超实数微积分初步
   既然超实数是“数”，那么，超实数“数轴”、“平面”
，等等基础概念，都不难定义了。
    进入21世纪，大学微积分教科书，使用超实数系统，变得简单多了。请看本文附件。
    简单就是“美”！
袁萌  陈启清   7月15日
附件：
Hyperreal Calculus
Abstract
This project deals with doing calculus not by using epsilons and deltas, but by using a number system called the hyperreal numbers. The hyperreal numbers is an extension of the normal real numbers with both innitely small and innitely large numbers added. We will rst show how this system can be created,and then shows ome basic properties of the
hyperreal numbers. Then we will show how one can treat the topics of convergence, continuity, limits and dierentiation in this system and we will show that the two approaches give rise to the same denitions and results.
Contents
1 Construction of the hyperreal numbers 3
1.1 Intuitive construction . . . . . . . 3
1.2 Ultralters . . .  3
1.3 Formal construction .  . 4
1.4 Innitely small and large numbers . . .. 5
1.5 Enlarging sets . . . . . . . . . . . 5
1.6 Extending functions . . . . .. . 6
2 The transfer principle 6 2.1 Stating the transfer principle . . . . . . . . . 6
2.2 Using the transfer principle . . . . . .7
3 Properties of the hyperreals 8
3.1 Terminology and notation . . . . . . . . . . . . . . . . . . . . . . 8
3.2 Arithmetic of hyperreals . .  9
3.3 Halo（） 9 3.4 Shadows . . . . . . . . . . . . . . . . . . . . .. 10
4  Convergence   11
4.1 Convergence in hyperreal calculus. . . . . . . . . . . . . . . . . . 11
4.2 Monotone convergence . . . . . . . . . . . . . . .. . 12
5 Continuity 13
5.1 Continuity in hyperreal calculus 13
5.2 Examples . .. 14
5.3 Theorems about continuity . .15
5.4 Uniform continuity . . . . . 16
6 Limits and derivatives 17 6.1 Limits in hyperreal calculus . . . .   17
6.2 Dierentiation in hyperreal calculus. . 18
6.3 Examples 18
6.4 Increments . . . . . . . . . . . . . 19
6.5 Theorems about derivatives . . . . . 19
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1 Construction of the hyperreal numbers 1.1 
Intuitive construction We want to construct the hyperreal numbers as sequences of real numbers hrni = hr1,r2,...i, and the idea is to let sequences where limn→∞rn = 0 represent innitely small numbers, or innitesimals, and let sequences where limn→∞rn =∞ represent innitely large numbers. However, if we simply let each hyperreal number be dened as a sequence of real numbers, and let addition and multiplication be dened as elementwise additionandmultiplicationofsequences, wehavetheproblemthatthisstructure is not a eld, since h1,0,1,0,...i
h0,1,0,1,...i=h0,0,0,0,...i. The way we solve this is by introducing an equivalence relation on the set of real-valued sequences. We want to identify two sequences if the set of indices for which the sequences agree is a large subset of N, for a certain technical meaning of large. Let us rst discuss some properties we should expect this concept of largeness to have. • N itself must be large, since a sequence must be equivalent with itself. • If a set contains a large set, it should be large itself. • The empty set ∅ should not be large. • We want our relation to be transitive, so if the sequences r and s agree on a large set, and s and t agree on a large set, we want r and t to agree on a large set.
1.2 Ultralters Our model of a large set is a mathematical structure called an ultralter. Denition 1.1 (Ultralters). We dene an ultralter on N, F, to be a set of subsets of N such that: • If X ∈ F and X ⊆ Y ⊆ N, then Y ∈ F. That is, F is closed under supersets. • If X ∈F and Y ∈F, then X ∩Y ∈F. F is closed under intersections. • N∈F, but ∅6∈F. • For any subset A of N, F contains exactly one of A and N\A. We say that an ultralter is free if it contains no nite subsets of N. Note that a free ultralter will contain all conite subsets of N (sets with nite complement) due to the last property of an ultralter. Theorem 1.2. There exists a free ultralter on N. Proof. See [Kei76, p. 49]. 
