The Pacific
The West
Central Canada
The Atlantic
The North
Number of seats
Population
70
12.43
13.06
18.34
18.48
60
58.88
61.52
9.48
6.61
50
0.9
0.32
GDP
12.8
18.32
56.22
5.56
0.48
40
PERCENTAGE (%)
30
20
10
0
The Pacific
The West
Central Can
CANADA POLIT
State the contrapositive version of (G2) in Denition 5.1.9.
Answer. If N is any real number such that G < N then there exists an element a A that
satises a < N .
5.1.12 It follows from Exercise 5.1.11 that the set B = cfw_a : a A is bounded from above
and
5.1.2 Suppose that A is bounded. Then there exists a positive real number M such that
|a| M for all a A. Equivalently,
M a M
(a A)
It follows that if A is bounded then A is bounded above by M and is bounded below by
M .
Conversely, suppose that A is bound
1. Prove that countability is preserved under the equivalence relation equinumerous. That
is, prove that if A is equinumerous with B and A is countable then so is B (and of the same
type nite or countably innite). (Hint: Theorem 4.6.5 takes care of the ca
1. (Pigeon hole principle) Suppose that A and B are nite sets, that |A| > |B |, and that
f : A B is any function. Prove that there exist x1 = x2 in A such that f (x1 ) = f (x2 ).
Proof #1. Suppose that the conclusion does not hold. Then f : A B is 11. Let
4.6.7 Assume rst that both A and B are nite sets of cardinalities m and n respectively.
By hypothesis, there exist bijections f1 : Nm A and g : Nn B . In addition, it is
m
easy to see that the translation T : Nn Nn dened by T (x) = x + m is a bijection
m
1. Assume that f : Nn +1 Nm+1 is 11, that f (n + 1) = k for some 1 k m and that
f (l) = m + 1 for some 1 l n . Dene g : Nn Nm
f (x) , x = l
g (x) =
k, x=l
Prove that Nm is indeed a codomain for g .
Proof. Let x denote an arbitrary element of Nn . There a
14. In each problem a function f : A B is given. Prove that f : A B is a bijection by
nding a function g : B A that satises the conditions (g f )(x) = x for all x in A and
(f g )(y ) = y for all y in B . Important. Be sure to verify that A is a codomain f
4.4.4 (Q2): This is false. Let A = cfw_1, B = cfw_2, 3, C = cfw_4 and dene f : A B by
f (1) = 2, and g : B C by g (2) = g (3) = 4. Then f and g f are 11, but g is not 11.
(Q3): This is true, and remains true even if we remove the condition that g be 11. A
4.3.3 (b)
f (A1 A2 ) = cfw_y B : (x A1 A2 )(y = f (x)
= cfw_y B : (x A1 )(y = f (x) (x A2 )(y = f (x)
= cfw_y B : y f (A1 ) y f (A2 ) = cfw_y B : y f (A1 ) f (A2 )
= f (A1 ) f (A2 )
(b) (Element chasing) f (A1 A2 ) f (A1 ) f (A2 ): Let y denote any elemen
1. Refer to our Math Department function from class and ll in the table below. (To t
everything in, youll probably need to make your own larger copy of the table.)
f 1 [f (A)]
A
f (AC )
[f (A)]C
Geom. (G)
cfw_1,2,3,4,5,6
cfw_1,2,4,5,6,7,8,9
G cfw_BK
Appl.
4.2.3 Let y Nm be arbitrary. By denition, y = k + m for some 1 k n. But then
n
T (k ) = k + m = y . Since y was arbitrary, the translation T is onto.
4.2.4 Let 1 y 7 be arbitrary. Then 0 y 1 6 and therefore 0 (y 1)/3 2.
If x = (y 1)/3, then x belongs to t
(Jonahs question) The sets do not have to be equinumerous. Take A to be the set of rational
numbers and B to be the set of irrational numbers.
