_n))\\ &\kern6pt\vdots \end{align*} My question is: can we state a version of the recursion theorem for this set $$A$$, according to this partial order?

For instance, an example which comes to mind is the binomial coefficients, $$c(n,k) = \begin{cases} \hfil 1 \hfil & \text{if (n,k)\in A_1}\\ \hfil c(n-1,k-1) + c(n-1,k) & \text{otherwise}. \end{cases}$$ Here we have $$A=\{(n,m)\in\mathbb N^2 : m\leqslant n\}$$, and we are ordering it in the following way:

• for any $$n\in\mathbb N$$, $$(n,0) \preccurlyeq a$$ for all $$a\in A$$
• for any $$n\in\mathbb N$$, $$(n,n) \preccurlyeq a$$ for all $$a\in A$$
• for any $$n,k,\ell\in\mathbb N$$, $$(n,k) \preccurlyeq (n+1,\ell)$$.

This captures the natural order implicit in Pascal's triangle:

Note that $$A_1$$ contains precisely base cases, and then the rest of the sets are such that we can define $$c(n,k)$$ at the $$A_k$$th level using its values at the previous levels.

Another example is remainder upon division:

$$r(n,d) = \begin{cases} \hfil n \hfil & \text{if 0\leqslant n\leqslant d}\\ r(n-d,d) & \text{if n > d}\\ r(n+d,d) & \text{if n < 0} \end{cases}$$

Here the set is $$A=\mathbb Z\times \mathbb N$$, and the order is

• $$(n,d)\preccurlyeq a$$ if $$0\leqslant n\leqslant d$$ for any $$d$$
• $$(n,d)\preccurlyeq (x,y)$$ if $$|ny|\leqslant|xd|$$.

If it is to be formulated similarly, I was thinking the theorem would look something like this:

Suppose $$(A,\preccurlyeq)$$ is a poset where $$\preccurlyeq$$ is well-founded, and $$\{A_1, A_2,\dots\}$$ is the partition of "minimal sets" induced by $$\preccurlyeq$$. Let $$X$$ be any set, let $$x_a\in X$$ for each $$a\in A_1$$, and let $$g\colon A\times X^? \to X$$ be a function. Then there exists a unique function $$f\colon A\to X$$ such that $$f(a) = \begin{cases} \hfil x_a\hfil & \text{if a\in A_1}\\ g(a, f(a_1),f(a_2),\dots) & \text{if a\in A_k and A_1\cup\cdots\cup A_{k-1} = \{a_1,a_2,\dots\},} \end{cases}$$

i.e., the base cases are all for $$a\in A_1$$, and the inductive case is allowed to use the value of $$f$$ at any of the $$a$$ in the previous sets $$A_1,\dots,A_{k-1}$$. But this formulation doesn't feel as succinct as it could be, and also the arity of $$g$$ varies depending on which $$A_k$$ our $$a$$ lies in. It also doesn't feel like it captures cleanly the examples I had in mind; I can't clearly tell you what $$g$$ to take.

I appreciate any help with formulating this theorem. I want to have a general theorem which justifies "definition by recursion" for any well-founded poset.

• In general it is not necessarily the case that every element of $A$ is in some $A_k$; you may have to continue the process transfinitely. (For instance, if $A$ is well-ordered, each $A_k$ is a singleton and $A$ might be longer than $\omega$.) – Eric Wofsey Nov 25 '20 at 0:27
• @EricWofsey You are right about this, but we can assume $A$ is countable for the sake of the question. – Luke Collins Nov 25 '20 at 0:34

There is no need to get a total order associated to your partial order to make sense of recursion. Indeed, you don't even need to have a partial order! Here is a simple way to formulate recursion with respect to any well-founded relation.

Theorem: Let $$X$$ be a set and let $$<$$ be a well-founded relation on $$X$$ (i.e., for any nonempty subset $$A\subseteq X$$, there exists $$a\in A$$ such that $$b\not< a$$ for all $$b\in A$$). Let $$F:X\times V\to V$$ be any (class) function. Then there exists a unique function $$f$$ on $$X$$ such that for each $$x\in X$$, $$f(x)=F(x,f|_{I(x)})$$ where $$I(x)=\{y\in X:y.

In other words, if you have a rule for defining $$f(x)$$ given $$x$$ and the values of $$f(y)$$ for all $$y, then this defines a unique function on all of $$X$$. For instance, you can use this on Pascal's triangle with $$X=\{(a,b)\in\mathbb{N}^2:b\leq a\}$$ where you say $$(a,b)<(c,d)$$ iff $$c=a+1$$ and $$d$$ is either $$b$$ or $$b+1$$ (i.e., $$(a,b)$$ is one of the two spots above $$(c,d)$$ in the triangle). The rule $$F$$ would then be to define $$f(0,0)=1$$, and otherwise define $$f(c,d)$$ to be the sum of all the values $$f(a,b)$$ such that $$(a,b)<(c,d)$$.

