Chapter 2 ELEMENTARY SET THEORY

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1 Chapter 2 ELEMENTARY SET THEORY 2.1 Itroductio We adopt the aive as opposed to axiomatic poit of view for set theory ad regard the otios of a set as primitive ad well-uderstood without formal defiitios. We just assume that a set A is a collectio of objects characterized by some defiig property that allows us to thi of the objects as a whole etity. The defiig property has to be such that it must be clear whether a give object belogs to the set or ot. The objects possessig the property are called elemets or members of the set. We deote sets by commo capital letters A, B, C, etc. ad elemets or objects of the sets by lower case letters a, b, c,etc. For example, we write A = {a, b, c} to idicate that A is a collectio of elemets a, b, c. If x is ay elemet, if A is a set, ad if x belogs to A we write x A. If x does t belog to A we write x / A Descriptio of a Set To describe a particular set, we have to idicate the property that characterizes its elemets. To start, lets cosider a set which has o members. Sice a set is determied by its elemets, there is oly oe such set which is called the empty set, or the ull set, which is deoted by. Ay set A, cosistig of oe or more elemets is said to be o-empty or o-void. Defiitio 34 A o-void set A is said to be fiite if A cotais distict elemets, where is some positive iteger. Such a set A is said to be of order. The ull set is defied to be fiite with order zero. A set cosistig of exactly oe elemet, say A = {a}, is called a sigleto. ifiite. If a set A is ot fiite, the we say that A is To describe a fiite set, oe way to proceed is by listig or eumeratig its members. For example, S = {1, 3, 5} meas that S is the set whose elemets are the itegers 1, 3 ad 5. Note that we use braces to deote the set ad that the order of the elemets plays o role i describig the set. Thus the sets S = {1, 3, 5} ad S = {5, 3, 1} are to be viewed as exactly the same set.

2 20 Elemetary Set Theory But i geeral, ad especially for ifiite sets, we do ot describe a set by listig all its members. A coveiet method of characterizig sets is as follows. Suppose that for each elemets x of a set A there is a propositioal fuctio P x which is either true or false. We may the defie a set C which cosists of all elemets x A such that P x is true, ad we may write C = {x A : Px is true}. For brevity of otatio, we will omit the words is true ad, whe it is clear which set x belogs to, we will also omit the referece set A. Thus we will sometimes write {x : P x}. From a give set, we ca form ew sets, called subsets of the give set. I geeral we say that a set A is a subset of a set B wheever every elemet of A belogs also to B. More formally Defiitio 35 Give two sets A ad B, we say that A is a subset of B, or that A is cotaied i B, addeoteita B or B A, iff x : x A x B. We say that A is a proper subset of B, or that A is strictly cotaied i B, ad deote it A B, iff [A B B A] is true. Defiitio 36 We say that a set A is equal to B ad write A = B iff x A x B, i.e., A ad B have the same elemets. Equivaletly A = B [A B B A]. We say that A is differet from B, ad we write A B, if A ad B are ot equal. If x ad y deote the same elemet of a set we say that they are equal ad we write x = y. If x ad y deote distict elemets of a set, we write x y. Exercise 37 Show that if A is ay set, the A A ad A. Exercise 38 Show that A = B [A B B A]. Defiitio 39 Let X be a set ad let A X.The complemet of subset A with respect to X is the set of elemets of X which do ot belog to A. We deote the complemet of A with respect to X by A C X = {x X : x/ A}. Whe it is clear that the complemet is with respect to X, we simply say the complemet of A, ad write A C = {x X : x/ A}. I every discussio ivolvig sets, we will always have a give fixed set i mid from which we tae elemets ad subsets. We will call this set the uiversal set, ad we will usually deote it by X. Throughout the remaider of the preset chapter, X will deote always a arbitrary o-empty fixed set. Exercise 40 Prove the followig statemets. Let A,B, Cbe subsets of X. The 1. if A B ad B C the A C.

