311 Chap 2

2.13 (Formal definition of ordered pair)

(Formal definition of ordered pair).LetAandBbe sets, leta∈A,b∈B. Then we define the ordered pair(a,b)by(a,b) ={{a},{a,b}}. The following proposition shows that (a,b) behaves as "a pair". Notethat (1,2) may also mean an open interval from 1 to 2. As some wordsin various languages have double or multiple meanings (such wordsare called "homonyms"), so do some mathematical notations. Themeaning of (a,b) will depend on the context or will be specified bywords if needed.

Pairs and Cartesian products.

As mentioned in Remark 2.3, allmathematical notions can be defined in terms of set theory. As anexample, let us take up challenge of defining a pair (a,b), like theone used to denote the coordinates of a point on a plane. Note that(a,b)not={a,b}since (a,b)not= (b,a) foranot=b.

2.15(Cartesian Product)

A×Bof setsAandBis defined byA×B={(a,b) :a∈Aandb∈B}Note: writing (a,b)∈A×Bimplies thata∈Aandb∈B.We defineA^1=A,andA^2,A^3,...recursively byA^(n+1)=A^(n)×A.Forsimplicity, we denote ((a,b),c) inA3=A2×Aby (a,b,c).HenceA^3isthe set of all triples of elements ofA.Similarly we definen-tuples aselements ofAn.Since every point in space can be uniquely identified (in a givensystem of coordinates) by itsx,y, andz−coordinates, which are realnumbers, the infinite 3-dimensional space is often denoted byR^3.

2.5 Disjoint

Sets A and B are disjoint if A∩B=∅.

2.7 Example

Every subsetBofA={1,2,3}is determined by asequence of Yes's and/or No's answering the following three questions:•Does1belong toB?•Does2belong toB?•Does3belong toB?For example,Y,N,Ycorresponds toB={1,3} ⊆AandN,N,Ncorresponds to∅⊆A. Since there are2×2×2 = 8possible sequenceslike that,|2^A|= 8.More generally, one can prove that|2^A|= 2^|A|.That fact motivatesthe term "power set" and the notation for it

2.9 Proposition (Distributive Rules)

IfA,B,Care sets, then(1)A∩(B∪C) = (A∩B)∪(A∩C)(2)A∪(B∩C) = (A∪B)∩(A∪C)To prove that two sets,XandYcoincide (i.e. are equal), one usuallyneeds to show separately thatX⊆YandY⊆X.Proof of (1):Proof ofA∩(B∪C)⊆(A∩B)∪(A∩C) :Letx∈A∩(B∪C). Thenx∈Aand, furthermore,x∈Borx∈C.Hence,x∈A∩Bin the first case andx∈A∩Cin the other. Therefore,x∈A∩B∪A∩Cfor all suchx.Proof ofA∩(B∪C)⊃(A∩B)∪(A∩C) : in class or HW.

2.8 Proposition (Basic properties of sets)

IfA,B,Care sets, then(1)∅∩A=∅;∅∪A=A(2)A∩B⊆A(3)A⊆A∪B(4)A∪B=B∪A;A∩B=B∩A("commutative" property of∪,∩)(5)A∪(B∪C) = (A∪B)∪C. (associative property of∪). Thesame for∩.(6)A∪A=A∩A=A(7)IfA⊆B, thenA∪C⊆B∪CandA∩C⊆B∩C("Propositions" are "little" theorems. We will comment more aboutthem later.)Proof.The proofs are very easy. Let us prove (2) for example: Letx∈A∩B.Thenx∈Aandx∈B. Hence, in particular,x∈A.Inother words, every element ofA∩Bbelongs toA.

2.12 Proposition (de Morgan's laws)

IfA,BandXare sets then(1)X−(A∪B) = (X−A)∩(X−B)(2)X−(A∩B) = (X−A)∪(X−B)The proofs of these properties can be visualized by use of Venn dia-grams.There is no well established comprehensive order of precedence inset theory. However,∩,∪and = correspond to∧,∨, and⇔.Hence,these set theory operations should be interpreted in the order writtenabove. For example, the first distributive rule above can be written asA∩(B∪C) =A∩B∪A∩C.According to this analogy between set theory and logic, the set sub-traction should precede∩, but that rule is not universally followed. Toavoid ambiguities, use parentheses instead.

2.11 Complement

IfXandAare sets, the complement,ofAinX, denoted byX−A,is defined byX−A={x∈X:x not a ∈A}X−Ais alternatively written asX\A. X−Ais also called theset differencebut that can be ambiguous as the difference of sets mayrefer toX−Aor toA−X.

