Internal Set Theory and an Intuitive Development of the Calculus

INTERNAL SET THEORY AND AN INTUITIVE DEVELOPMENT OF THE CALCULUS

SIMON LAZARUS

Abstract. In this paper, we present Internal Set Theory (IST) as a means to more intuitively develop mathematics while still maintaining complete rigor. We begin by stating the axioms of IST and demonstrating several of their central consequences. We then use these results to rigorously introduce the notions of inﬁnitesimal numbers and inﬁnite closeness. Lastly, we develop the calculus in a nonstandard fashion based on these intuitive notions.

Contents 1. Introduction 1 2. Fundamentals of Internal Set Theory 3 3. Standard Relations 7 4. Inﬁnitesimal and Unlimited Numbers 9 5. The Calculus 12 Continuity 12 Di↵erentiatiation 14 Integration 19 6. Appendix 24 Acknowledgments 27 References 27

1. Introduction The vast majority of modern mathematics is based in principle upon Zermelo- Fraenkel Set Theory with the Axiom of Choice (ZFC). The axioms of ZFC stipulate the rules of set formation based upon the (undefined) binary predicate ,where x y intuitively states “x is an element of the set y.” Using and the2 logical symbols2 (for all), (there exists), (and), (or), (not), =2 (implies), and (is8 equivalent9 to; if and only if)^ under the_ rules¬ of first-order) logic and the axioms() of ZFC, one can create definitions, state theorems, and write proofs. Although these methods are quite powerful, in many cases the strict requirements of ZFC can lead to definitions and proofs that are less intuitive than we may wish them to be. For this reason, we introduce Internal Set Theory (IST) to allow for many more intuitive definitions and proofs to be perfectly rigorous. IST is an extension of ZFC: it introduces the (undefined) unary predicate “standard” (we say

Date:September3,2012. 1 2 SIMON LAZARUS

“standard(x)” to mean that the set x is standard) and adds to the axioms of ZFC three axiom schemata that allow us to work with formulas involving this predicate. We assume the reader is familiar with the basic structure of a well-formed formula in first-order logic1. We recall that a variable x is said to be free in a formula A if A does not quantify x with “ x” or “ x.” For example, the formula x R has x as 8 9 2 a free variable, while the formula x Q(x R) has no free variables; x is bound. A formula with no free variables is8 said2 to be2 a statement. Every theorem of ZFC is a statement formulated from (i) variables quantified by or and (ii) some or all of the symbols , , ,= , , and . Formulas of8 ZFC9 are created in a similar manner, but^ not_ ¬ all of) a formula’s() variables2 need to be bound. In much the same way, formulas of IST are created using free and bound variables, some or all of the above symbols, as well as possibly the predicate “standard.” If A is a formula of IST in which “standard” does not appear, then A is clearly also a formula of ZFC. In this case, we say that A is an internal formula. If B is a formula of IST in which “standard” does appear, we say that B is external. IST merely adds new terminology to ZFC and gives axioms to explain how it can be used. In this way, any statement that is provable in ZFC will necessarily be provable in IST. However, the axioms of ZFC apply only to internal concepts. In particular, given any set x and any external formula with at least one free variable, the Axiom of Specification cannot be used to form a subset of x whose elements are exactly those of x which satisfy the given formula—it can only do so for internal formulas. We are therefore in need of new axiom schemata that allow us to form and work with sets based upon external formulas. These are the Principles of Transfer, Idealization, and Standardization. After stating these axiom schemata and a few of their important consequences, we will rigorously develop the notions of infinitesimal and unlimited numbers and infinite closeness, which we will use to develop the calculus in a fashion that we find much more intuitive than the usual presentation. One may wonder: even if IST does provide easier methods with which to state definitions and prove theorems, why should we believe anything IST begets? That is, just because we can show using the new methods of IST that some internal Theorem A holds true, does that necessarily mean Theorem A follows from ZFC alone? The answer to this question is a resounding yes. In particular, we have the following result.

Conservation Theorem (Powell). Let A be an internal theorem. If A is provable in IST, then A is provable in ZFC.

A proof of the Conservation Theorem is omitted in this paper. Within the proof given in [1], Nelson presents a method of reducing any proof of A using IST to a proof of A using ZFC alone. Therefore, we can have full conﬁdence in the fruits of IST: any proof of an internal theorem that employs IST immediately implies a proof of that same theorem that employs only ZFC. Additionally, this theorem guarantees that IST is consistent if ZFC is consistent. With this in mind, we begin our formal discussion of Internal Set Theory.

1Essentially, this means the reader will neither accept nor produce combinations of symbols that amount only to nonsense. INTERNAL SET THEORY AND AN INTUITIVE DEVELOPMENT OF THE CALCULUS 3

2. Fundamentals of Internal Set Theory Before stating the three axiom schemata mentioned in the previous section, we introduce some notation. First, if y is any set, we say finite(y)ifthereisno bijection between y and any proper subset of y. Now let A be any formula with free variable(s) including x. Then we introduce the following notation. Notation. (i) We use finxA(x) to mean x(finite(x)= A(x)) and we use finxA(x) to mean8 x(finite(x) A(x)).8 ) 9 (ii) We use 9stxA(x) to mean^ x(standard(x)= A(x)) and we use stxA(x) to mean8 x(standard(x) 8A(x)). ) 9 (iii) We use 9st finxA(x) to^ mean st(x)(finite(x)= A(x)) and we use st finxA(8x) to mean stx(finite(8x) A(x)). ) 9 9 ^ We can now state the three axiom schemata of Internal Set Theory.