3 Operatio o sets X C = 3. C = X 4. A C C = A 5. A B B C A C 6. A = B A C = B C Next we eed to cosider sets whose elemets are sets themselves. For example, if A,B ad C are subsets of X, the the collectio A = {A,B,C } is a set, whose elemets are A,B ad C. We usually call a set whose elemets are subsets of X a family of subsets of X or a collectio of subsets of X. I terms of otatio, this cocludes our hierarchical system, where lowercase letters refer to elemets of X, upper case letter to subsets of X ad script letters to families of subsets of X. Lets ote that the empty set is a subset of X. It is possible the to form a oempty set whose oly elemet is the empty set, i.e. { }. I this case, { } is a sigleto. We see that { }ad { }. There is a special family of subsets of X to which we have give a special ame. Defiitio 41 Let A be ay subset i X. We defie the power class of A or the power set of A to be the family of all subsets of A. We deote the power set of A by P A. Specifically P A ={B : B A} The power class of the empty set is P ={ },i.e. the sigleto of. The power class of a sigleto is P {a} ={,{a}}.notice that the power set of A always cotais A ad. I geeral, if A is a fiite set with elemets, the P A cotais 2 elemets. This is why sometimes the power set of A is improperly deoted as 2 A. Exercise 42 Prove that if A is a fiite set with elemets, the P A cotais 2 elemets. 2.2 Operatio o sets Uio ad Itersectio Defiitio 43 Let A ad B be two subsets of X.We defie the uio of sets A ad B, deoted by A B, as the set of all elemets that are i A or B; i.e. A B = {x X :[x A x B]}.

4 22 Elemetary Set Theory We also defie the itersectio of sets A ad B,deoted by A B, as the set of all elemets that are i both A ad B; i.e. A B = {x X :[x A x B]}. Both defiitios ca be exteded to fiite or ifiite collectio of sets. Defiitio 44 Two sets are said to be disjoit if they do t have ay elemet i commo, i. e.. if A B =. Exercise 45 Prove the followig properties 1. A A C = ; A A C = X. 2. Commutative laws: A B = B A; A B = B A. 3. Associative law: A B C =A B C = A B C; A B C =A B C = A B C. 4. Distributive law: A B C =A B A C; A B C =A B A C. 5. A B [A B =B ] [A B =A] 6. De Morga s laws A B C = A C B C A B C = A C B C From the associative law established i the previous exercise, there is o ambiguity i writig A B C. Extedig this cocept, let be ay positive iteger ad let A 1,A 2..., A deote subsets of X. The set A 1 A 2... A is defied to be the set ofall x X which belog to at least oe ofthe subsets A i, ad we write i=1 A i = A 1 A 2... A = {x X : x A i for some i =1,...,} Usig a similar argumet we defie the itersectio of subsets of X as i=1 A i = A 1 A 2... A = {x X : x A i for all i =1,..., }

5 Relatios 23 Theorem 46 Let A 1,A 2..., A be subsets of X. The [ i=1 A i ] C = i=1 A C i [ i=1 A i ] C = i=1 A C i Proof. Is left as a exercise. Use your results from the previous exercise Relative Differece Defiitio 47 Let A ad B be two subsets of X. The relative differece of B ad A, deoted as B A, also ow as the relative complemet of A i B is defied as the set of elemets i B that do ot belog to A. HeceB A = {x X : x B x/ A}. Exercise 48 Prove the followig statemets. Let A ad B be two subsets of X. The 1. A B = A B C. 2. A B C = A C B. 3. A B A B B A. 4. B B A=A A B. 5. B A = B B A = Exercise 49 Let B ada 1,A 2..., A be subsets of X. The 1. B [ i=1 A i ]= i=1 B A i 2. B [ i=1 A i ]= 2.3 Relatios Ordered pairs i=1 B A i So far the order of the elemets has ot bee relevat. Thus the set {a, b} has bee cosidered equivalet to the set {b, a}. There are times however where the order is importat. Whe we wish to idicate that a set of two elemets a ad b is ordered, we eclose the elemets i paretheses: a, b. The a is called the first elemet ad b the secod elemet. A set of such two elemets is called a ordered pair, ad is defied i the followig way. Defiitio 50 A ordered pair a, b is a set whose members are {a} ad {a, b}. That is a, b ={{a}, {a, b}}. From the defiitio, we ca derive a mai property of ordered pairs.

