Set and basic set operations
Set
Asetis a collection of elements. LetX be a set. Ifx is an element inX thenx∈X, otherwisex∈/X. A set with no element is called anempty setand denoted by∅.
A⊆B (Ais asubsetofB): All elements ofAare member ofB. Note that∅ ⊆Afor any setA.
A⊂B (Aisstrict subsetofB): A⊆B and there is at least onex such thatx∈B butx∈/A.
A=B (AisequaltoB): IfA⊆BandB⊆AthenA=B. That is, all elements inA andB are the same.
Ac (ComplementofA): Suppose all sets under consideration are subset of a fixed set U(which can be context specific). Ac={x∈U|x∈/A}.
Basic set operations
UnionofAandB:A∪B={x|x∈Aorx∈B} IntersectionofAandB:A∩B={x|x∈Aandx∈B}.
Aset minusB:A−B={x|x∈Aandx∈/B}.
Useful results
1. Distributive laws: (i)A∩(B∪C) = (A∩B)∪(A∩C), (ii) A∪(B∩C) = (A∪B)∩(A∪C)
2. De Morgan’s laws: (i) (A∪B)c=Ac∩Bc, (ii) (A∩B)c=Ac∪Bc
Set and basic set operations
Proof of Distributive law(1-i)
We want to showA∩(B∪C) = (A∩B)∪(A∩C). We prove this by showing A∩(B∪C)⊆(A∩B)∪(A∩C) and (A∩B)∪(A∩C)⊆A∩(B∪C).
First, lets show thatLHS⊆RHS. Take anyx∈A∩(B∪C). This meansx∈Aand x∈B∪C. Moreoverx∈B∪C means either (i)x∈B, or (ii)x∈C.
If (i) is the case, thenx∈Bandx∈Atogether means thatx∈A∩B. Hence x∈(A∩B)∪(A∩C) because (A∩B)∪(A∩C) is a superset of (A∪B).
On the other hand, if (ii) is the case, thenx∈Candx∈Ameans thatx∈A∩C.
Hence, once again,x∈(A∩B)∪(A∩C) because (A∩B)∪(A∩C) is a superset of (A∩C).
Thus no matter which of case (i) or (ii) is true, we getx∈(A∩B)∪(A∩C). Since we started from an arbitrary memberxofA∩(B∪C), we can claim that each member ofA∩(B∪C) is also member of (A∩B)∪(A∩C). Hence
A∩(B∪C)⊆(A∩B)∪(A∩C).
Next, lets show thatRHS⊆LHS. Take anyy∈(A∩B)∪(A∩C). This means either, (i)y∈(A∩B) or (ii)y∈(A∩C). We shall show that in both casey∈A∩(B∪C).
If (i) is true theny∈(A∩B). It meansy∈Aandy∈B. Buty∈B makes y∈B∪C, becauseB∪C is a superset ofB. Thus we have,y∈Aandy∈B∪C, together which meansy∈A∩(B∪C).
We can follow the same steps for case (ii). Herey∈(A∩C), meansy∈Aand y∈C. Buty∈C makesy∈B∪C, becauseB∪Cis a superset ofC. Once again y∈Aandy∈B∪Ctogether meansy∈A∩(B∪C).
Irrespective of case (i) or (ii), we obtain thaty is a member ofA∩(B∪C). Thus (A∩B)∪(A∩C)⊆A∩(B∪C).
Logic
Proposition
Apropositionis a statement which is either true or false. LetPbe a proposition. It can have two ‘truth values’: True, False.
∼PisnegationofP.∼Pis a statement that is ‘opposite’ toP. ∼Ptakes truth value True whenPis False and∼Ptakes truth value False whenP is True.
For any 2 propositionsPandQ, there are 4 possible assignments of truth values.
(1)Pis True andQ is True (2)Pis True andQ is False.
(3)Pis False andQis True (4)Pis False andQis False.
We can combine 2 propositions.
(P &Q) is a new proposition. This combined proposition means bothP andQ. It is true in case 1 and false in cases 2, 3 and 4.
