Ex 1.1.Collection of lines with constant slope and changing intercepts or, vice-versa does the job.

Set Theory Ex 1.1.Collection of lines with constant slope and changing intercepts (or, vice-versa) does the job.That is, F=fLrjr2Rg; where,Lr=f(x; y)2R 2 jx+y=rgare lines with slope1 and interceptr. Other similar sets can be constructed which can be indexed byR.Ex 1.2.A few subsethood relations are NZQR NQ + R + R NQ

R

R Similar relations can be obtained by comparing elements.Ex 1.3.The results forNare given. Other results can be constructed in a similar manner.N[Z=Z N\Z=N N[Q=Q N\Q=N N[Q + =Q + N\Q + =N N[Q

=fx2Qjx <0 orx2Ng N\Q

=; N[Q

=Q

N\Q

=N N[R + =R + N\R + =N N[R

=fx2Rjx <0 orx2Ng N\R

=; N[R

=R

N\R

=N A few observations that can be made are as follows. The reader can make as many observations as they want to.

1.ABimpliesA[B=BandA\B=A.

• Some of the (newly formed) sets are merely a result of union/intersection of known sets.

1 Solutions Manual for An Introduction to Metric Spaces, 1e Dhananjay Gopal, Aniruddha Deshmukh (All Chapters) Chapter

• Stuvia.com - The Marketplace to Buy and Sell your Study Material

Downloaded by: tutorsection | [email protected]

Distribution of this document is illegal Want to earn $1.236 extra per year? 1 / 3

Stuvia.com - The Marketplace to Buy and Sell your Study Material2CHAPTER 1. SET THEORY Ex 1.4.IfFis a pair-wise disjoint family indexed by , then T 2 A6=;would imply the existence of some elementxin all ofA's contradicting the pair-wise disjointness. Thus, any pair-wise disjoint family is disjoint.The familyff1;2g;f2;3g;f1;3ggis disjoint but not pair-wise disjoint, sincef1;2g \ f2;3g=f2g 6=;.Ex 1.5.Example in Solution to exercise 1.4 does the job.Ex 1.6.Prove the two-way subsethood for each of the equalities using the basic principles of logic and properties of conjunction and disjunction operators.Ex 1.7.Prove the two-way subsethood for each of the equalities using the basic principles of logic and properties of conjunction and disjunction operators.Ex 1.8.P(;) =f;g.Ex 1.9.The three sets cannot be equal since the elements in all three sets are dierent.R 3 contains 3-tuples, while (RR)RandR(RR) contains 2-tuple. Also, the rst entry in a 2-tuple of (RR)Ris again a 2-tuple, while for elements inR(RR), the second entry is a 2-tuple.Ex 1.10.Q 2;

A=;, because if not so, the choice functionf:; !

S 2; Awill be such thatf()2A.This is possible only when we have such's. Many also argue that since;does not have any elements, this is vacously true. However, even in such an argument, the empty product is not dened. If there is a diculty to the reader in understanding this, the reader may take it as a convention.Ex 1.11.(x1; x2; ; xn)L(y1; y2; ; yn) if and only if (x16=y1)x1R1y1) and (x1=y1andx26=y2)x2R2y2) and and (x1=y1; x2=y2; ; xn1=yn1)xnRnyn) Ex 1.12.

: Refexive, Anti-symmetric, Transitive

: Refexive, Anti-symmetric, Transitive

= : Refexive, Symmetric, Anti-symmetric, Transitive

j: Refexive, Anti-symmetric, Transitive

: Refexive, Symmetric, Transitive

Ex 1.13.Generalising modulo relation onZforn2N, we havem1m2(modn) if and only if njm1m2.The equivalence classes for this relation aref[0];[1];[2]; ;[n1]g, and an integermis in an equiv- alence classjif and only if the remainder after dividingmbynisr.Ex 1.14.The equivalence classes for = relation onXfor eachx2Xare [x] =fxg.The equivalence classes for(modulo relation) are as in solution to exercise 1.13.Ex 1.15.In the setX=ff1g;f2g;f3g;f1;2g;f2;3g;f1;3ggequipped withrelation, there is no least element.Ex 1.16.In the example of solution to exercise 1.15, the setsf1g;f2g;f3gare minimal elements and the setsf1;2g;f2;3g;f1;3gare maximal elements..Ex 1.17.In general, it is not true. ForX=f1;2;3g, the POSET (P(X);) has a least element, namely;, but the subsetff1g;f2g;f3ggofP(X) has no least element. Stuvia.com - The Marketplace to Buy and Sell your Study Material