6 24 Elemetary Set Theory Theorem 51 x, y=a, b x = a ad y = b Proof. Lets first prove the coverse. If x = a ad y = b the a, b ={{a}, {a, b}} = {{x}, {x, y}} =x, y Now lets assume x, y =a, b. The by defiitio we have {{x}, {x, y}} = {{a}, {a, b}}. Cosider two cases, depedig o whether x = y or x y. If x = y the {x} = {x, y},so x, y ={{x}}. Sice x, y =a, b the we have {{x}} = {{a}, {a, b}}. The set o the left had side has oly oe member {x}.thus the set o the right must also have oe member, so {a} = {a, b}, cocludig that a = b. The {{x}} = {{a}}, so {x} = {a} ad x = a. Thus x = y = a = b. If x y, the it must be a b otherwise we would be arrivig to a cotradictio. From the assumptio we ow that x, y =a, b, so {x} {{a}, {a, b}} implyig that {x} = {a} or {x} = {a, b}. I either case we have a {x}, ad so a = x. By the same argumet we have {x, y} {{a}, {a, b}} ad give that x y it has to be {x, y} = {a, b}. Now x = a, x y ad y {a, b} implies y = b. Exercise 52 Exted the defiitio of a ordered pair to a fiite set of elemets. Call it -dimesioal vector or -tuple Cartesia Product Defiitio 53 The Cartesia product of two sets A ad B, writte A B, is the set of all ordered pairs a, b such that a A ad b B. That is A B = {a, b :a A, ad b B} Notatio We will write A 2 = A A Properties It is possible to derive some importat properties, which we eumerate below. 1. A B C D =A C B D. 2. A B C =A C B C. Exercise 54 Prove the two properties stated above.

7 Relatios Biary Relatios A biary relatio betwee two objects a ad b is a coditio ivolvig a ad b that is either true or false. For example, the relatio less tha is a relatio betwee positive itegers. Whe cosiderig a relatio betwee two objects, i some cases it is ecessary to ow which object comes first. For istace, if Mar is taller tha Paul is true the Paul is taller tha Mar is false. Thus, it is atural for the formal defiitio of a relatio to deped o the cocept of a ordered pair. Defiitio 55 LetAadBbesets. Arelatio betwee A ad Bis ay subset RofA B. We say that a A ad b B are related byrifa, b R, ad we deote this by sayig arb. If B = A, the we spea of a relatio R A A beig a relatio o A. Defiitio 56 Let R be a relatio betwee two sets A ad B. The domai of R, deoted DR, is the set of all first elemets of members of R. Formally, DR = {a A : b B a, b R}. The rage of R, deoted RR, is the set of all secod elemets of R. Formally, RR ={b B : a A a, b R}.Clearly from the defiitio we obtai DR A, ad RR B. Defiitio 57 A biary relatio R o a set A ca have the followig properties: Completeess: a, b A, arb bra. Reflexive property: a A, ara. Irreflexive property: a A, ara. Symmetric property: a, b A, arb bra. Atisymmetric property: a, b A ad a b, arb bra. Trasitive property: a, b, c A, arb brc arc. Example The relatio defied o S = {1, 2, 3, 4, 5} is complete, reflexive ad atisymmetric. 2. Let S be the set of all people who live i the US, ad suppose that two people x ad y are related by R if x lives withi a mile of y. The R is ot complete, is reflexive ad symmetric, but ot trasitive. Exercise 59 Let R be the relatio to be taller ad youger tha o the set S of all the Eco 897 studets. What properties does R have? We are specifically iterested i two types of relatios which come up frequetly i ecoomics ad mathematics, amely equivalece ad order relatios.

8 26 Elemetary Set Theory Equivalece relatios Defiitio 60 A relatio R o a set S is a equivalece relatio if it is reflexive, symmetric ad trasitive. A equivalece relatio is ofte deoted by = or. Note that completeess is ot a requiremet for equivalece. This type of relatio is sigled out because it possess the properties aturally associated with the idea of equality. Example 61 Let S be the set of all people who live i the US, ad suppose that two people x ad y are related by R if x has the same age as y, the R is reflexive, trasitive ad symmetric. Hece R is a equivalece relatio. Exercise 62 Let S be the set of all lies i a plae. Determie which of the followig are equivalece relatios xryif xisparalleltoy xryif xisperpedicular to y Give a equivalece relatio R o a set S, it is atural to group together all the elemets that are related to a particular elemet. More precisely Defiitio 63 Let R A 2 be a equivalece relatio o a set A, ad let x be a elemet of A. The equivalece class of x with respect to R is the set E x = {y A : yrx}. Exercise 64 Prove the followig statemets: 1 Let S be a equivalece relatio defied o a set A. The everytwo differet equivalece classes must be disjoit. 2 Let S be a equivalece relatio defied o a set A. The everyelemet of A belogs to at most oe equivalece class. Thus we see that a equivalece relatio R o a set S breas S ito disjoit subsets i a atural way. This operatio o a set is ow as a partitio Partitio Defiitio 65 Let A be a subset of X. A partitio oempty subsets of A such that of A is a collectio A of 1. Each x A belogs to some subset S A. 2. For all S, T A, if S T, the S T =.