(P orQ) is another proposition. This combined proposition means eitherP orQ or both. So it is true in cases 1, 2, 3 and false in case of 4.
Proposition with quantifiers
Most of the time we would deal with Proposition given an underlying ‘state’. LetAbe the set of all relevant states. Instead of checking justP, we shall checkP(x) that is validity ofPatx.
For all: Suppose a propositionP is true at all statesx∈A. Our new proposition with quantifier is,P(x) is True∀x.
There exists: If there is at least onex∈Asuch thatP is True at statex. Our new proposition with quantifier is,∃xsuch thatP(x) is True (there exists).
Logic
Negation of proposition with quantifiers
Negation of [P(x) is True∀x]: ∃xsuch thatP(x) is False Negation of [∃xsuch thatP(x) is True]:P(x) is False∀x Logical implication
PimpliesQ (denoted byP⇒Q) : This itself is a Proposition. We check validity of propositionsP(x) andQ(x) for allx∈A. The PropositionP⇒Qis True if for all x∈Aeither of the following three hold:
(1)P(x) is True andQ(x) is True, (2)P(x) is False andQ(x) is True (4)P(x) is False andQ(x) is False.
The PropositionP⇒Q is False if there exists at least onex∈Asuch thatP(x) is True andQ(x) is False.
Note that the set{x|P(x) is True}is a subset of{x|Q(x) is True}. We also say thatQis necessary forP. OrP is sufficient forQ.
Q⇒Pis called the converse of the statementP⇒Q. There is no logical connection between these two statements. However if both are True, then we denote that by P⇔Q.
Useful observation
[P⇒Q] and [∼Q⇒∼P] are the same statement Reading: SB, Appendix A1
Function, Inverse function and Composition of function
Definitions of function and related concepts
Take two setsAandB. Afunctionf is a rule that maps each element ofAto exactly one element inB.
That isf(a)∈B∀a∈A. We denote this byf :A−→B. Ais called thedomainoff.
Note 1: More than one elements ofAcan be mapped to the same element inB. Note 2: Some members ofB may not be reached byf.
Take anyA0⊆A. IfA0is the restricted domain, thenimage/RangeA0underf is the setf(A0) ={b∈B|f(a) =b,a∈A0}.
Example: Indifference curve Definitions of inverse of a function
Take a functionf :A−→B. Inverse off, denoted byf−1, is the route back fromB toA.
For eachb∈f(A),f−1(b) ={a∈A|f(a) =b}.
Note thatf−1is not necessarily a function. (QN: Why?)
Take anyB0⊆B.PreimageofB0underf isf−1(B0) ={a∈A|f(a) =b,b∈B0}.
Composition of functions
We cancomposetwo functions to get a new function in the following manner.
Supposef :A−→Bandg:B−→C. Theng◦f is the composite function which maps each element ofAto exactly one element inC, through a temporary stop atB.
g◦f(a) =g(f(a)) ∀a∈A.
Set operations on image and preimage
Results: Set operations on preimage 1.B1⊆B2⊆B=⇒f−1(B1)⊆f−1(B2) 2.B1,B2⊆B=⇒
(i)f−1(B1)∪f−1(B2) =f−1(B1∪B2) (ii)f−1(B1)∩f−1(B2) =f−1(B1∩B2) (iii)f−1(B1)−f−1(B2) =f−1(B1−B2) Are these results the same for image?
3.A1⊆A2⊆A=⇒f(A1)⊆f(A2) 4.A1,A2⊆A=⇒
(i)f(A1)∪f(A2) =f(A1∪A2) (ii)f(A1)∩f(A2)⊇f(A1∩A2) (iii)f(A1)−f(A2)⊆f(A1−A2)
Type of function
Definitions: Types of function
f :A−→Bisinjective (one-to-one)iff(a) =f(a0) =⇒a=a0.
f :A−→Bissurjective (onto)if for everyb∈B there existsa∈As.t. b=f(a).
A function that is both one-to-one and onto is calledbijective.
Useful results 5.Iff is bijective then
(i)f−1is a function fromBtoA (ii)f−1is bijective.
