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The Natural Numbers The natural numbers As mentioned earlier in the course, the natural numbers can be constructed using the axioms of set theory. In this note we want to discuss the necessary details of this construction. Recall that Axiom VII, often called the axiom of infinity, guarantees the existence of at least one inductive set – that is, a set X with the property that (i) ∅ ∈ X, and (ii) A ∪ {A} ∈ X whenever A ∈ X. Notice that such an inductive set X must contain the sets ∅, {∅}, {∅, {∅}}, {∅, {∅}, {∅, {∅}}}, ···. However, it may also contain a lot of other elements. In order to get rid of all conceivable extraneous elements in X we can form the smallest possible subset of X which is still inductive, called its inductive core: \ C(X) = {I | I ⊆ X and I is and inductive set} = {A ∈ X | A ∈ I for every inductive set I with I ⊆ X}. We would like to define the natural numbers N to be this inductive core C(X). For this definition to be meaningful, however, we need to check if one arrives at the same inductive core regardless of the inductive set one starts out with. To this end, suppose that Y is also an inductive set. We wish to show that C(X) = C(Y ). First, observe that X ∪ Y is an inductive set, too. Also, since the sets involved in the intersection defining C(X ∪ Y ) are all those sets involved in the intersection defining C(X) plus possibly more, we conclude that C(X ∪ Y ) ⊆ C(X). Conversely, let A ∈ C(X). In order to show that A ∈ C(X ∪ Y ), we let I be an arbitrary inductive subset of X ∪ Y and argue that A ∈ I. But that follows immediately upon observing that I ∩ X is an inductive subset of X and must therefore contain A. We have now shown that C(X ∪ Y ) = C(X). Similarly, C(X ∪ Y ) = C(Y ). Hence C(X) = C(Y ). This little discussion enables us to make our definition: Definition. Let X be any inductive set. We define the set of natural numbers as N = C(X). Next, let us give the elements of N shorter names. We shall write 0, 1, 2, 3, ··· for the sets ∅, {∅}, {∅, {∅}}, {∅, {∅}, {∅, {∅}}}, ···. That is to say, we write 0 for ∅, 1 for {0}, 2 for {0, 1}, 3 for {0, 1, 2}, and so on. We will also use the convenient abbreviation A + 1 = A ∪ {A}. Hence, 1 = 0 + 1, 2 = 1 + 1, 3 = 2 + 1, and so on. This does not mean, however, that we already know how to add natural numbers. We will define addition and multiplication of natural numbers later. At this point, A + 1 simply stands for A ∪ {A}. Notice that the principle of mathematical induction follows immediately from the definition of N: Theorem [Mathematical induction]. Let a predicate P (n) be given. Suppose that P (0) is true and that P (n + 1) is true whenever P (n) is true. Then P (n) is true for all n ∈ N. 1 Proof. Let A = {n ∈ N | P (n) is true }. Then 0 ∈ A and n + 1 ∈ A whenever n ∈ A. Hence, A is an inductive set, so that N = C(A) ⊆ A ⊆ N. That is A = N. We now put an ordering on the natural numbers: Definition. For m, n ∈ N we define m < n to mean m ∈ n. For example, 3 < 5 since 3 ∈ 5 = {0, 1, 2, 3, 4}. Exercises. Use mathematical induction to prove the following statements for all m, n, k ∈ N. [Solutions are sketched at the end of this note.] 1. m < n + 1 if and only if either m < n or m = n. 2. m ⊆ n if and only if either m < n or m = n. 3. If m < n, then m 6= n. 4. Writing m ≤ n for “m < n or m = n”, we get: (i) n ≤ n. (ii) If m ≤ n and n ≤ m, then m = n. (iii) If m ≤ n and n ≤ k, then m ≤ k. (iv) If m < n and n < k, then m < k. 5. If m < n, then m + 1 < n + 1. 