9 Relatios 27 Example 66 Let S be the set of all studets i a particular uiversity. For x ad y i S, defie xry iff x ad y were bor i the same caledar year. The R is a equivalece relatio, ad a typical equivalece class is the set of all studets who were bor i the same year as studet z. Exercise 67 Cosider the previous example. If y E z, does this mea that y ad z are the same age?. Would it have bee the same to defie the equivalece relatio as the set of all studets who were bor i a particular year, say i 1980? Not oly does a equivalece relatio o a set S determie a partitio of S, but the partitio ca be used to determie the relatio. We formalize this as a theorem Theorem 68 Let R be a equivalece relatio o a set S. The A = {E x : x S} is a partitio of S. The relatio belogs to the same piece as is the same as R. Coversely, if P is a partitio of S, let R be defied by xry iff x ad y are i the same piece of the partitio. The R is a equivalece relatio ad the correspodig partitio ito equivalece classes is the same as P. Proof. Sice R is reflexive, each elemet x S belogs to some equivalece class E x A. Moreover, from a previous exercise, we ow that two equivalece classes must be either equal or disjoit. Hece all the differet equivalece classes have emptyitersectio, maig A a partitio o S. Now suppose that P is the relatio belogs to the same piece equivalece class as. The xp y x, y E z for some z S xr zad yrz for some z S xry Thus P ad R are the same. Coversely, suppose that P is a partitio of S ad let R be defied by xry iff x ad y are i the same piece of the partitio. Clearly, R is reflexive ad symmetric. To see that R is trasitive, suppose that xry ad yrz. We ow that y A for some A P, therefore x A, z A ad hece xrz. Fially, the equivalece classes of R correspod to the pieces of P because of the way R was defied Order relatios Defiitio 69 A biary relatio is called a quasi-order or preorder if it is reflexive ad trasitive. Example 70 Let S be the set of all studets i a particular uiversity. Let R be the relatio is at least as tall as ad as old as. The R is ot complete, but is reflexive ad trasitive. The it is a preorder.

10 28 Elemetary Set Theory Defiitio 71 A biary relatio is called a order if it is, reflexive, trasitive, ad atisymmetric. It is called a strict order if it is irreflexive, trasitive ad atisymmetric. Exercise 72 Is the preorder of the previous example a order? Justify. Defiitio 73 A order that is also complete is called a complete or total order. Notatio To stress that some orders may ot be complete, we sometimes call them partial orders. A order is ofte deoted by or. The otatio a<bidicates a strict order where a b ad a b. It is equivalet to write y >x istead of x <y. A example of partial order is set iclusio;for each set X, the relatio is a partial order i the power set PX. The reaso for qualifyig partial is that sometimes the order betwee two elemets is ot defied, i.e., there ca be M, N PX such that either M N or N M. Defiitio 74 A ordered set is a set S i which a order is defied. Agai, the order ca be either partial or complete. For the case of a complete ordered set,the followig statemet ca be show Exercise 75 Let S be a complete ordered set. The x, y S, oe ad oly oe of the statemets x<y, x = y, y < x is true. 2.4 The Set of Natural Numbers. Iductio Lets assume the existece of a set N cosistig of the positive itegers, also called atural umbers. Thus N = {1, 2, 3, 4...} This set ca be characterized from the followig set of Axioms, called the Peao axioms i hoor of the Italia mathematicia Giuseppe Peao, who developed this approach i the late 19th cetury. We suppose that there exist a set P whose elemets are called atural umbers ad a relatio o P called successor with the followig properties: P1. There exists a atural umber, deoted by 1, that is ot the successor of ay other atural umber. Note that ay complete relatio is also reflexive. This maes our defiitio of complete orderig somehow redudat, give that oe requiremet is implied by aother. Nevertheless this way of defiig a order is customary at least i ecoomicsad we decided to stic with the widely adopted defiitio.

