Now, with all the definitions we ve made, we re ready see where all the math stuff we took for granted, like numbers, come from.

The Natural Numbers Now, with all the definitions we ve made, we re ready see where all the math stuff we took for granted, like numbers, come from. Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 1 / 11

The Natural Numbers Now, with all the definitions we ve made, we re ready see where all the math stuff we took for granted, like numbers, come from. Our first goal is to construct the set of natural numbers N = {0, 1, 2, 3, 4,...}. This is going to come about in a rather hairy way, but in the end you ll see why. Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 1 / 11

The Natural Numbers Now, with all the definitions we ve made, we re ready see where all the math stuff we took for granted, like numbers, come from. Our first goal is to construct the set of natural numbers N = {0, 1, 2, 3, 4,...}. This is going to come about in a rather hairy way, but in the end you ll see why. The first thing we need, of course, is a few more definitions: Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 1 / 11

The Natural Numbers Now, with all the definitions we ve made, we re ready see where all the math stuff we took for granted, like numbers, come from. Our first goal is to construct the set of natural numbers N = {0, 1, 2, 3, 4,...}. This is going to come about in a rather hairy way, but in the end you ll see why. The first thing we need, of course, is a few more definitions: Definition Given a set x, the successor of x, denoted S(x) is the set S(x) = x {x}. Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 1 / 11

Since we re now ready to start getting some things that are familiar to us, we will start calling the empty set 0, but at this point, this is just another name for. Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 2 / 11

Since we re now ready to start getting some things that are familiar to us, we will start calling the empty set 0, but at this point, this is just another name for. Definition A set I is called inductive if 0 I and if x I, then S(x) I. Just think for a moment how wild an inductive set I must be. Just from the fact that 0 = I, by definition, this means S({ }) = { } = { } I. Again by definition S({ }) = { } {{ }} = {, { }} I. And this keeps going and going. Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 2 / 11

ZF Nothing in our list of axioms so far tells us that an inductive set actually exists. First off, all sets that exist by our axioms so far must be finite. Clearly, inductive sets at least have some sort of infinite properties. So we need a new axiom. Axiom (6. The Axiom of Infinity) There exists an inductive set I. Notice if we have another inductive set J, then 0 J also. But if we can find even one x J such that x / I, then all the successors of that x are in J and not in I. What we want to get our hands on is the smallest possible inductive set. Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 3 / 11

Let P(J) be the property J is an inductive set. By the axiom of comprehension and the axiom of infinity ω = {x I J(P(J) x J)} is a set. Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 4 / 11

Let P(J) be the property J is an inductive set. By the axiom of comprehension and the axiom of infinity ω = {x I J(P(J) x J)} is a set. So ω is a set that is a subset of every possible inductive set. Since we want the smallest possible inductive set, all we have to prove now is: Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 4 / 11

Let P(J) be the property J is an inductive set. By the axiom of comprehension and the axiom of infinity ω = {x I J(P(J) x J)} is a set. So ω is a set that is a subset of every possible inductive set. Since we want the smallest possible inductive set, all we have to prove now is: Proposition ω is an inductive set Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 4 / 11

Proof. 0 ω since for all sets J such that J is inductive, by definition of inductive sets 0 J. Another way to say this is: J(P(J) 0 J) is true. Since ω = {x I J(P(J) x J)} this means 0 ω. Now suppose x ω. We want to prove that S(x) ω. Since we assumed x ω, this means J(P(J) x J) is true. This says, for any inductive set J, we know x J. By definition of J being an inductive set, we know S(x) J. Another way to say this is J(P(J) S(x) J), but this means exactly that S(x) ω Therefore ω is inductive. Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 5 / 11

The point of this is to notice that we constructed ω as a smallest inductive set. That is, Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 6 / 11

The point of this is to notice that we constructed ω as a smallest inductive set. That is, Proposition (The Principal of Mathematical Induction) If S ω and S is an inductive set, then S = ω Proof. Suppose S ω is an inductive set. Take any element x ω. This means, by definition of ω, x satisfies the property: if J is an inductive set, then x J. We assumed S is an inductive set, so x S. We ve shown x ω x S. This says ω S. Since we assumed already S ω, by the axiom of extension, S = ω. Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 6 / 11

Now consider the following construction: Starting with = 0, define the set n + 1 = S(n) = n {n}. Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 7 / 11

Now consider the following construction: Starting with = 0, define the set n + 1 = S(n) = n {n}. Of course, you re thinking n is a number not a set. For us, everything is a set. Writing this out 0 = 1 = S(0) = { } = { } = {0} 2 = S(1) = 1 {1} = {0} {{0}} = {0, {0}} = {0, 1} Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 7 / 11

Now consider the following construction: Starting with = 0, define the set n + 1 = S(n) = n {n}. Of course, you re thinking n is a number not a set. For us, everything is a set. Writing this out 0 = 1 = S(0) = { } = { } = {0} 2 = S(1) = 1 {1} = {0} {{0}} = {0, {0}} = {0, 1} 3 = S(2) = 2 {2} = {0, 1} {{0, 1}} = {0, 1, {0, 1}} = {0, 1, 2} Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 7 / 11

Now consider the following construction: Starting with = 0, define the set n + 1 = S(n) = n {n}. Of course, you re thinking n is a number not a set. For us, everything is a set. Writing this out 0 = 1 = S(0) = { } = { } = {0} 2 = S(1) = 1 {1} = {0} {{0}} = {0, {0}} = {0, 1} 3 = S(2) = 2 {2} = {0, 1} {{0, 1}} = {0, 1, {0, 1}} = {0, 1, 2} n + 1 = S(n) = {0, 1, 2,..., n} Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 7 / 11