Transfer Principle. Let A(x, t1,...,tn) be an internal formula with free variables x, t1,...,tn and no other free variables. Then (T) stt ... stt ( stxA(x, t ,...,t )= xA(x, t ,...,t )). 8 1 8 n 8 1 n )8 1 n Idealization Principle. Let A(x, y) be an internal formula with free variables including x and y.Then (I) st ﬁnz x y zA(x, y) x styA(x, y). 8 9 8 2 () 9 8 Standardization Principle. Let A(z) be a (not necessarily internal) formula with free variable(s) including z.Then (S) stx sty stz(z y z x A(z)). 8 9 8 2 () 2 ^ For the remainder of this section, we develop the intuitive meaning of these axiom schemata as we employ them. By replacing A with A and employing contraposition in (T) and (I) above, we can arrive at the logical¬duals of (T) and (I):

(T1) stt ... stt ( xA(x, t ,...,t )= stxA(x, t ,...,t )). 8 1 8 n 9 1 n )9 1 n (I1) st ﬁnz x y zA(x, y) x styA(x, y). 9 8 9 2 () 8 9 We now discuss some important consequences of (T) and (T1). First, we observe that if A is any internal formula whose only free variable is x and there exists a unique x such that A(x) holds true, then that x must be standard, as (T1) asserts the existence of a standard x satisfying A. Thus, one way of thinking about the standard sets is that any set that can be uniquely described within ZFC is standard. For example, the set of natural numbers N, the function sin, and the real number e are all standard sets since they are all sets which are uniquely described within ZFC. (The reader should recall that every entity in ZFC is a set, including numbers and functions.) More generally, (T1) tells us that any set which is uniquely determined by some ﬁnite number of other standard sets will itself be standard. We o↵er a few examples of this below. 4 SIMON LAZARUS

Examples 2.1. (i) If A is a standard set, then the power set of A (the set of all subsets of A) is standard since it is uniquely determined by A. (ii) If x and y are standard sets, then the ordered pair (x, y) is standard since it is uniquely determined by x and y. (iii) If f is a standard function and a is a standard element of the domain of f, then f(a) is standard since it is uniquely determined by f and a. (iv) If f and g are standard functions such that f g is a function, then f g is standard since it is uniquely determined by f and g. Remark 2.2. We note that restricted quantifiers maintain their restrictions through applications of (T) and (T1). For example, if X and Y are sets and B is an internal formula whose only free variables are x, t1,...,tn, then for all standard t1,...,tn we have stx XB(x)= x XB(x), and letting ! denote unique existence, we also have8 2!xB(x)= )8!stxB2(x). (Here, !stxB(x)9 means there exists a unique x satisfying B9 (x) and x)9is standard; it doesn’t9 mean the weaker statement “there exists a unique standard x satisfying B(x).”) Restrictions on quantifiers behave similarly through (I) and (I1); for example, if A is an internal formula with free variables including x and y,wehave st finz Y x X y zA(x, y) x X sty YA(x, y). 8 ✓ 9 2 8 2 () 9 2 8 2 We leave the proof of these simple facts to the reader, and we shall employ these results as needed in this paper. Other commonly employed restrictions on quantifiers include “finite” (as in finxB(x)) as well as any combinations of the above (as in !finx XB(x)). 8 9 2 Before carrying on, we introduce some new concepts and prove a related theorem that will guide our intuition in IST. Notation. Let A be an internal statement. We use Ast to mean the statement obtained by changing each instance in A of to st and each instance in A of to st. We call Ast the relativization of A to the8 standard8 sets. 9 9 Remark 2.3. In a similar manner to 2.2, we note that the relativization to the standard sets of a statement involving restricted quantifiers is merely that statement with the same restrictions on each quantifier plus the restriction “standard” on each quantifier. For example, if X and Y are sets and B is a quantifier-free internal formula whose only free variables are x and y, then the relativization to the standard sets of finx X !y YB(x, y) would be st finx X !sty YB(x, y). As before,8 the proof2 is9 simple2 and left to the reader.8 2 9 2 Definition 2.4. Let A be a statement. We say A is in prenex normal form if A is of the form Q1x1 ...QnxnB(x1,...,xn), where (i) B is an internal formula whose only free variables are x1,...,xn such that neither nor appears in B and (ii) for every i 1,...,n , Q is either or . 8 9 2{ } i 8 9 The following is a simple but tedious exercise in first-order logic. Exercise 2.5. Every internal statement can be rewritten in prenex normal form. With these tools in hand, we can begin to demonstrate some of the theorem- proving power of (T). INTERNAL SET THEORY AND AN INTUITIVE DEVELOPMENT OF THE CALCULUS 5

Relativization Theorem. Let A be an internal statement. Then A is true if and only if Ast is true.

Proof. By 2.5, we can write A as Q1x1 ...QnxnB(x1,...,xn), where B is a quantifier free internal formula whose only free variables are x1,...,xn and each Qi is st st st either or .Thus,A is Q1 x1 ...Qn xnB(x1,...,xn). We note that the con- verses8 of (T)9 and (T1) are obviously true, as (i) a formula A being true for all x means it is true for all standard x and (ii) the existence of a standard x satisfying a formula A clearly implies the existence of some x satisfying A. First, suppose A is true. Suppose Q1 is . Then by the converse of (T), we have Qstx Q x ...Q x B(x ,...,x ). Similarly,8 if Q is , we obtain the same result 1 1 2 2 n n 1 n 1 9 through (T1). Working rightwards towards B(x1,...,xn) in this fashion, we obtain st st st st st Q1 x1 ...Qn xnB(x1,...,xn), namely A . Now suppose instead A is true. If Qn st st st st st is , we use (T) to obtain Q1 x1 ...Qn 1xn 1QnxnB(x1,...,xn). If Qn is , we8 use the converse of (T1) to obtain the same result. Working leftwards towards9 st Q1 in this fashion, we obtain Q1x1 ...QnxnB(x1,...,xn), namely A. ⇤ Although the Relativization Theorem is a near-immediate consequence of (T), it is one of the central results of IST. The Relativization Theorem demonstrates two important facts. First, every theorem of ZFC is true when relativized to the standard sets. Second, one can prove any theorem of ZFC by merely proving its relativization to the standard sets. This latter task is often far simpler than the former. We now move on to some consequences of (I). Theorem 2.6. Let X be a set. Then X is standard and finite if and only if each element of X is standard. Proof. In (I1), let A(x, y) be the formula x X x = y.Thenwehave 62 _ st finz x y z(x X x = y) x sty(x X x = y). 9 8 9 2 62 _ () 8 9 62 _ This is the same as saying (*) st finz(X z) x X(standard(x)). 9 ✓ () 8 2 First, suppose X is standard and finite. Then we can choose X as the z in the left-hand side of (*), so every x in X is standard. Now suppose instead that each x in X is standard. Then by (*), st finz(X z). Let }(z) be the power set of z (the set of all subsets of z). Since z is9 standard,✓}(z) is standard by 2.1. Also, since z is finite, }(z) is finite. Thus, by the previous part of this proof, every element of }(z) is standard. So since X }(z) (as X z), X is standard. Lastly, since X z and z is finite, X is finite. 2 ✓ ✓ ⇤ The following remarkable theorem gives us further intuition into the nature of the standard sets. Theorem 2.7. There exists a finite set S such that X S whenever X is a standard set. 2 Proof. Let A(x, y) be the formula (finite(x) y x). Now consider the left-hand side of (I). Let z be any standard finite set. Then^ 2 clearly x y zA(x, y), as taking x = z we have y z(finite(z) y z), which is true. Thus,9 8 the2 left-hand side of (I) holds, so (I) tells8 us2 x sty(finite(^ x2) y x). This is equivalent to finx sty(y x). 9 8 ^ 2 9 8 2 So we can take S as any such x. ⇤ 6 SIMON LAZARUS

