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
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
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
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
(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
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
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]
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
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
3.6.7. We verify that is an equivalence relation by showing it is reexive, symmetric, and
transitive.
Reexive. By denition (x, y ) (x, y ) provided y y = 3(x x). Since both sides of this
equation are 0, this is clearly the case. Therefore, is reexive.
Sym
(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_
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.)
A
f (AC )
[f (A)]C
f 1 [f (A)]
Geom.
Appl.
Dyn. Sys.
Algeb.
Discrete
Untenured
Tenured
Old
Math 230
Instructor: Dr. Irl C. Bivens
Sets and Proofs
Spring, 12
Oce: Chambers 3040, 7048942317
Oce Hours: Mon: 2:304:00 (with some exceptions of 2:303:45 due to Departmental
Meetings), Tue: 2:004:00, Wed: 2:304:00, Thur: 2:004:00, Fri: 2:303:30.
Textboo