11 The Set of Natural Numbers. Iductio 29 P2. Every atural umber has a uique successor. If m P, the we let m deote the successor of m. P3. Every atural umber except 1 is the successor of exactly oe atural umber. P4. If M is a set of atural umbers such that 1. 1 M ad 2. P, if M,the M,the M = P. Axioms P1 to P3 express the ituitive otio that 1 is the first atural umber ad that we ca progress through the atural umbers i successio oe at a time. Axiom P4 is the equivalet of what we will call later the priciple of mathematical iductio. Exercise 76 Cosider the relatio successor o N. What are its properties? Usig the previous axioms, we ca defie what additio ad multiplicatio meas. Basically D1. P, defie +1=. D2. Let, m P. If m = ad + is defied, the defie + m = + = +. D3. P, defie 1=. D4. Let, m P. If m = ad is defied, the defie m = = +. Note that i D2 ad D4 the existece of is assured by axiom P3. Based o the defiitio of sum, we ca ow itroduce a strict order < ad a order relatios o N Defiitio 77 Let N be the set of atural umbers. For x, y N we say that x<y iff N : x + = y.we also say that x y iff [x <y x = y] is true. Exercise 78 Verify that < is ideed a strict order o N.

12 30 Elemetary Set Theory The Well Orderig Property ad the Priciple of Mathematical Iductio There is a additioal property of N that expresses i a precise way the idea that each oempty subset of N must have a least first elemet. Theorem 79 well-orderig property of N IfS is a oempty subset of N, the there exists a elemet m S such that m for all S. Proof. Lets cosider two cases: if 1 S, ad if 1 / S. I the first case, we ow from axiom P1 that 1 is ot the successor of ay elemet of N. The for every elemet of S it must be 1. So m exists ad is 1. If 1 / S the proof goes by cotradictio. Suppose that there is o such least elemet m. We will costruct the set M = { N : / S {1, 2,..., 1,} S = }. This is the largest set of cosecutive atural umbers startig at 1 that do ot itersect with S. Clearly 1 M by assumptio. Now if l M, the l = l +1 must also belog to M. For if ot the l S ad the it would be the least elemet of S. Therefore from axiom P4 M = N, implyig that S =, cotradictig the assumptio that S is oempty. Oe importat tool to be used whe provig theorems about the atural umbers comes from axiom P4 ad is ow as the priciple of mathematical iductio. It eables us to coclude that a give statemet about atural umbers is true for all atural umbers without havig to verify it for each umber oe at a time. Theorem 80 Priciple of Mathematical Iductio Let P be a statemet that is either true or false for each N. The P is true for all N provided that 1. P1 is true, ad 2. For each N, ifp is true the P +1 is also true. Proof. The proof is trivial if we rely o axiom P4. Lets call S to the set S = { N : P is true} From the assumptios we ow that a 1 S ad b N, S +1 S. From axiom P4 we have that S = N, implyig that P holds for all. Exercise 81 Tae the well orderig property as a axiom ad prove the Priciple of Mathematical Iductio without relyig o axiom P4. Hit: the proof goes by cotradictio. Costruct a auxiliary set S = { N : P is false} ad use the well orderig property to guaratee the existece of a least elemet m S. It is customary to refer to the verificatio of part 1 of the theorem as the basis for iductio ad part 2 as the iductio step.the assumptio that P is true i verifyig part 2 is ow as the iductio hypothesis. It is essetial that both parts be verified to have a valid proof usig mathematical iductio.

13 The Set of Natural Numbers. Iductio 31 Exercise 82 Prove the followig theorem: Let m N ad let P be a statemet that is either true or false for each m. The P is true for all m provided that 1. Pm is true ad 2. For each N,if P is true the P +1 is true Applicatios A useful ad importat theorem is prove by iductio. Write x! for xx 1x! for x 1, set 0! = 1, ad write for the ratio for all x. x x! x! Observe that + 1 = +1 for further details o combiatory umbers ad their use, chec the appedix of this chapter. Theorem 83 The Biomial Theorem: a + b = =0 a b. Proof. The formula is trivially valid for =1. Basis of iductio Suppose the formula of the biomial theorem holds for iductive step. Show the that it also holds for +1. a + b +1 = a + ba + b =a + b = =0 = = = = =0 0 a +1b + a +1b a =0 { l=1 a +1 + l=1 =0 l l=1 a +1 b + =0 l 1 +1 l a b +1 l 1 a b by assumptio a l+1 b l } which is ideed the correspodig expressio for +1. chg of var l=+1 a l+1 b l + a l+1 b l b +1 b +1 Exercise 84 Prove or give a couterexample of the followig statemets 1. For every positive iteger, is a eve umber. 2. N, i=1 i = N, = i=1 i N, i=1 i3 = = 4 i=1 i2 5. Assumig x 1, 1+x + x x = 1 x+1 1 x, N

