Math 310 Class Notes 5: the Recursion Theorem for N

Math 310 Class Notes 5: The Recursion Theorem for N

In this set of notes we will tighten a loose end that has been left dangling. Namely, in Class Notes 3 we used the word ”assign” when we defined the notion of + for N (see Definition 2 of the notes). The word ”assign” actually means we have a function that does the assignment. To be precise, we have the following theorem. Theorem 1. (The recursion theorem) Let c be a fixed natural number and let f : N → N be a function. Then there is a unique function g : N → N such that g(1) = c and g(x0) = f(g(x)) for all x ∈ N. Before proving this theorem, let us see a few applications. Application 1. Consider the function f : N → N given by f : x 7→ x0.

Fix a number m ∈ N. Then by the recursion theorem there is a function g such that g(1) = m0 and g(x0) = f(g(x)) for all x ∈ N. Let us now deﬁne m + n := g(n) for all n ∈ N. Theorem 2. Fix m ∈ N. We have m + n0 = (m + n)0 for all n ∈ N. Proof. m + n0 = g(n0) = f(g(n)) = f(m + n) = (m + n)0. Slightly harder to verify is the following. Theorem 3. Fix m ∈ N. We have m0 + n = (m + n)0 for all n ∈ N. Proof. m + n = g(n) for all n ∈ N, where g(1) = m0 and g(n0) = f(g(n)). Likewise, m0 + n = h(n) for all n ∈ N, where h(1) = (m0)0 and h(n0) = f(h(n)). Let S be the set of all n ∈ N such that m0 + n = (m + n)0. We ﬁrst show 1 ∈ S. Now m0 + 1 = h(1) = (m0)0 while m + 1 = g(1) = m0, so that m0 + 1 = (m0)0 = (m + 1)0. Thus 1 ∈ S. Next suppose n ∈ S. We establish that n0 ∈ S. That n ∈ S means m0 + n = (m + n)0, which is the induction hypothesis. Therefore, by the preceding theorem and the induction hypothesis we have m0 + n0 = (m0 + n)0 = ((m + n)0)0 = (m + n0)0. Therefore, n0 ∈ N and so S = N. That is m0 + n = (m + n)0 for all n ∈ N. 1 2

This theorem is what we have seen in class. The advantage of this approach via the recursion formula is that we can take care of m + n for all m and n at one stroke. Application 2. Fix m ∈ N. Consider the function f : N → N given by f : x 7→ x + m. Then by the recursion theorem there is a function g : N → N such that g(1) = m and g(x0) = f(g(x)). We deﬁne m · n := g(n) for all n ∈ N. Exercise. Prove that m · n0 = m · n + m and m0 · n = m · n + n for all n ∈ N. Application 3. Fix a ∈ N. Consider the function f : N → N given by f : x 7→ ax. Then by the recursion theorem there is a function g : N → N such that g(1) = a and g(x0) = f(g(x)). We deﬁne an := g(n).

Exerxise. By the mathematical induction, prove that am+n = am · an and (am)n = amn. Now let us look at the proof of the recursion theorem. Proof. Existence of g. By deﬁnition, a function g : N → N is a special sort of subset of N × N. Recall that g(1) = c means (1, c) ∈ g and g(x0) = f(g(x)) means that if we let y = g(x), i.e., let (x, y) ∈ g, then (x0, f(y)) ∈ g as this just says g(x0) = f(g(x)) = f(y). Motivated by this, let us begin to construct the function g by consid- ering the set S, where S := {A ⊂ N × N : (1, c) ∈ A and (x, y) ∈ A implies (x0, f(y)) ∈ A}. S is not empty since N × N ∈ S. Consider the set

g := ∩A∈S A, which consists of elements common to all A ∈ S. Now (1, c) ∈ g since it is in all A ∈ S. Moreover, if (x, y) ∈ g then (x, y) ∈ A for all A ∈ S. But then (x0, f(y)) ∈ A for all A ∈ S so that 3

(x0, f(y)) ∈ g. We conclude therefore that g is in fact an element of S; it is the smallest element of S. We now prove that g is a function from N to N. First we show that the domain of g is N. Let T be the subset of n ∈ N such that there is a z ∈ N for which (n, z) ∈ g. Now 1 ∈ T since (1, c) ∈ g. Suppose n ∈ T . Then (n, z) ∈ g for some z ∈ N. This gives (n0, f(z)) ∈ g. Hence n0 ∈ T . It follows by mathematical induction that T = N. That is, the domain of g is N. Secondly, we show that each domain element of g is mapped to a unique target element. To this end, consider the set U := {x ∈ N : (x, y), (x, z) ∈ g imply y = z}. We show 1 ∈ U. Suppose this is not the case. That is, suppose we have (1, z), in addition (1, c) ∈ g, where z =6 c. Then the set g \{(1, z)} would be a member of S (WHY?). But this contradicts the fact that g is already the smallest element of S. Thus, 1 ∈ U. We show that if x ∈ U then x0 ∈ U. Suppose this is not the case. Then there is an m ∈ U so that m0 ∈/ U. Since m ∈ U, there is a unique n ∈ N with (m, n) ∈ g. As a consequence (m0, f(n)) ∈ g. Since m0 ∈/ U, there is an (m0, z) ∈ g, in addition to (m0, f(n)) ∈ g, where z =6 f(n). But then g \ {(m0, z)} is an element of S that is even smaller than g (WHY?), which is absurd. The mathematical induction then says that U = N. In conclusion, g is a function. Uniqueness of g. Suppose we have two functions g1 and g2 satisfying g1(1) = c = g2(1) 0 0 and g1(x ) = f(g1(x)) and g2(x ) = f(g2(x)) for all x ∈ N. Let us consider the set

V := {n ∈ N : g1(n) = g2(n)}.

1 ∈ V because g1(1) = g2(1) = c. If n ∈ V , then g1(n) = g2(n), so that 0 0 g1(n ) = f(g1(n)) = f(g2(n)) = g2(n ), which implies n0 ∈ V . Hence V = N by the mathematical induction. That is, g1 = g2.