Remark 2.8. For any set A,thereexistsafinitesetF A such that for all standard x in A we have x F . This follows from 2.7; letting ✓S be any finite set containing every standard set2 as one of its elements, we obtain such a set F by taking A S. \ It is worth noting that although there is a finite set which contains every standard set in it as an element, there is no set whose elements are precisely all of the standard sets; that is, every such finite set S will have nonstandard elements. (By 2.6, this also means that each such set S is not standard.) For suppose there were a finite set T whose elements were precisely all of the standard sets. Then T would be standard by 2.6, so we would have T T , a contradiction to the Axiom of Foundation. Another way of thinking2 about this is as follows. Although there are many possible sets S given above, we cannot take their intersection to achieve a “smallest set” containing every standard set, as we cannot take the intersection of these sets whatsoever. This is because the intersection of any collection of sets is defined only upon the set of all sets we wish to intersect - and there is no set “ S : finite(S) stx(x S) ” because the formula “finite(S) stx(x S)” is not internal{ (meaning^8 we can’t2 apply} the Axiom of Specification^8 to it, even2 if we considered only S in some given universal set U). Similarly, we cannot simply form a set “ x : standard(x) ,” as the formula standard(x) is external. { Since the Axiom} of Specification does not apply to external formulas, if we wish to create a subset of a given set corresponding to a particular external formula, we can only use the Standardization Principle (S) to do so. However, the method in which we can do so is somewhat indirect. (S) only allows us to create a subset whose standard elements satisfy the given formula: a subset formed by means of (S) may contain nonstandard elements which do not satisfy the formula. However, the subset (formed by (S)) of a given set x corresponding to any given formula is indeed unique; this follows from the lemma below. Lemma 2.9. Let X and Y be standard sets. Then X = Y if and only if X and Y have the same standard elements. Proof. If X = Y ,thenX and Y have the same elements and hence have the same standard elements. So suppose X and Y have the same standard elements, i.e. suppose stx(x X x Y ). Note that the formula (x X x Y ) has free variables8 of2 only ()x, X, and2 Y ,whereX and Y are standard.2 Thus,() by2 (T) we have x(x X x Y ); namely, X = Y . 8 2 () 2 ⇤ Now returning to (S), let x be a standard set and A be a formula (internal or external) with free variable(s) including z. Then (S) asserts that there exists a standard set y such that any standard set z is in y precisely when it is in x and it satisfies A(z). We therefore have stz y(z x), so since x and y are standard, by (T) we have z y(z x8). Thus,2 y2is a standard subset of x whose standard elements are8 those2 of x2 which satisfy A. Now suppose w is also a standard subset of x whose standard elements are those which satisfy A.Then clearly stz(z y z w), as the standard elements of both z and w are precisely8 the standard2 () elements2 of x satisfying A. So by 2.9, w = y.Thus,the set y produced by (S) is the unique standard subset of x whose standard elements satisfy A. This result is summarized below. Notation. Let x be a set and A be a formula with free variable(s) including z. We denote the unique standard subset of x whose standard elements are those of x which satisfy A by S z x : A(z) . { 2 } INTERNAL SET THEORY AND AN INTUITIVE DEVELOPMENT OF THE CALCULUS 7

As a ﬁnal note in this section, we discuss the notion of a standard function. (Recall that a standard function is merely a standard set which is a function.) Theorem 2.10. Let A be a (possibly external) formula with free variables including x and y and let X and Y be standard sets. Suppose that for each standard x in X there exists at least one corresponding standard y in Y such that A(x,y) holds. Then there exists a standard function y˜ : X Y such that for all standard x in X, A(x, y˜(x)) holds. ! Proof. First, suppose for each standard x in X there exists a unique standard y in Y such that A(x, y). Let Z = S (x, y) X Y : A(x, y) . Then by the above, we st st st { 2 ⇥ } have x X y Y y0 Y [(x, y) Z ((x, y0) Z = y0 = y)] since by 2.6 (x,8 y) is2 standard9 2 precisely8 2 when x and2 y ^are standard.2 So) since X, Y , and Z are standard, by (T) in accordance with 2.2 we have

