# Section 2.1

## 1.1

Find the union ${\displaystyle A_{1}\cup A_{2}}$ and the intersection ${\displaystyle A_{1}\cap A_{2}}$ of the two sets ${\displaystyle A_{1}}$ and ${\displaystyle A_{2}}$ where

(a) ${\displaystyle A_{1}=\{0,1,2\},A_{2}=\{2,3,4\}\,}$

(b) ${\displaystyle A_{1}=\{x:0

(c) ${\displaystyle A_{1}=\{(x,y):0

Solution (c):

${\displaystyle A_{1}\cup A_{2}=\{(x,y):0

${\displaystyle A_{1}\cap A_{2}=\{(x,y):1

Rest are similar.

## 1.3

List all the possible arrangements of the four letters m,a,r and y. Let ${\displaystyle A_{1}}$ be the collection of the arrangements in which y is in the last position. Let ${\displaystyle A_{2}}$ be the collection of the arrangements in which m is in the first position. Find the union and intersection of ${\displaystyle A_{1}}$ and ${\displaystyle A_{2}}$.

Solution: The arrangements are: mary, mayr, mray, mrya, myra, myar, amry, amyr, aymr, ayrm, arym, army, ryma, ryam, ramy, raym, rmya, rmay, yrma, yram, yamr, yarm, ymra, ymar.

${\displaystyle A_{1}=\{mary,mray,amry,army,ramy,rmay\}}$

${\displaystyle A_{2}=\{mary,mayr,mray,mrya,myra,myar\}}$. The rest is obvious.

## 1.5

If a sequence of sets ${\displaystyle A_{1},A_{2},A_{3}\cdots }$ is such that ${\displaystyle A_{k}\subset A_{k+1},k=1,2,3,\cdots }$ the sequence is said to be a nondecreasing sequence. Give an example of this kind of sequence of sets.

Solution: Let ${\displaystyle A_{n}=(0,n)}$ where n = 1, 2, 3...

## 1.6

If a sequence of sets ${\displaystyle A_{1},A_{2},A_{3}\cdots }$ is such that ${\displaystyle A_{k}\supset A_{k+1},k=1,2,3,\cdots }$ the sequence is said to be a nonincreasing sequence. Give an example of this kind of sequence of sets.

Solution: Let ${\displaystyle A_{n}=(0,{\frac {1}{n}})}$ where n=1,2,3...

## 1.7

If ${\displaystyle A_{1},A_{2},A_{3}\cdots }$ are sets such that ${\displaystyle A_{k}\subset A_{k+1}k=1,2,3,\cdots ,\lim _{k\to \infty }A_{k}}$ is defined as the union ${\displaystyle A_{1}\cup A_{2}\cup A_{3}\cup \cdots }$. Find ${\displaystyle \lim _{k\to \infty }A_{k}}$ if

(a) ${\displaystyle A_{k}=\{x:1/k\leq x\leq 3-1/k\},k=1,2,3\cdots }$

(b) ${\displaystyle A_{k}=\{(x,y):1/k\leq x^{2}+y^{2}\leq 4-1/k\},k=1,2,3\cdots }$

Solution: (b) ${\displaystyle \lim _{k\to \infty }A_{k}=\{(x,y):1

## 1.8

If ${\displaystyle A_{1},A_{2},A_{3}\cdots }$ are sets such that ${\displaystyle A_{k}\supset A_{k+1}k=1,2,3,\cdots ,\lim _{k\to \infty }A_{k}}$ is defined as the intersection ${\displaystyle A_{1}\cap A_{2}\cap A_{3}\cup \cdots }$. Find ${\displaystyle \lim _{k\to \infty }A_{k}}$ if

(a) ${\displaystyle A_{k}=\{x:2-1/k

(b) ${\displaystyle A_{k}=\{x:2

(c) ${\displaystyle A_{k}=\{(x,y):0\leq x^{2}+y^{2}\leq 1/k\},k=1,2,3\cdots }$

Solution: (c) ${\displaystyle \lim _{k\to \infty }A_{k}=\{(0,0)\}}$

## 1.9

For every one dimensional set A let ${\displaystyle Q(A)=\sum _{A}f(x)}$ where ${\displaystyle f(x)={\begin{cases}{\frac {2}{3}}({\frac {1}{3}})^{x},&x=0,1,2,\cdots \\0,&otherwise\end{cases}}}$. If ${\displaystyle A_{1}=\{x:x=0,1,2,3\}\,}$ and ${\displaystyle A_{2}=\{x:x=0,1,2,\cdots \}}$, find ${\displaystyle Q(A_{1})}$ and ${\displaystyle Q(A_{2})}$.