2.1 Subset

LetAandBbe sets. We say thatAiscontained inB, orAis a subsetofB, writtenA⊆B, if wheneverx∈A, thenx∈B. As an alternative, we sometimes say thatBcontainsA, writtenB⊇A.Note that∅⊆Afor all setsA.

2.14 Proposition 2.14 (Fundamental property of ordered pairs)(Fundamental property of ordered pairs)

Letaandcbe inA; andbanddbe inB. Then(a,b) = (c,d)if and only ifa=candb=d.Proof.Since the⇐implication is obvious, we only need to prove⇒.Leta,c∈Aandb,d∈Bbe such that (a,b) = (c,d). Observe that(a,b) is a set of 1 or 2 elements depending on whetheraequalsbornot. Indeed, ifa=bthen(1)(a,b) ={{a},{a,b}}={{a},{a}}={{a}}and ifanot=bthen (a,b) contains two different elements. (Why?)For the proof, assume first thata=b. Then (a,b) = (c,d) is a 1element set and, hence,c=d.Furthermore, by (1),{a}={c}. Hencea=cand the implication follows.Ifanot=bthen (a,b) contains 2 elements: a 1 element set,{a},anda 2 element set,{a,b}.By assumption (a,b) = (c,d). Therefore (c,d)is also a 2 element set, i.e.{c,d} not={c}. Since a one element setcannot equal a two element set (recall that two sets are equal if theyhave the same elements), our assumptions imply that{a}={c}and{a,b}={c,d}.Hencea=candb=d.

2.3 Remark

Remark 2.3.(1) These axioms require that a set cannot be its own element. Hence,wild sets are "illegal." In particular, one cannot call the collection ofall sets "a set"! (As you can see, we used an informal word "collection"to refer to all sets).(2) All mathematical knowledge of human civilization can be derivedfrom these axioms.(3) In particular all mathematical notions (except sets) can be rigor-ously defined on the basis of set theory.(4) We do not know if these axioms are consistent (i.e. do not lead toa contradiction, analogous to Russell's paradox). However, most math-ematicians believe that they are consistent, since no contradiction hasbeen found in the last 100 years.(5) There are limits to what can be proved from any consistent set ofaxioms. In particular we will see that the validity of some statementscannot be either proved or disproved- c.f. Section 16.

2.6 Power set

The power setof a setA, denoted2^AorP(A), is the set of all subsets ofA. That is,2^A={X:X⊆A}.

2.4 Union and intersection

The unionA∪Bof two setsAandBis defined byA∪B={x:x∈Aorx∈B}. Recall that 'or'in mathematics means 'and/or.' The intersectionA∩Bis defined byA∩B={x:x∈Aandx∈B}.Sometimes infinite unions are useful. For example, the set of all realnumbersxsuch that sin(x)>0 can be written as{x∈R: sin(x)>0}or, more explicitly, as⋃k∈Z(2πk,2πk+π),meaning a union of intervals (2πk,2πk+π) for all possible integralvalues ofk.Here and throughout the notes, the symbolsRandZdenote the setof all real numbers and the set of all integers (whole numbers). Notethat the symbolsRandZmay be used to denote various other thingsat various times.

2.2 Equality of Sets

We say thatAequalsB, writtenA=B, ifA⊆BandA⊇B.Hence{apple, horse, 2}={horse, 2, apple}={horse, 2, apple, horse}.Sets can be elements of other sets, e.g.A={{3,5,7},1}is a set of twoelements, one of which is a set itself. IfB={1,3,5,7}, thenAnot=B,sinceBhas 4 elements. Also,Anot={{1,3,5},7},since 7 is not an element ofA.We say tha tA is a proper subset of BifA⊆BbutAnot=B. We denote this byA⊆notequaltoB. It is important to distinguish between the symbols∈and⊂.Thefollowing are true:∅⊆{1,2}, 1∈{1,2},∅∈{∅,5}.The following areFALSE:∅∈{1,2}, 1⊆{1,2}.IfAis a set then we often define its subsets by writing{x∈A:xsatisfies some condition}, e.g.{x∈R:x^(2)<4}is the set of all real numbersxsuch thatx^(2)<4.The number of elements of a setAis called itscardinalityand it isdenoted by|A|.For example,|{3,{3}}|= 2,|R|=∞.

2.10 Remark

You cannot prove Propositions 2.8 and 2.9 "by exam-ple", because the claim is that the statements hold for allsetsA,B,C.For example, verifyingA∩(B∪C) = (A∩B)∪(A∩C)forA={a,b,c}, B={d,e}, C={a,d}does not constitute a proof of Proposition 2.9(1).However, "there exists" statements can be proved by example. Forexample, consider the problem:Prove that there exist setsA,Bsuch thatA∪B6=A∩B.Proof: TakeA={a,b}, B=∅. ThenA∪B={a,b}andA∩B=∅are different.