x X y Y y0 Y [(x, y) Z ((x, y0) Z = y0 = y)]; 8 2 9 2 8 2 2 ^ 2 ) that is, x X !y Y [(x, y) Z]. Thus, Z is a (standard) function from X to Y . Also, by8 the2 definition9 2 of Z, the2 hypothesis at the beginning of this paragraph, and 2.6 we know that whenever x is standard we have A(x, Z(x)). So we can choose Z as oury ˜. Now consider the case where y is not necessarily unique given x. Let }(Y )be the power set of Y . Then by 2.1 }(Y ) is standard since Y is standard. Recall that given any x X, S y Y : A(x, y) is a unique standard subset of Y .Thus, certainly for2 all standard{ 2 x X there} exists a unique standard z }(Y )such that z = S y Y : A(x, y) 2. By the first part of this proof, then,2 there exists a standard{ function2 z ˜ : X } }(Y ) such that for all standard x X we have z˜(x)=S y Y : A(x, y) . Now! given any standard x in X, S y Y2 : A(x, y) is nonempty,{ as2 by hypothesis} sty Y (A(x, y)). Thus, for all standard{ 2 x in X,˜z}(x) is a nonempty standard set. 9 2 This means we can apply the relativized-to-the-standard-sets version of the Ax- iom of Choice to obtain that there exists a standard function C such that stx X, we have C(˜z(x)) z˜(x). Thus, C z˜ maps each standard element x of X8 to some2 element of S y 2Y : A(x, y) . (That is, for all standard x in X, A(x, (C z˜)(x)) holds.) Now{ since2 C andz ˜ are} standard, by 2.1 C z˜ is standard. Also, by the above, it is certainly true that stx X((C z˜)(x) Y ). So since X, Y , and C z˜ are all standard, by (T) in accordance8 2 with 2.2, we2 know x X((C z˜)(x) Y). So to summarize: (i) C z˜ is standard, (ii) x X((C8 z˜2)(x) Y ), and2 (iii) stx XA(x, (C z˜)(x)). This means C z˜ su8ces2 as a desired y ˜.2 8 2 ⇤ Remark 2.11. Let A and B be sets such that A is standard. Then we can define a unique standard function f : A B solely by defining the (standard) values which f takes on at each standard x in! A. For let g : A B be any standard function such that stx A(g(x)=f(x)). Then since A, g,! and f are standard, by (T) we have x 8A(g(2x)=f(x)); namely, g = f. 8 2 3. Standard Relations The purpose of this section is to introduce a means of defining properties based on external formulas that we will frequently employ in our development of the calculus. In ZFC we may define properties of sets through relations. For example, 8 SIMON LAZARUS letting RR denote the set of functions from R to R, we may define the relation “f : R R is bounded above by M R on [a, b]” by the set ! 2

A := (f,M,a,b) RR R R R : x [a, b](f(x) M) . { 2 ⇥ ⇥ ⇥ 8 2  }

In this way, given any function g : R R and any Z, c, d R, we can immediately determine whether or not g is bounded! above by Z on2 [c, d]: this is true if (g, Z, c, d) A; it is false if (g, Z, c, d) A. However,2 the above method of defining62 relations only works using internal formulas (such as x [a, b](f(x) M) in the above example). If we wish to define a relation via an external8 2 formula, we must employ (S) to create a standard relation. In general, if for standard sets A1,...,An we wish to define a property B that any a A ,...,a A may or may not have, we do so via the standard set 1 2 1 n 2 n S (a ,...,a ) A ... A : B(a ,...,a ) { 1 n 2 1 ⇥ ⇥ n 1 n } which we shall call B¯. Then given any c1 A1,...,cn An,weusemembership in B¯ to determine whether or not c ,...,c2 have property2 B:if(c ,...,c ) B¯, 1 n 1 n 2 we say c1,...,cn have property B; if not, we say they do not. (Note that every relation of ZFC is a standard relation, as it is a uniquely defined set and hence is a standard set, and its standard elements satisfy the given formula because all of its elements satisfy it.) At this point, we note that given any standard sets A1,...,An we can define a unique standard relation on A1 ... An solely by its standard elements. This follows from 2.9, as a standard relation⇥ ⇥ is a standard set. That is: we can define what it means for any c1 A1,...,cn An to have some property B simply by saying “for all standard (c 2,...,c ) A 2 ... A ,wesayc ,...,c have property 1 n 2 1 ⇥ ⇥ n 1 n B if B(c1,...,cn) holds true.” (The reader should note that since (c1,...,cn)is a finite set, it will be standard precisely when each of c1,...,cn is standard.) In the future, whenever we define a relation by its standard elements we shall use a sentence such as “given c A ,...,c A ,wesayc ,...,c have property B if 1 2 1 n 2 n 1 n (for c1,...,cn standard) we have B(c1,...,cn).” The reader should note that although (by the above methods) we can easily define a standard relation on A1 ... An corresponding to a property B and can easily check whether or not some⇥ given⇥standard c A ,...,c A have property 1 2 1 n 2 n B, it is not always possible to know whether any given c1 A1,...,cn An have property B. This is because nonstandard elements of the relation2 S (a ,...,a2 ) { 1 n 2 A1 ... An : B(a1,...,an) need not satisfy B(c1,...,cn); see Section 2. Despite⇥ ⇥ this fact, proving} theorems related to properties defined by standard relations is quite easy. For example, let ↵ and be standard relations on A1 ... A . Suppose some Theorem 1 says a A ... a A [(a ,...,a ) ↵⇥ =⇥ n 8 1 2 1 8 n 2 n 1 n 2 ) (a1,...,an) ]; that is, if a1 A1 ...an An have property ↵ then they have property .Thenby(T),since2 ↵2 and are2 standard, we may prove Theorem 1 by st st merely proving a1 A1 ... an An[(a1,...,an) ↵ = (a1,...,an) ]. By similar arguments,8 2 we find8 that whenever2 we are faced2 with) a theorem relating2 to properties (defined by standard relations) of some finite number of sets, in our proof of that theorem we may immediately assume that those sets are standard. We shall do so frequently in Section 5. INTERNAL SET THEORY AND AN INTUITIVE DEVELOPMENT OF THE CALCULUS 9