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1.3 Formal construction Let F be a xed free ultralter on N. We dene a relation ≡ on the set of real-valued sequences RN by letting hrni≡hsni ⇐⇒ {n ∈N| rn = sn}∈F. Proposition 1.3 (Equivalence). The relation ≡ is an equivalence relation on RN. Proof. We check all needed properties of an equivalence relation. Reexivity Since the set {n ∈N| rn = rn}= N, and N∈F, ≡ is reexive. Symmetry The sets {n ∈N| rn = sn} and {n ∈N| sn = rn} are the same, so if one belongs to F, so does the other. Transitivity Assume that hrni≡hsni and hsni≡htni. Then both {n ∈ N | rn = sn}∈F and {n ∈N| sn = tn}∈F. Since {n ∈N| rn = sn}∩{n ∈ N| sn = tn}⊆{n ∈N| rn = tn}, and F is closed under intersections and supersets, {n ∈N| rn = tn}∈F, and so hrni≡htni, as desired.   Since ≡ is an equivalence relation, we can dene the set of hyperreal numbers ∗R as the set of real-valued sequences modulo the equivalence relation ≡. In symbols, ∗R ={[r]| r ∈RN}= RN/ ≡ . We dene addition and multiplication of elements in ∗R by doing elementwise addition and multiplication in the related sequences, more formally as [r]+[s]=[hrni]+[hsni]=[hrn +sni] [r]•[s]=[hrni]•[hsni]=[hrn •sni]. We dene the ordering relation < by letting [r] < [s] ⇐⇒ {n ∈N| rn < sn}∈F. At this point, let us introduce some notation to make our arguments easier to read. For two sequences hrni and hsni, we denote the agreement set {n ∈N| rn = sn} byJr = sK. We can apply the same notation to other relations, so for example we haveJr < sK={n ∈N| rn < sn}. Proposition 1.4. The operations + and•are well-dened, and so is the relation <. Proof. We rst show that + is well-dened. If we have that hrni ≡ hr0 ni and hsni ≡ hs0ni, thenJr = r0K∈ F andJs = s0K∈ F, which means thatJr = r0K∩Js = s0K∈F. What we now need to show is thatJr + s = r0 + s0K∈F. If, for some k ∈ N, both rk = r0 k and sk = s0 k, then rk + sk = r0 k + s0 k, hence if k ∈Jr = r0K∩Js = s0K, then k ∈Jr + s = r0 + s0K, which shows that Jr = r0K∩Js = s0K⊆Jr + s = r0 + s0K. SinceJr = r0K∩Js = s0K∈ F, so is Jr +s = r0 +s0K. So if r ≡ r0 and s ≡ s0, r +s ≡ r0 +s0, which shows that the operation is well-dened. Showing that • is well-dened is similar. 3
We will now show that < is well-dened, which means that we need to show that if hrni ≡ hr0 ni and hsni ≡ hs0ni, then ifJr < sK∈ F, thenJr0 < s0K∈ F.Firstly, assume that Jr = r0K∈F and thatJs = s0K∈F. Then, we need to provethat if Jr < sK∈F thenJr0 < s0K∈F. So let us assume thatJr < sK∈F, and then prove thatJr0 < s0K∈F. By our assumptions, we have thatJr = r0K∩Js = s0K∩Jr < sK∈ F. Ifk ∈Jr = r0K∩Js = s0K∩Jr < sK, then rk = r0 k, sk = s0 k and rk < sk, and therefore r0 k < s0 k, so k ∈Jr0 < s0K. So,Jr = r0K∩Js = s0K∩Jr < sK⊆Jr0 < s0K,and since F is closed under supersets, we conclude thatJr0 < s0K∈F, whichshows that < is well-dened.   1.4 Innitely small and large numbers One of the main reasons for constructing the hyperreals is that we want to have access to innitely large and innitely small numbers, and now we can prove their existence. Theorem 1.5. There exists a number ε ∈ ∗R such that 0 < ε < r for any positive real number r, and there exists a number ω ∈ ∗R such that ω > r for any real number r. Proof. First, we need to talk about real numbers in ∗R. The way to do this is that given a real number r ∈R, we can identify this with a hyperreal number ∗r ∈ ∗R as ∗r =hr,r,...i. We will generally omit the ∗-decoration, and simply refer to this number as r. Now, let us turn to the actual proof. Let ε = h1, 1 2,...i = h1 ni . For anypositive real number r, the set {n ∈ N | 1 n > r} must be nite, and therefore {n ∈N| 1 n < r}isconite,andhencebelongstoourfreeultralterF. Therefore, we can conclude that ε < r. Also, since {n ∈N|0 < 1 n}= N∈F, it must bethe case that 0 < ε. So the number ε is a hyperreal number which is greater than 0, but smaller than any positive real number. Let ω =[h1,2,...i]=[hni]. For any real number r, the set{n ∈N| r ≥ n}is nite, and hence{n ∈N| r < n}is conite, and belongs toF, which means that ω > r. This proves that ω is a hyperreal number greater than any real number. 
1.5 Enlarging sets For a given subset A of R we can dene an “enlarged” subset ∗A of ∗R by saying that a hyperreal number r is an element in ∗A if and only if the set of n such that rn is an element in A is large. Formally this can be dened as [r]∈∗A ⇐⇒ {n ∈N| rn ∈ A}∈F. Again, we need to check that this is well-dened. Using theJ...Knotation,let Jr ∈ AK={n ∈N| rn ∈ A}. We have that Jr = r0K∩Jr ∈ AK⊆Jr0 ∈ AK,so if r ≡ r0 andJr ∈ AK∈F, thenJr0 ∈ AK∈F, which shows that enlargements are well-dened.
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