Solution: ${\displaystyle Q(A_{1})={\frac {80}{81}}}$ and ${\displaystyle Q(A_{2})=1}$ using the formulae of sum of a geometric series.

## 1.10

For every one dimensional set A for which the integral exists, let ${\displaystyle Q(A)=\int _{A}f(x)}$ where ${\displaystyle f(x)={\begin{cases}6x(1-x),&0, otherwise let ${\displaystyle Q(A)}$ be undefined. If ${\displaystyle A_{1}=\{x:{\frac {1}{4}} and ${\displaystyle A_{3}=\{x:0, find ${\displaystyle Q(A_{1}),Q(A_{2})}$ and ${\displaystyle Q(A_{3})}$.

Solution: ${\displaystyle Q(A_{1})}$ can be found by integrating f(x) from ${\displaystyle {\frac {1}{4}}}$ to ${\displaystyle {\frac {3}{4}}}$, ${\displaystyle Q(A_{2})=0}$ as integral over a set of measure zero is 0 and ${\displaystyle Q(A_{3})}$ can be found by integrating f(x) from 0 to 1.

## 1.11

Let ${\displaystyle Q(A)=\int _{A}\int (x^{2}+y^{2})dxdy}$ for every two dimensional set A for which the integral exists; otherwise let ${\displaystyle Q(A)}$ be undefined. If ${\displaystyle A_{1}=\{(x,y):-1\leq x\leq 1,-1\leq y\leq 1\},A_{2}=\{(x,y):-1\leq x=y\leq 1\},}$ and ${\displaystyle A_{3}=\{(x,y):x^{2}+y^{2}\leq 1\}}$, find ${\displaystyle Q(A_{1}),Q(A_{2})}$ and ${\displaystyle Q(A_{3})}$.

Solution: ${\displaystyle Q(A_{1})=\int _{-1}^{1}\int _{-1}^{1}(x^{2}+y^{2})dxdy={\frac {8}{3}}}$, ${\displaystyle Q(A_{2})}$ is zero since it represents volume of a sheet and ${\displaystyle Q(A_{3})=\int _{0}^{2\pi }\int _{0}^{1}(r^{2})rdrd\theta }$ if we introduce polar coordinates. This evaluates to ${\displaystyle {\frac {\pi }{2}}}$.

## 1.15

To join a certain club, a person must be either a statistician or a mathematician or both. Of the 25 members in this club, 19 are statisticians and 16 are mathematicians. How many persons in the club are both a statistician and a mathematician?

Soluion: The number of persons which are statisticians or mathematicians = number of statisticians + number of mathematicians - number of persons which are both statisticians and mathematicians. This can be proved directly using properties of the counting measure. The general proof is as follows: Let A and B be two finite sets. Endow the natural numbers with the counting measure which we shall denote by n. We shall denote union by + and intersection by . and the context will make it clear whether we mean actual addition, multiplication or union, intersection. Now A-B (Here - represents the relative complement, AB and B-A are disjoint and as n is sigma additive so

1. n(A+B)=n(A-B)+n(B-A)+n(AB)
2. n(A)=n(A-B)+n(AB) which implies n(A-B)=n(A)-n(AB)
3. n(B)=n(B-A)+n(AB) which implies n(B-A)=n(B)-n(AB)

Substituting n(A-B) and n(B-A) from (2) and (3) into (1) gives us n(A+B)=n(A)+n(B)-n(AB). So the answer is 10.

## 1.16

After a hard fought football game, it was reported that of the 11 starting players, 8 hurt a hip, 6 hurt an arm, 5 hurt a knee, 3 hurt both a hip and an arm, 2 hurt both a hip and a knee, 1 hurt both an arm and a knee, and no one hurt all three. Comment on the accuracy of the report.

Solution: We shall first establish that in the language of the previous problem:

${\displaystyle n(A+B+C)=n(A)+n(B)+n(C)-n(AB)-n(BC)-n(CA)+n(ABC)\,}$.

Now,

${\displaystyle n(A+B+C)=n(A+B)+n(C)-n((A+B)C)\,}$ ${\displaystyle =n(A)+n(B)+n(C)-n(AB)-n(AC+BC)\,}$ ${\displaystyle =n(A)+n(B)+n(C)-n(AB)-{n(AC)+n(BC)-n(ACBC)}\,}$ ${\displaystyle =n(A)+n(B)+n(C)-n(AB)-n(BC)-n(CA)+n(ABC)\,}$

Now if we let A, B and C to be the set of players with injured hips, knees and arms respectively, we would find the contradiction on substituting the appropriate terms in the above formula. Hence the report is not accurate.