4. Infinitesimal and Unlimited Numbers In this section, we introduce several of the notions upon which we will develop the calculus. We begin with the following definition. Definition 4.1. Let x be a real number. We say x is infinitesimal if for all standard ✏>0wehave x <✏.Wesayx is limited if there exists a standard real number y such that x | y|.Ifx is not limited, we say x is unlimited. | | We next introduce external induction, which will be quite useful to us both in proofs and in terms of our intuition. Recall the induction theorem of ZFC: if A is an internal formula with free variables including n such that (i) A(0) holds and (ii) k N(A(k)= A(k + 1)), then n NA(n). (Here, we implicitly assume that8 2 all of the other) free variables of A 8have2 already been quantified, so we can talk, for example, of A(0) being true or false.) External induction is the IST analogue of ZFC induction; it allows for external formulas. However, the External Induction Theorem has two weaker hypotheses and a weaker conclusion than the ZFC induction theorem. External Induction Theorem. Let A be a (possibly external) formula with free variables including n.Suppose(i) A(0) holds and (ii) for all standard k N we 2 have A(k)= A(k + 1). Then for all standard n N we have A(n). ) 2 Proof. Let B = S n N : A(n) . Since 0 is standard (since it is uniquely described in ZFC) and since{ by2 hypothesis} A(0) holds, by definition 0 B. Now suppose for 2 some standard k N that k B.Sincek is standard and in B, by definition we have A(k), meaning by2 hypothesis2 we have A(k + 1). Also, since k is standard, k +1is standard (as it is uniquely determined by k), so since we have A(k+1), by definition (k +1) B. In summary, we know stk N(k B = (k +1) B). By applying 2 8 2 2 ) 2 (T), we obtain k N(k B = (k + 1) B). Thus, by the ZFC induction 8 2 2 ) 2 theorem, we have n N(n B). So by definition n N(standard(n)= A(n)); 8 2 2 8 2 ) that is, stn N we have A(n). 8 2 ⇤ Whenever one works with external notions, one should be careful not to apply ZFC induction in place of external induction. For example, as in the above proof we know that 0 is standard and that whenever k N is standard k +1 willbe standard. An inappropriate application of ZFC induction2 would therefore tell us that every natural number n is standard. This is not the case. By 2.7, let S be any finite set containing each standard set as one of its elements. Thus, N S is finite, \ where stn N we have n N S.SosinceN is infinite, there must exist infinitely 8 2 2 \ many nonstandard elements of N. This does not contradict the External Induction Theorem, as that theorem merely gives us the obvious fact stn N(standard(n)). Now returning to unlimited and infinitesimal numbers,8 we note2 that if x is a standard real number, then x is limited. This is because x is standard (since it is uniquely determined by x) and x x . In fact, within| the| natural numbers we can make a stronger observation.| || |

Observation 4.2. Let m N. Then m is standard if and only if m is limited. 2 Proof. One direction of implication was proved above. For the other direction, suppose m is limited. Then by definition sty R(m y). This means y (the smallest integer greater than or equal to9 y)2 is standard since it is uniquelyd e 10 SIMON LAZARUS determined by the standard number y. We now claim st (1) n N x N(x n = standard(x)). 8 2 8 2  ) If (1) is true, then our desired result would follow with y in place of n and m in place of x. So we now prove (1) by external induction.d Ase a base case, clearly x N(x 1= standard(x)), as the only such natural number x is 1, which 8 2  ) is standard. As an induction hypothesis, suppose for some standard k N that 2 x N(x k = standard(x)). Then k + 1 is standard, so clearly we have 8 2  ) x N(x k +1 = standard(x)). By external induction, then, (1) holds. 8 2  ) ⇤ Thus, we now see that the conclusion of external induction is precisely that A(n) holds for all limited n N. Additionally, we now know there exist infinitely many unlimited natural numbers;2 this follows from 4.2 and the fact that there exist infinitely many nonstandard elements of N. We next discuss the existence of infinitesimals. Observation 4.3. (i) If y R is standard and infinitesimal, then y =0. However, 2 (ii) there exists a (nonstandard) x R such that x is infinitesimal but x =0. 2 6 Proof. (i) Suppose y is infinitesimal and y = 0. Then st✏>0( y <✏). Now suppose y is standard. Then we have y >6 0 and y 8is standard| | (since y is uniquely determined by a standard y), so| choosing| y |as| our ✏,wefind y < |y|,a contradiction. Thus, if y is infinitesimal and y = 0,| then| y is not standard.| | | | 6 (ii) We know st finz R x R y z(x =0 (y>0= x 0 if z contains any positive real numbers or we can choose x =1ifz does{ 2 not contain} any positive numbers. So by (I) in accordance with 2.2, we find x R sty R(x =0 (y>0= x 0 and st✏>0( x <✏). Now let A be the open interval (0, x ). Then there| | exist an uncountable8 | | number of elements of A. Also, for all z A| |and for all standard ✏>0, we have z < x <✏, so every element of A is infinitesimal.2 | | | | Remark 4.4. Let x and y be infinitesimal. Then st>0wehave x <and y <. Now let ✏>0 be standard. Then ✏/2 > 08 and ✏/2 is standard| | since it is uniquely| | determined by ✏. So choosing as ✏/2, we find x + y <✏/2+✏/2, which gives x + y <✏by the Triangle Inequality. Thus, if x and| | y are| | infinitesimal, then x + y |is infinitesimal.| Using the notion of infinitesimals, we now introduce the principal notion upon which we will develop the calculus. Definition 4.5. Let x and y be real numbers. We say x and y are infinitely close if x y is infinitesimal. In this case, we write x y. ' Under this definition, we see that a real number ↵ is infinitesimal precisely when ↵ 0. Thus, an equivalent way of saying x y is ↵ 0(x = y + ↵). We therefore' observe immediately that every real number' is9 infinitely' close to itself, as 0 is infinitesimal. Additionally, for all real numbers x and y, x y implies y x. For suppose x y.Then ↵ 0(x = y + ↵), meaning y = x '↵.Butsince'↵ is ' 9 ' INTERNAL SET THEORY AND AN INTUITIVE DEVELOPMENT OF THE CALCULUS 11 infinitesimal, clearly ↵ is infinitesimal, so y x. Lastly, if x, y, z R such that x y and y z,then x z. For suppose 'x y and y z.Then2 ↵, 0 such' that x ='y + ↵ and y'= z + ; namely, x ='z + ↵ + .' But by 4.4,9 ↵ + ' is infinitesimal, so by definition x z. In summary: is an equivalence relation. Despite this fact, one must still' be careful when' working with : although it is an equivalence relation, it is still an external notion, so we can' only employ external induction to any formulas involving . For example, if we let ↵>0be ' infinitesimal and x R,weknowx x + ↵ and whenever k N is standard we have x x + k↵ =2 x x +(k +' 1)↵. However, external induction2 only tells ' ) ' us x x + n↵ when n is limited. In fact, it is not true that given any n N we have 'x x + n↵; as this would mean we could choose n suciently large such2 that x x +' 1, which cannot be true since (x + 1) x = 1 and 1 > 0 is standard. The following' theorem reinforces the impossibility of such a situation. Theorem 4.6. Let x be a limited real number. Then there exists a unique standard y R such that x y. 2 ' Proof. Since x is limited, by definition there exists a standard M 0 such that x M. Let A = S z R : z x . Then by definition A is standard and | st|zA(z x), meaning{ 2stz A(z }M)sincex x M.SosinceM and A are8 standard,2  by (T) we have8 2z A(z M); that is,|A|is bounded above by M. Note that M x since M 8 2x x because M x . Also, M is standard since M is standard. Thus, by definition| | M A,soAis| | nonempty. Therefore, A has a least upper bound; call it l.Thenl is standard2 since it is uniquely determined by the standard set A. Now suppose x l.Theneither(i) st✏>0(x l ✏) or (ii) st>0(l x ). Suppose first that6' (i) is true. Then l 9+ ✏ x. Also, l + ✏ is standard9 since it is uniquely determined by the standard numbers l and ✏. So by definition l + ✏ A; but l + ✏>l, a contradiction to the fact that l is an upper bound for A.So2 suppose instead that (ii) is true. Then x l , and l is standard since l and are standard. Now by the definition of Awe know stz A(z x). Thus, certainly stz A(z l ). So since A and l are standard,8 2 by (T) we have z A(z8 l 2 ). This means l is an upper bound for A;butl | 0(|st(x) x <✏). Suppose st(x) [a, b]. Then either (i) st(x) >bor (ii) st(8 x) 0. Also, st(x) b is standard since st(x) and b are standard. Thus, by definition 12 SIMON LAZARUS st(x) x < st(x) b, so clearly st(x) xb,a contradiction| | to the fact that x [a, b]. Similarly, (ii) leads to a contradiction. Thus, st(x) [a, b]. 2 2 ⇤ As a last note in this section, we present (in the form of exercises) several simple facts about limited and infinitesimal numbers and standard parts of numbers. Sev- eral of these results will be needed to develop the calculus. In the exercises below, x, y, a, and b are real numbers. For the sake of illustration, we prove (ix). Exercises 4.8. (i) If a is limited and x a,thenx is limited. (ii) If x and y are limited,' then x + y and xy are limited. (iii) If x is limited and x a,thenst(x) = a. (iv) If x a and y b,then6' x + y a +6 b. (v) If x ' 0 and y ' 0, then xy 0.' (vi) If x 6' 0 and y 6'is limited, then6' xy 0. (vii) Suppose' a and b are limited. If x ' a and y b,thenxy ab. ' 1 ' ' (viii) Suppose x = 0. Then x 0 if and only if x is unlimited. 6 ' 1 1 (ix) Suppose a 0. If x a,then x a . (x) Suppose x 6'and y are' limited. Then' st(x + y)=st(x)+st(y) and st(xy)=st(x)st(y). Also, if x 0, then st( 1 )= 1 . 6' x st(x) (xi) Suppose a and b are standard and a