(iii)f−1(f(a)) =a∀a∈Aandf(f−1(b)) =b∀b∈B.
6.Iff is one-to-one then 4−(ii) and 4−(iii) hold with equality.
7.Iff is one-to-one thenf is a bijective function fromAtof(A).
Definition: Cardinality of set
Cardinality is total number of elements in a set. Two setsAandB have the same cardinality, denoted by|A|=|B|, if there exists a bijective function fromAtoB.
Reading: SB 13.5
Vector
Definitions of Cartesian product
Take two setsAandB. The Cartesian product ofAandB is, A×B={(x,y)|x∈A,y∈B}.
A typical element ofA×Bis an ordered pair (x,y) wherexis fromAandy fromB.
Similarly we can define Cartesian product ofnsets,A1×A2×. . .×An. A special case:A1=A2. . .=An, is often denoted byAn.
We shall focus onRn, whereRis the set of all real numbers.
A vector inRn:x= (x1,x2, . . . ,xn) wherexk∈ R ∀n.
Some special vectors:0= (0, . . . ,0),ej= (0,0, . . . , 1
|{z}
j th position
, . . . ,0)
Basic operations on vector
Sum:x+y= (x1,x2, . . . ,xn) + (y1,y2, . . . ,yn) = (x1+y1,x2+y2, . . . ,xn+yn)
−x= (−x1,−x2, . . . ,−xn). Hence,
Difference:x−y= (x1,x2, . . . ,xn)−(y1,y2, . . . ,yn) = (x1−y1,x2−y2, . . . ,xn−yn) Scalar multiplication: Letα∈ R,αx= (αx1, αx2, . . . , αxn)
Useful observations α, α1, α2∈ Randx,y∈ Rn 1.x+y=y+x
2. (α1+α2)x=α1x+α2x 3.α(x+y) =αx+αy
Inner product and Norm
Definitions of Inner product of two vectors
This operation assigns a number to each pair of vectors. This is also connected to angle between two vectors (we will skip this geometric interpretation for now).
x·y=x1y1+x2y2. . .xnyn=Pn i=1xiyi
Useful observations:α∈ Randx,y,z∈ Rn 1.x·y=y·x
2.x·(y+z) =x·y+x·z 3.x·(αy) =α(x·y) Definitions of Norm
Norm is the length of a vector. It is the distance between0andx.
||x||=p (x·x) =
q Pn
i=1xi2
Useful observations:α∈ Randx∈ Rn 1.||x|| ≥0 and||x||= 0⇔x= 0 2.||αx||=|α| ||x||
Two important inequalities:x,y∈ Rn
1. Cauchy-Scwarz inequality: (i)|x·y| ≤ ||x|| ||y||, (ii) equality hold iffx=αy for someα∈ R
2. Triangle inequality: (i)||x+y|| ≤ ||x||+||y||, (ii) equality hold iffx=αy for someα >0
Proof of Cauchy-Scwarz and Triangle inequalities
Proof of Cauchy-Scwarz:
(i) Definez=x+ty. We know thatz·z≥0 and equality hold ifz= 0
z·z= (x+ty)·(x+ty) =x·x+t2(y·y) + 2t(x·y) =||x||2+t2||y||2+ 2t(x·y)≥0 Chooset=− x·y
||y||2 and rearrange the above inequality to obtain Cauchy-Scwarz inequality.
Qn: Why do use this particulart?
(ii) Equality holds⇒x+ty= 0,⇒x=−ty.
Ifx=αy, check that Cauchy-Scwarz holds with equality.
Proof of Triangle inequality:
(i)||x+y||2= (x+y)·(x+y) =||x||2+||y||2+ 2(x·y)≤ ||x||2+||y||2+ 2(||x|| ||y||) The last inequality follows from Cauchy-Scwarz
Taking square-root, we obtain Triangle inequality.
(ii) Equality holds⇒Cauchy-Scwarz hold with equality⇒x=αy. Ifx=αy, check that Triangle inequality holds with equality.
Reading: SB, Chapter 10