14 32 Elemetary Set Theory 6. N, >3 2! 7. N, > N, >3 2! 2.5 Appedix: Biomial Coefficiets Biomial coefficiets are used i may scieces for coutig evets ad computig probabilities, ad they costitute a useful mathematical tool i differet areas i Ecoomics. Defiitio 85 The factorial of a atural umber,!, is computed as! = The factorial of 0 is defied to be 0! = 1. Factorial umbers are useful for computig permutatios of distiguishable elemets. For example, suppose that you have to compute the umber of possible arragemets of 10 idividuals i a lie. For the first positio you have te possible choices. Oce the first positio is tae, you have 9 remaiig optios for the secod place. So for the first two positios you have 10 9 = 90 possibilities. For the third positio there are oly 8 possible guys to pic ad so o. So for a lie of 10 guys we have = 10! optios. A very used property of permutatios is that! = 1!. This is trivial to show by costructio from the defiitio ad is therefore left as a exercise. The ituitio is the same as before. To compute the possible rearragemets of elemets we have optios for the first place, ad for each of those choices we have all the permutatios of the remaiig 1 elemets. The permutatio of elemets is the basis of biomial coefficiets, also called combiatios. The combiatio of elemets, tae i groups of is deoted by, ad is computed as Defiitio 86 =!!!. The ratioale for this defiitio is the followig. Arrage the elemets i a lie. We will say that we obtai the same combiatio of elemets wheever the same elemets occupy the first positios i the lie. This is very importat, whe we cosider combiatios, the orderig of the elemets is immaterial. We already ow that the total umber of permutatios of the elemets is!. However, from all those! arragemets, may of them correspod to the same combiatio, meaig that the same objects occupy the first positios. If each combiatio is couted N times by computig!, we the ow that the true umber of combiatios will be!. N The remaiig poit is to fid N. To do this, just tae a give arragemet ad compute all the permutatios that give the same combiatio of elemets. We have that for a give positio of the last elemets, we ca rearrage the first

15 Appedix: Biomial Coefficiets 33 elemets i! ways; ad i a similar fashio, for a give positio of the first elemets, we ca rearrage the last elemets i! ways. Therefore the total umber of rearragemets that give the same combiatio is N =!!. This leads to the fial result that =!!!. Exercise 87 Computig the factorial of a umber may be cumbersome if is large try 1000! for example, but show that whe computig combiatorial umbers this difficulty may be overcome by calculatig = if is ot too large. Just to covice yourself, compute! Example 88 Suppose you pic four cards from the top of a 52 card pile. What is the chace of your gettig a poer of aces? The probability is computed as the umber of favorable evets divided by the total umber of evets. The latter is simply 52 = 52!. The former is simply 1, sice there is oly oe combiatio that comprises a 48! 4! 1 poer of aces. Therefore the chace is 52 4 = ,725 Our first use of biomial coefficiets will be to provide a guess for the biomial formula, that we have prove by iductio above. I geeral, this is a limitatio of the proofs by iductio. We have to come up with the result before the proof begis. To fid a guess of the biomial formula we ow that a + b = a + b a + b a + b... a + b }{{} times by the distributive law of multiplicatio, we ow that each term of the biomial sum is calculated as the multiplicatio of elemets, some of them beig a ad beig the rest b. It is therefore obvious that the biomial sum will have the form a + b = =0 α a b. The oly problem is to fid the α. But that is easy if we use combiatorial umbers. We ca thi of the term a b asalistof ordered elemets, with 4 a s ad b s. For example if =1, the we ow that there is oly oe a, either i positio 1,2, 3,..., 1, ad the rest are b s. I geeral, for a fixed particular all we have to do is to fid i how may differet ways ca we rearrage the a s ad b s. This is the same as computig i how may differet positios ca we fid a s ad b s, which is the umber of combiatios of the available positios tae i groups of. So α =, ad the biomial formula is computed as a + b = =0 a b

16 34 Elemetary Set Theory 2.6 Refereces S.R. Lay, Aalysis with a Itroductio to Proof. Chapter 2. Third Editio. Pretice Hall. A. Matozzi, Lecture Notes Eco 897 Uiversity of Pesylvaia Summer A. N. Mitchell ad C. J. Herget, Applied Algebra ad Fuctioal Aalysis. Dover Publicatios. H.L. Royde, Real Aalysis. Third Editio. Pretice Hall.