Theorem 5.3. Let a, b R such that af(st(y)). Then since f(st(y)) f(z)+↵,wehave ↵ 2↵ f(z) f(st(y)) > 0. But since z, st(y), and f are standard, f(z) f(st(| |y)) is positive and standard, a contradiction to the fact that ↵ 0. Thus, stx [a, b](f(x) f(st(y)). Since st(y) [a, b], this clearly implies stc' [a, b] st8x 2[a, b](f(x) f(c)). Letting C be the2 standard relation corresponding9 2 to functions8 2 continuous on [a, b], we can summarize our result as: stf sta stb[(f,a,b) C = stc [a, b] stx [a, b](f(x) f(c))], which is identical8 8 to 8stf sta stb2 stc )9[a, b] st2x [a,8 b]((f,a,b2 ) C = f(x) f(c)). So by (T), we8 have8 f8 a 9b c2 [a, b8] x 2 [a, b]((f,a,b) 2 C =) f(x)  f(c)), 8 8 8 9 2 8 2 2 )  which is identical to the desired result. ⇤ Obviously, under the conditions of 5.3 there also exists d [a, b] such that for all x [a, b]wehavef(x) f(d). 2 2 Theorem 5.4. Let f :[a, b] R be such that f(a) < 0 0but f(st(xi)) is standard (since f is standard), a contradiction| || to ↵ 0.| Thus, f(st(| x )) 0.| ' i Now note that xi 1 F A, as xi 1 A would imply xi 1 xi, a contradiction. 2 \ 2 By 4.8 (xii), we know a st(xi 1) st(xi) b since a xi 1 xi b.       Now suppose xi 1 xi.Thensincexi st(xi) and xi 1 st(xi 1), we have 6' st ' ' st(xi) st(xi 1); that is, ✏>0( st(xi) st(xi 1) ✏), meaning we certainly 6' 9 | | 14 SIMON LAZARUS have st(xi) st(xi 1) >✏/2. Thus, xi 1 st(xi 1) < st(xi 1)+✏/2 < st(xi) xi. ' ' So by 4.8 (xi), xi 1 < st(xi 1)+✏/2