Notice every element in the sequence {0, 1, 2,...} is either 0 which is in ω or is a successor of some other element already in ω. So this sequence is secretly the set ω. Therefore we define N = ω. We will use the two names N and ω interchangeably. The important thing to remember, regardless if you call it N or ω, is that it is the smallest inductive set. The reason the previous proposition is called the principal of mathematical induction is the following: At many times in our mathematical lives we will be given a statement P(n) and be asked to prove it is true for all n N. Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 8 / 11

Notice every element in the sequence {0, 1, 2,...} is either 0 which is in ω or is a successor of some other element already in ω. So this sequence is secretly the set ω. Therefore we define N = ω. We will use the two names N and ω interchangeably. The important thing to remember, regardless if you call it N or ω, is that it is the smallest inductive set. The reason the previous proposition is called the principal of mathematical induction is the following: At many times in our mathematical lives we will be given a statement P(n) and be asked to prove it is true for all n N. Notice to prove P(n) n N we are required to do an infinite number of proofs. So how do we accomplish this? Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 8 / 11

Theorem (Mathematical Induction) Suppose for each n N, P(n) is a statement. If P(0) is true and P(k) P(k + 1), then for all n N, P(n) is true. Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 9 / 11

Theorem (Mathematical Induction) Suppose for each n N, P(n) is a statement. If P(0) is true and P(k) P(k + 1), then for all n N, P(n) is true. Proof. Let S = {n N P(n)}. Since P(0) is true, 0 S. Now take any k S. By definition of S, P(k) is then true. By hypothesis, P(k) P(k + 1), since P(k) is true we therefore know P(k + 1) is true. This says, k + 1 S. Since k + 1 is the successor of k, we have shown S is inductive. Since S N, by the Principal of Mathematical Induction S = N, and therefore P(n) is true for all n N. Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 9 / 11

Doing proofs by mathematical induction come up a lot. For example, suppose, while making a sandwich, you recognized that 1 = 2 1 1 Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 10 / 11

Doing proofs by mathematical induction come up a lot. For example, suppose, while making a sandwich, you recognized that 1 = 2 1 1, 1 + 2 = 2 2 1 Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 10 / 11

Doing proofs by mathematical induction come up a lot. For example, suppose, while making a sandwich, you recognized that 1 = 2 1 1, 1 + 2 = 2 2 1, 1 + 2 + 2 2 = 2 3 1 Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 10 / 11

Doing proofs by mathematical induction come up a lot. For example, suppose, while making a sandwich, you recognized that 1 = 2 1 1, 1 + 2 = 2 2 1, 1 + 2 + 2 2 = 2 3 1, 1 + 2 + 2 2 + 2 3 = 2 4 1, etc. Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 10 / 11

Doing proofs by mathematical induction come up a lot. For example, suppose, while making a sandwich, you recognized that 1 = 2 1 1, 1 + 2 = 2 2 1, 1 + 2 + 2 2 = 2 3 1, 1 + 2 + 2 2 + 2 3 = 2 4 1, etc. You guess that for all n 2 0 + 2 1 + + 2 n = 2 n+1 1. Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 10 / 11

Doing proofs by mathematical induction come up a lot. For example, suppose, while making a sandwich, you recognized that 1 = 2 1 1, 1 + 2 = 2 2 1, 1 + 2 + 2 2 = 2 3 1, 1 + 2 + 2 2 + 2 3 = 2 4 1, etc. You guess that for all n 2 0 + 2 1 + + 2 n = 2 n+1 1. How would you prove this is true for every n? Maybe it s not true for n = 38 for example. Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 10 / 11

Doing proofs by mathematical induction come up a lot. For example, suppose, while making a sandwich, you recognized that 1 = 2 1 1, 1 + 2 = 2 2 1, 1 + 2 + 2 2 = 2 3 1, 1 + 2 + 2 2 + 2 3 = 2 4 1, etc. You guess that for all n 2 0 + 2 1 + + 2 n = 2 n+1 1. How would you prove this is true for every n? Maybe it s not true for n = 38 for example. We use mathematical induction. Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 10 / 11

Proposition For all n N, 2 0 + 2 1 + + 2 n = 2 n+1 1. Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 11 / 11

Proposition For all n N, 2 0 + 2 1 + + 2 n = 2 n+1 1. Proof. Let P(n) be the statement 2 0 + 2 1 + + 2 n = 2 n+1 1 is true. Now clearly 2 0 = 2 0+1 1, so P(0) is true. Next suppose that for some k N, P(k) is true. Notice that 2 0 + 2 1 + + 2 k+1 = 2 0 + 2(2 0 + 2 1 + + 2 k ). Since we assumed P(k) is true, we know 2 0 + 2(2 0 + 2 1 + + 2 k ) = 2 0 + 2(2 k+1 1) = 2 (k+1)+1 1. So we ve shown 2 0 + 2 1 + + 2 k+1 = 2 (k+1)+1 1 Therefore P(k + 1) is also true. We ve shown P(0) and P(k) P(k + 1). Therefore, by mathematical induction, P(n) is true for every n N. Math 144 Summer 2012 (UCR) Pro-Notes July 6, 2012 11 / 11