Suppose that A is a nonempty set of real numbers that has a maximum value M . Prove
that A has a LUB that is equ
14. In each problem a function f : A B is given. Prove that f : A B is a bijection by
nding a function g : B A that satises the conditions (g f )(x) = x for all x in A and
(f g )(y ) = y for all y in B . Important. Be sure to verify that A is a codomain f
6. (16 points) Let f : R [1, 1] be dened by f (x) = sin x and dene a collection F
of subsets of R by F = cfw_A = f 1 (cfw_) : 1 1. (Here, f 1 (cfw_) refers to the
preimage of the singleton cfw_ under f . Dont confuse f 1 with the bijection
arcsin : [1, 1]
Seat Alloc PopulationGDP
The Pacific
The West
Central Ca
The Atlanti
The North
12.43
18.34
58.88
9.48
0.9
13.06
12.8
18.48
25.27
61.52
56.22
Percentage
of Canada's about The number of seat in the House of Common,Distributio
6.61
5.56
0.32
0.48
70
61.52
58
Seat Alloc PopulationGDP
The Pacific
The West
Central Ca
The Atlanti
The North
12.43
18.34
58.88
9.48
70
0.9
13.06
12.8
Percentage
of
18.48
25.27
61.52
56.22
6.61
5.56
0.32
0.48
Canada's about The number of seat in the House of Comm
61.52
60
58.88
56.22
5
1. (Inspired by Nancy Brown) Suppose that (A1 , B1 , f1 ) and (A2 , B2 , f2 ) are two functions
and that B1 = B2 and f1 = f2 . Must it be the case that A1 = A2 ? If so, then prove it. If
not, provide a counterexample.
2. (Inspired by Nancy Brown) Suppose
1. Prove that two functions (A1 , B1 , f1 ) and (A2 , B2 , f2 ) are equal if A1 = A2 = A, B1 =
B2 = B , and (x A)(f1 (x) = f2 (x). (Compare with Denition 4.1.9 on page 124.)
2. Let h(x) = x2 with domain (, 2] and codomain R, and g (x) = 6 x with domain
(2
Assume the following postulates.
P1: Either Real Analysis is dicult or not many students take Real Analysis.
Symbolically P1 becomes ? where r: Real Analysis is dicult. and s: Many students
take Real Analysis.
P2: If Math 230 is easy then Real Analysis is
7. (24 points) Refer to the Sally Problem and in each part determine if the statement is
a logical conclusion. If the statement is a logical conclusion, give a proof, and if it is not,
provide a counterexample.
(a) If Sally takes Math 230 then Sally does
1. In each part nd the aw(s) in the syntax.
(a) (x)(x A (y B x > y )
(b) (x)(x B (y A y > x)
(c) (x A) > (y B )
(d) x A > B
(e) (x A) > (x B )
(f) (x A) (x A x B )
(g) (f )(f C S )
2. One proposed answer to 1.3.7 was (x)(x2 + x = 1 x < 0). Explain why thi
Theorem 5.3.14 The empty set and the real numbers are the only subsets of the real
numbers that are both open and closed.
Proof. We give a proof by contradiction. Suppose A is a subset of the real numbers that
has the following four properties: (i) A = ,
6. (18 points) Suppose a collection of disks that are black on one side and white on the
other are scattered over the surface of a table. A game is to be played in which there are
two types of moves (i) and (ii): (i) all the disks are ipped over ; (ii) (i
1. (Inspired by Nancy Brown) Suppose that (A1 , B1 , f1 ) and (A2 , B2 , f2 ) are two functions
and that B1 = B2 and f1 = f2 . Must it be the case that A1 = A2 ? If so, then prove it. If
not, provide a counterexample.
Solution. Yes, A1 = A2 . We rst prove
1. Prove that two functions (A1 , B1 , f1 ) and (A2 , B2 , f2 ) are equal if A1 = A2 = A, B1 =
B2 = B , and (x A)(f1 (x) = f2 (x). (Compare with Denition 4.1.9 on page 124.)
Solution. To prove that the ordered triples are equal, we must show that A1 = A2
The Algebra of Sets
Commutative Laws. Both and are commutative.
Associative Laws. Both and are associative.
Distributive Laws. distributes across and vice versa.
Double Negation Law. (AC )C = A
DeMorgans Laws. (A B )C = AC B C and (A B )C = AC B C
Idempot
(z, z ) = cfw_0.
Prove that
z R+
Proof. Step 1. We rst show that cfw_0
(z, z ). The set cfw_0 contains only the single
z R+
element 0. Since z < 0 < z for every positive number z , it follows that 0
(z, z ).
z R+
Hence, cfw_0
(z, z ).
z R+
(z, z ) cfw_