I f : I R !f 0 : I R[f D = f 0 R], 8 8 ! 9 ! 2 ) 2 or I f : I R[f D = !f 0 : I R(f 0 R)]. 8 8 ! 2 )9 ! 2 In other words, given any open interval I and any function f that is di↵erentiable on I, there exists a unique function f 0 : I R such that (for f and I standard) we st f(y) f(x)! have x I y I(y x = f 0(x)). We call f 0 the derivative of f 8 2 8 2 ⇠ ) y x ' on I, or merely the derivative of f. Before moving on, we find it appropriate to point the reader to the appendix, wherein we demonstrate that our definitions of continuity and di↵erentiability of f at a are the same as their conventional definitions. We now create an alternative, equivalent definition of di↵erentiability. By sub- stituting the definition of and multiplying by x a in the definition of “f is di↵erentiable at a,” we obtain' st c R x R[x a = ↵ 0(f(x) f(a)=(x a)(c + ↵))]. 9 2 8 2 ⇠ )9 ' Clearly, this occurs precisely when st c R x R[x a = ↵ 0(f(x) f(a)=(x a)(c + ↵))], 9 2 8 2 ' )9 ' as x = a implies f(x) f(a)=0=(x a)(c + ↵). Note that x a occurs if and only if h 0(x = a +h). Thus, our definition is equivalent to ' 9 ' st c R x R[ h 0(x = a + h)= ↵ 0(f(x) f(a)=(x a)(c + ↵))]. 9 2 8 2 9 ' )9 ' This is logically identical to st c R x R h 0 ↵ 0[x = a + h = f(x) f(a)=(x a)(c + ↵)], 9 2 8 2 8 ' 9 ' ) which is trivially the same as st c R h 0 ↵ 0 x R[x = a + h = f(x) f(a)=(x a)(c + ↵)], 9 2 8 ' 9 ' 8 2 ) or st c R h 0 ↵ 0 x = a + h[f(x) f(a)=(x a)(c + ↵)]. 9 2 8 ' 9 ' 8 So replacing x with a + h, our definition becomes st c R h 0 ↵ 0[f(a + h) f(a)=h(c + ↵)]. 9 2 8 ' 9 ' where, as usual, c is the derivative of f at a. We summarize this result in the equivalent definition of di↵erentiability given below.

Deﬁnition 5.7. Let I R be an open interval, f : I R, and a I.Wesayf is ✓ ! 2 di↵erentiable at a if (for f and a standard) there exists a standard c R such that for all h 0thereexists↵ 0 such that f(a + h)=f(a)+ch + ↵h2. In this case, ' ' we say c is the derivative of f at a and denote it by f 0(a). Using this deﬁnition, it becomes very easy to prove many basic properties of the derivative.

Lemma 5.8. Let I R be an open interval, f : I R,anda I.Iff is di↵erentiable at a, then✓ f is continuous at a. ! 2 16 SIMON LAZARUS

Proof. By (T), we can assume f and a are standard. Suppose x a.Thenthere exists h 0 such that x = a + h, so by deﬁnition there exists ↵' 0 such that ' ' f(a + h)=f(a)+(f 0(a)h + ↵h). Now by 4.8 (vi) and (iv) we have f 0(a)h + ↵h 0 ' since f 0(a) is standard and hence limited. That is, there exists 0 such that f(x)=f(a)+, meaning f(x) f(a). ' ' ⇤ Theorem 5.9. Let f and g be di↵erentiable at a. Then f + g and fg are di↵erentiable at a with

(i) (f + g)0(a)=f 0(a)+g0(a) and (ii) (fg)0(a)=f(a)g0(a)+g(a)f 0(a). Additionally, if g(a) =0, then 1 is di↵erentiable at a with 6 g 1 g0(a) (iii) ( g )0(a)= [g(a)]2 . Proof. We can assume f, g, and a are standard. Let h 0. Then there exist ' ↵, 0 such that f(a + h)=f(a)+f 0(a)h + ↵h and g(a + h)=g(a)+g0(a)h + h. ' (i) This means (f + g)(a + h)=(f + g)(a)+(f 0(a)+g0(a))h +(↵ + )h.Butby 4.4, ↵ + 0, so by deﬁnition we have (f + g)0(a)=f 0(a)+g0(a). (ii) By simple' multiplication, we have 2 (fg)(a + h)=f(a)g(a)+f(a)g0(a)h + f(a)h + g(a)f 0(a)h + f 0(a)g0(a)h 2 2 2 +f 0(a)h + g(a)↵h + g0(a)↵h + ↵h . By factoring out an h from all but the ﬁrst term and rearranging, we obtain

(fg)(a + h)=[f(a) + f 0(a)g0(a)h + f 0(a)h + g(a)↵ + g0(a)↵h + ↵h]h

+f(a)g(a)+(f(a)g0(a)+g(a)f 0(a))h.

Since f(a), g(a), f 0(a) and g0(a) are standard, they are limited. So since ↵, , and h are all infinitesimal, by 4.8 (vi) and (iv) the entire term in brackets in the above equation is infinitesimal. That is, 0 such that (fg)0(a)=f(a)g(a)+ 9 ' (f(a)g0(a)+g(a)f 0(a))h + h, as desired. (iii) For this part, we employ our first definition of the derivative of g at a. g(x) g(a) Let x a.Then g0(a). Since g(a) = 0 and g(a) is standard, by 4.6 ⇠ x a ' 6 g(a) 0. Since g(x) g(a) (as g is continuous at a because it is di↵erentiable at a) and6' since g(a) is limited' (as it is standard), g(x) is limited by 4.8 (i). By 4.8 (vii), then, g(x)g(a) [g(a)]2.Sinceg(a) 0, by 4.8 (v) [g(a)]2 0. Thus, by 4.8 (ix) we have ' 6' 6' 1 1 . g(x)g(a) ' [g(a)]2 2 1 Also, since [g(a)] 0, by 4.8 (viii) 2 is limited. So since g0(a) is also limited 6' [g(a)] g(a) g(x) (as it is standard) and xa g0(a), by 4.8 (vii) we have ' 1 1 g(x) g(a) 1 g(a) g(x) 1 = g0(a). x a g(x)g(a) · x a ' [g(a)]2 · ⇤ Theorem 5.10. Suppose f is di↵erentiable at g(a) and g is di↵erentiable at a. Then f g is di↵erentiable at a with (f g)0(a)=f 0(g(a)) g0(a). · Proof. We can assume f, g, and a are standard. Let h 0. Then there exists ↵ 0 ' ' such that g(a + h)=g(a)+g0(a)h + ↵h.Thus,f(g(a + h)) = f(g(a)+g0(a)h + ↵h). INTERNAL SET THEORY AND AN INTUITIVE DEVELOPMENT OF THE CALCULUS 17