Proof of Theorem: Say that a subset $$I\subseteq X$$ is an initial segment if $$x\in I$$ and $$y implies $$y\in I$$. Note that if $$I$$ is an initial segment of $$X$$, then there exists at most one function $$f$$ on $$I$$ satisfying $$f(x)=F(x,f|_{I(x)})$$ for each $$x\in I$$ (given two such functions, by well-foundedness there is a minimal element on which they disagree, which then gives a contradiction). Call such a function good. Now let $$S$$ be the set of all such good functions defined on initial segments of $$X$$ (this is a set by Replacement), and let $$f=\bigcup S$$. Note that if $$g,h\in S$$ then $$g$$ and $$h$$ agree on $$\operatorname{dom}(g)\cap\operatorname{dom}(h)$$ since there is at most one good function on that set, and so $$f$$ is a function. The domain of $$f$$ is an initial segment of $$X$$ (since it is a union of initial segments), and $$f$$ is good since if $$x\in\operatorname{dom}(f)$$, there is some $$g\in S$$ such that $$x\in\operatorname{dom}(g)$$ and then $$f$$ and $$g$$ agree on $$I(x)$$ and also on $$x$$.

To complete the proof, it remains to be shown that $$f$$ is in fact defined on all of $$X$$. Suppose it is not; then there is a minimal element $$x\in X$$ where $$f$$ is not defined. This means $$f$$ is defined on $$I(x)$$. But now we can extend $$f$$ to a good function $$f'$$ on $$\operatorname{dom}(f)\cup\{x\}$$ by defining $$f'(x)=F(x,f|_{I(x)})$$. This $$f'$$ is then in $$S$$, contradicting the definition of $$f$$. Thus $$f$$ must be defined on all of $$X$$.

That said, your idea of partitioning your set is a natural and useful one, and is called the rank with respect to a well-founded relation. Here is one way to define it.

Definition: Let $$X$$ be a set and let $$<$$ be a well-founded relation on $$X$$. We then recursively define a function $$\operatorname{rank}:X\to Ord$$ as follows: $$\operatorname{rank}(x)=\sup\{\operatorname{rank}(y)+1:y. (Note that such a recursive definition makes sense using the Theorem above.)

So, $$\operatorname{rank}(x)=0$$ if $$x$$ is minimal, $$\operatorname{rank}(x)=1$$ if $$x$$ is not minimal but all its predecessors are minimal, and so on. Your sets $$A_k$$ are exactly the elements of rank $$k$$. In general, though, the rank of an element may not be finite (in other words, you would need to continue your process transfinitely in order to exhaust your set). For instance, if $$X$$ is well-ordered, then the rank function is just the unique isomorphism from $$X$$ to an ordinal.

Now, how can you use rank to make recursive definitions? Well, just observe that if you define $$x\prec y$$ to mean that $$\operatorname{rank}(x)<\operatorname{rank}(y)$$, then $$\prec$$ is also a well-founded relation on $$X$$ (indeed, this uses only the fact that $$\operatorname{rank}$$ is some function to the ordinals!). So, using the Theorem, you can recursively define functions on $$X$$ by defining $$f(x)$$ in terms of the values of $$f(y)$$ for all $$y$$ of lower rank than $$x$$.

• This is amazing! Could you recommend a reference on this stuff? I can't seem to find much online about order theory. – Luke Collins Nov 25 '20 at 1:00
• I don't really know a reference, but I imagine any decent introduction to axiomatic set theory will discuss this in some form. In particular, this general notion of recursion and rank are very important in set theory when applied to the relation $\in$. – Eric Wofsey Nov 25 '20 at 1:03
• To double check that I understand the notion of rank correctly, for instance, in $\{1-\frac 1n : n\in\mathbb N\}\cup\{1\},$ we have $\operatorname{rank}(1) = \omega+1$? – Luke Collins Nov 25 '20 at 1:33
• Not quite: the rank would be $\omega$, since all of the smaller numbers have finite rank. – Eric Wofsey Nov 25 '20 at 2:42
• Look, Eric, I'm not as active these days to get so much incoming reputation that I can afford to be giving you a bounty every two days. You're just being unreasonable! – Asaf Karagila Nov 25 '20 at 10:32