Downloaded by: tutorsection | [email protected]

Distribution of this document is illegal Want to earn $1.236 extra per year? 2 / 3

Stuvia.com - The Marketplace to Buy and Sell your Study Material3 Ex 1.18.(x; y)2((0;0);(1;1)) if and only if (0;0)(x; y), (x; y)(1;1), (x; y)6= (0;0) and (x; y)6= (1;1).(0;0)(x; y) if and only if 0x 2 +y 2 . Also, (x; y)6= (0;0) impliesx 2 +y 2 >0. This covers all ofR 2 except for (0;0).(x; y)(1;1) if and only ifx 2 =y 2

• and ifx

2 +y 2 = 2, then tan 1

y x

tan 1

• =

4 . This covers everything (strictly) inside the circle centered at (0;0) of radius p

• and the semi-circlebelowthe line

y=x.The required interval will be the intersection of the two regions obtained above.Ex 1.19.f(0) is not dened, and thereforefis not a function.Ex 1.20.1.f(x) =e x .

2.f(x) = (x1) (x2).

3.f(x) =x 2 .

4.f(x) =x.Ex 1.21.This is not possible since if the domain is non-empty, there is atleast one image making the codomain non-empty.Ex 1.22.f:f1;2g ! f1;2;3gandg:f1;2;3g ! f1;2gdened asf(1) = 1,f(2) = 2,g(1) = 1, g(2) =g(3) = 2. These two functions are as required in all three parts.Ex 1.23.ForX6=Y, if we have functionsf:X!Yandg:Y!X, then the two functions fg:Y!Yandgf:X!Xare well-dened but cannot be equal since the domains are not equal.

Ex 1.24.f:f1;2;3g ! f1;2gdened asf(1) = 1,f(2) =f(3) = 2.

f:f1;2g ! f1;2;3gdened asf(1) = 1,f(2) = 2.

Ex 1.25.SinceidXandidYare bijective, bothgfandfgare bijective. This implies bothfand gare bijective.We also havef 1 =f 1 idY=f 1 g(fg) = (f 1 f)g=idXg=g.Finally, (f 1 ) 1 f 1 =idYandf 1 (f 1 ) 1 =idX. Therefore,f= (f 1 ) 1 .Ex 1.26.Functions in solution 1.24 can be used in appropriate order to obtain required results.Ex 1.27.Ifa=bandc=d, both the intervals are singleton sets so that there is a bijection between them.Ifa=bandc6=d, then [a; b] is a singleton set while [c; d] contains at least two points (candd), so that any function from [a; b] to [c; d] is not onto and therefore not a bijection.Ifa6=bandc=d, using similar arguments, any function is not one-to-one and therefore not a bijection.Ex 1.28.f(x) =

dc ba

x+c using proper composition of scaling and translation of intervals.Ex 1.29.The required bijection is a composition of bijections between the intervals (a; b);(0;1); (0;1);[0;1]; and [0;1]![c; d] as mentioned in the text and exercises.Ifa=b, then (a; b) =;so that no bijection is possible.Ifa6=bandc=d, then [c; d] is a singleton set while (a; b) contains at least two distinct points, namely a+b 2 and a+b 4 so that again, no bijection is possible.

Ex 1.30.Forf:R!Rdened asf(x) =x

2 , letA1= [0;2] andA2= [2;2] so thatA1A2but f(A1) =f(A2) = [0;4]. Stuvia.com - The Marketplace to Buy and Sell your Study Material

Downloaded by: tutorsection | [email protected]

Distribution of this document is illegal Want to earn $1.236 extra per year?

• / 3