Since g0(a) is limited (as it is standard), by 4.8 (vi) and (iv) we know g0(a)h+↵h 0. Thus, by deﬁnition there exists 0 such that ' ' f(g(a)+[g0(a)h + ↵h]) = f(g(a)) + f 0(g(a))[g0(a)h + ↵h]+[g0(a)h + ↵h], meaning f(g(a + h)) = f(g(a)) + f 0(g(a))g0(a)h +[f 0(g(a))↵ + g0(a)+↵]h.So since g0(a) and f 0(g(a)) are limited (as they are standard), by 4.8 (vi) and (iv) the entire term in brackets in this equation is inﬁnitesimal. That is, there exists 0 ' such that (f g)(a + h)=(f g)(a)+f 0(g(a))g0(a)h + h, as desired. ⇤ Theorem 5.11. Let a, b, c, d R such that a0, we would have 0 = f 0(c)+↵,so h by 4.8 (xii) we would have st(0) st(f 0(c)+↵), or 0 f 0(c). Similarly, if instead f(c+h) f(c) we had chosen h<0, we would have 0 = f 0(c)+↵, giving 0 f 0(c).  h  Thus, f 0(c) = 0. ⇤ Obviously, the same result applies if c is a minimum point for f on (a, b). With this lemma in hand, 5.13 can be proven in its usual fashion.

Theorem 5.13. Let a, b R such that a 0, then f is increasing on I. (ii) f 0(x) 0, then f is nondecreasing on I. (iii) f 0(x)=0, then f is constant on I. Proof. By (T), we can assume f and I are standard, so I has standard endpoints. Let a, b be standard elements of I such that a 0, so f(b) >f(a)sinceb a>0. 18 SIMON LAZARUS

(ii) Let ↵>0 be inﬁnitesimal. For all x I,letg(x)=f(x)+↵x.Then 2 by 5.11 and 5.9, for all x I we have g0(x)=f 0(x)+↵>0, so g satisﬁes the hypotheses of (i). Thus, g(b2) >g(a), so 0 0) if and only if we have x I(f 0(x) > 0). The implication of the general8 2 case to the standard case is obvious.8 2 For the other direction of implication, note that f 0 is standard since f is standard (as f 0 is uniquely determined by f). st Thus, by (T) if we have x I(f 0(x) > 0) then we have x I(f 0(x) > 0). Similar results apply to parts8 (ii)2 and (iii) of 5.14. 8 2

Theorem 5.15. Let a, b R such that a 0) or x [a, b](f 0(x) < 0). 1 8 2 8 2 Then f is di↵erentiable on f([a, b]), and given any c [a, b] we have 2 1 1 (f )0(f(c)) = . f 0(c)

Proof. We prove the case where x [a, b](f 0(x) > 0); the other case is similar. So by 5.14, f is increasing on [a,8 b].2 We can assume f, a, b, and c are standard. Let x [a, b]. Then by 4.7 st(x) [a, b]. First, suppose x c.Thensincec is standard,2 st(x) c. Suppose st(x2) >c.Thensincef is increasing6' on [a, b], we have f(st(x)) >f6'(c). So since f, c, and st(x) are standard, by 4.6 f(st(x)) f(c). If instead st(x)

Integration.

Deﬁnition 5.16. Let a, b R such that a b and let f :[a, b] R.Wesayf is 2  ! integrable on [a, b] if (for f, a, and b standard) there exists I R such that 2

st f(ih)h = I 0 1 i :a ih

Remark 5.17. Let a

Theorem 5.18. Let a, b R be standard such that a0 be inﬁnitesimal. Suppose there exist standard m, M R 2 such that for all i in j Z : a jh < b we have m f(ih) M. Then { 2  }   i:a ih0, clearly there exists n N 0 such that n =min i N 0 : a ih . This gives (n 1)h

m(b a)=st(mh(N n)) st f(ih)h st(Mh(N n)) = M(b a).  0 1  ⇤ i:a ih

Through (T), we can prove a similar result when f, a, b, m and M aren’t necessarily standard: if f is integrable on [a, b] and there exist m, M R such that b 2 for all x in [a, b]wehavem f(x) M,thenm(b a) a f(t)dt M(b a). We now give an example of a function that is not integrable.  R Example 5.19. For all x R,letf(x)=1ifx is rational and f(x)=0ifx is 2 irrational. Let a, b R such that a0 such that a

st f(ih )h = b a>0butst f(ih )h =0. 0 1 11 0 2 21 ⇤ i:a ih

Theorem 5.20. Let a, b R such that a0 be inﬁnitesimal and let x [a, b] be standard. Since f is continuous on [a, b], we know f is continuous on [a,2 b], so by 5.3 max f(y) : y [a, b] 0 exists; call it M.Then| | M is standard since f, a, and b are{| standard.| 2 By the} Triangle Inequality, we therefore have

f(ih)h f(ih)h Mh M(x a),  | |  i:a ih

G(x)=st f(ih)h . 0 1 i:a ih

G(x) G(c)=st f(ih)h st f(ih)h 0 1 0 1 i:a ih

First, suppose x>c. Then by 4.8 (xi), since x h x and x and c are standard, we have x h>c. This means '

G(x) G(c)=st f(ih)h . 0 1 i:c ih

(x c)minf(t) st f(ih)h (x c) max f(t). c t x  0 1  c t x   i:c ihc = min f(t) max f(t) . 8 2 8 2 ) c t x  x c  c t x h     i Now suppose instead x

G(x) G(c)=st f(ih)h =st f(ih)h . 0 1 0 1 i:x ih

(c x) max f(t) st f(ih)h (c x) min f(t) . x t c  0 1  x t c ✓   ◆ i:x ihc = min f(t) max f(t) 8 2 8 2 ) c t x  x c  c t x h     i G(x) G(c) (2) x [a, b] c [a, b] xc.Since2c, x [a, b] and f is continuous2 on [a, b] and x ⇠ c,wemusthave y [c, x](f(y) 2f(c)). Thus, min f(t) f(c) max f(t'). So since f(c)is 8 2 ' c t x ' ' c t x     standard, by 4.8 (xii) and (1) above, we have G(x) G(c) f(c)=st min f(t) st st max f(t) = f(c), c t x  x c  c t x ✓   ◆ ✓ ◆ ✓   ◆ G(x) G(c) or in other words, f(c). By a similar argument, we obtain the exact x c ' same result if instead x