# Section 2.1Edit

## 1.1Edit

Find the union $A_1\cup A_2$ and the intersection $A_1\cap A_2$ of the two sets $A_1$ and $A_2$ where

(a) $A_1=\{0,1,2\},A_2=\{2,3,4\} \,$

(b) $A_1=\{x:0

(c) $A_1=\{(x,y):0

Solution (c):

$A_1\cup A_2=\{(x,y):0

$A_1\cap A_2=\{(x,y):1

Rest are similar.

## 1.3Edit

List all the possible arrangements of the four letters m,a,r and y. Let $A_1$ be the collection of the arrangements in which y is in the last position. Let $A_2$ be the collection of the arrangements in which m is in the first position. Find the union and intersection of $A_1$ and $A_2$.

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.

$A_1=\{mary,mray,amry,army,ramy,rmay\}$

$A_2=\{mary, mayr, mray, mrya, myra, myar\}$. The rest is obvious.

## 1.5Edit

If a sequence of sets $A_1,A_2,A_3\cdots$ is such that $A_k\subset A_{k+1}, k=1,2,3,\cdots$ the sequence is said to be a nondecreasing sequence. Give an example of this kind of sequence of sets.

Solution: Let $A_n=(0,n)$ where n = 1, 2, 3...

## 1.6Edit

If a sequence of sets $A_1,A_2,A_3\cdots$ is such that $A_k\supset A_{k+1}, k=1,2,3,\cdots$ the sequence is said to be a nonincreasing sequence. Give an example of this kind of sequence of sets.

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

## 1.7Edit

If $A_1,A_2,A_3\cdots$ are sets such that $A_k\subset A_{k+1} k=1,2,3,\cdots, \lim_{k \to \infty}A_k$ is defined as the union $A_1\cup A_2\cup A_3\cup\cdots$. Find $\lim_{k \to \infty}A_k$ if

(a) $A_k=\{x:1/k\le x\le 3-1/k\}, k=1,2,3\cdots$

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

Solution: (b) $\lim_{k \to \infty}A_k=\{(x,y):1< x^2+y^2<4\}$

## 1.8Edit

If $A_1,A_2,A_3\cdots$ are sets such that $A_k\supset A_{k+1} k=1,2,3,\cdots, \lim_{k \to \infty}A_k$ is defined as the intersection $A_1\cap A_2\cap A_3\cup\cdots$. Find $\lim_{k \to \infty}A_k$ if

(a) $A_k=\{x:2-1/k< x\le 2\}, k=1,2,3\cdots$

(b) $A_k=\{x:2

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

Solution: (c) $\lim_{k \to \infty}A_k=\{(0,0)\}$

## 1.9Edit

For every one dimensional set A let $Q(A)=\sum_A f(x)$ where $f(x)=\begin{cases}\frac{2}{3}(\frac{1}{3})^x, & x=0,1,2,\cdots \\ 0, & otherwise \end{cases}$. If $A_1=\{x:x=0,1,2,3\} \,$ and $A_2=\{x:x=0,1,2,\cdots\}$, find $Q(A_1)$ and $Q(A_2)$.

Solution: $Q(A_1)=\frac{80}{81}$ and $Q(A_2)=1$ using the formulae of sum of a geometric series.

## 1.10Edit

For every one dimensional set A for which the integral exists, let $Q(A)=\int_A f(x)$ where $f(x)=\begin{cases}6x(1-x), & 0, otherwise let $Q(A)$ be undefined. If $A_1=\{x:\frac{1}{4} and $A_3=\{x:0, find $Q(A_1), Q(A_2)$ and $Q(A_3)$.

Solution: $Q(A_1)$ can be found by integrating f(x) from $\frac{1}{4}$ to $\frac{3}{4}$, $Q(A_2)=0$ as integral over a set of measure zero is 0 and $Q(A_3)$ can be found by integrating f(x) from 0 to 1.

## 1.11Edit

Let $Q(A)=\int_A\int(x^2+y^2)dxdy$ for every two dimensional set A for which the integral exists; otherwise let $Q(A)$ be undefined. If $A_1=\{(x,y):-1\le x\le 1, -1\le y\le 1\}, A_2=\{(x,y):-1\le x=y\le 1\},$ and $A_3=\{(x,y):x^2+y^2\le 1\}$, find $Q(A_1), Q(A_2)$ and $Q(A_3)$.

Solution: $Q(A_1)= \int_{-1}^1\int_{-1}^1(x^2+y^2)dxdy=\frac{8}{3}$, $Q(A_2)$ is zero since it represents volume of a sheet and $Q(A_3)=\int_{0}^{2\pi}\int_{0}^1(r^2)rdrd\theta$ if we introduce polar coordinates. This evaluates to $\frac{\pi}{2}$.

## 1.15Edit

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.

## 1.16Edit

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:

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

Now,

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

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.