6. Exactly one of the following must be true: m < n, m = n, or n < m. 7. If m + 1 = n + 1, then m = n. Theorem [Recursion Theorem]. Let A be a set, a ∈ A, and r : N × A → A any function. Then there exists a function f : N → A such that (i) f(0) = a, and (ii) f(n + 1) = r(n, f(n)) for all n ∈ N. We say that the rule r together with the initial value a recursively define the sequence f. Proof. T Put f = {S ⊆ N×A | (0, a) ∈ S ∧∀n∀x[(n, x) ∈ S → (n+1, r(n, x)) ∈ S]} ⊆ N×A. Then (i) (0, a) ∈ f and (ii) ∀n∀x[(n, x) ∈ f → (n + 1, r(n, x)) ∈ f]. Therefore, once we have shown that f is a function it will automatically have properties (i) and (ii) of the theorem. (A quick induction argument shows that the domain of f is N.) We 2 will prove by induction on m that f passes the “vertical line test” and is therefore a function: (m, x1), (m, x2) ∈ f → x1 = x2. Basic Step: Suppose there are (0, x1), (0, x2) ∈ f with x1 6= x2. Then either x1 6= a or x2 6= a. Say, x1 6= a. Form the set S = f \{(0, x1)} ⊆ N × A. Observe that (i) (0, a) ∈ S, and (ii) that if (n, x) ∈ S, then (n, x) ∈ f so that (n + 1, r(n, x)) ∈ f. Hence, (n + 1, r(n, x)) ∈ S, because (n + 1, r(n, x)) 6= (0, x1) (since n + 1 6= 0). So, by definition of f, f ⊆ S. This contradicts the definition of S. Inductive Step: Suppose that (m + 1, x1), (m + 1, x2) ∈ f with x1 6= x2. If both x1 = r(m, z1) for some (m, z1) ∈ f and x2 = r(m, z2) for some (m, z2) ∈ f, then z1 = z2 by inductive hypothesis, so that x1 = x2 – which is not the case. We can therefore assume, without loss of generality, that x1 6= r(m, z) for all (m, z) ∈ f. Form the set S = f \{(m + 1, x1)} ⊆ N × A. As before, observe that (i) (0, a) ∈ S (since 0 6= m+1) and (ii) that if (n, x) ∈ S, then (n, x) ∈ f so that (n+1, r(n, x)) ∈ f. Hence, (n + 1, r(n, x)) ∈ S (otherwise m + 1 = n + 1 and x1 = r(n, x) = r(m, x) with (m, x) = (n, x) ∈ f, which is not allowed). The definition of f implies now that f ⊆ S, which in turn contradicts the definition of S. This completes the proof. We now use the recursion theorem to introduce addition and multiplication into our set of natural numbers. First we want to define what it means to add a natural number to a fixed m ∈ N. We do this by recursively defining a function fm : N → N. We require that (i) fm(0) = m; and (ii) fm(n + 1) = fm(n) + 1. The recursion theorem guarantees the existence of such a function fm. We will of course write m + n for fm(n). (Notice that fm(1) = fm(0) + 1 = m + 1 so that there is no ambiguity in using “m + 1” for both “m ∪ {m}” and “fm(1)”.) In this more suggestive notation, the above two items read: (i) m + 0 = m; and (ii) m + (n + 1) = (m + n) + 1. Exercises. 8. Show that (m + n) + k = m + (n + k) for all m, n, k ∈ N. 9. Show that m + n = n + m for all m, n ∈ N. 3 We also define multiplication of a fixed m ∈ N by a natural number via the recursion theorem by way of a function gm : N → N with the properties (i) gm(0) = 0; and (ii) gm(n + 1) = gm(n) + m. Writing m · n for gm(n), this recursive definition can be written as (i) m · 0 = 0; and (ii) m · (n + 1) = m · n + m. Here, we make the usual convention that multiplication is carried out before addition, unless specified otherwise. Exercises. Prove the following for all m, n, k ∈ N. 10. m · 1 = 1 · m = m. 11. (m + n) · k = m · k + n · k. 12. m · n = n · m. 13. (m · n) · k = m · (n · k). 14. If m < n, then m + k < n + k. 15. If m < n and k 6= 0, then m · k < n · k. 16. If m · k = n · k and k 6= 0, then m = n. 17. If m + k = n + k then m = n. 18. If m · n = 0, then either m = 0 or n = 0. 19. If m ≤ n, then there is exactly one d ∈ N such that m + d = n. 4 Solutions. 1. This follows straight from the definition of n + 1 = n ∪ {n} and the meaning of the symbol “<”. 2. “⇐” We will prove, by induction on n, that m < n implies m ⊆ n. The case n = 0 is vacuously true. Now suppose that m < n + 1. Then either m < n or m = n, by Exercise 1. Hence, by the inductive hypothesis, m ⊆ n, so that m ⊆ n ∪ {n} = n + 1. “⇒” Once more, we induct on n. If n = 0, then m = ∅ = n. Inductively, suppose that m ⊆ n + 1 = n ∪ {n}. If n 6∈ m, then m ⊆ n, so that by inductive hypothesis either m = n or m < n. Either way, this case implies that m ∈ n∪{n}, i.e.
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