Exam 1

Question 1

Every set contains the empty set and itself.

Answer 1

Theorem 2.2.2, page 42

Question 2

For sets X, A, and B, X is a subset of A intersect B iff X is a subset of A and X is a subset of B.

Answer 2

Exercise 2.3.8, pg. 44

Question 3

Definition of a subset

Answer 3

S is a subset of A if every element of S is in A pg. 42

Question 4

Definition of set equality

Answer 4

A = B if A is a subset of B and vice versa pg. 43

Question 5

If A is a subset of B, and B is a subset of C, then A is a subset of C.

Answer 5

Exercise2.24, pg. 42

Question 6

Definition of proper subset

Answer 6

B is a subset of X and B neq X

Question 7

Definition of “the union over alpha in the lambda of the B-alpha’s”

Answer 7

The set of elements x such that x is an element of any B-alpha for all alphas in lambda

Question 8

Definition of “the intersection over alpha in the lambda of the B-alphas”

Answer 8

The set of elements x such that x is an element of every B-alpha for all alphas in lambda

Question 9

A union (B intersect C)=(A union B) intersect (A union C)

Answer 9

Theorem 2.4.2, pg. 48

Question 10

A intersect (B union C)=(A intersect B) union (A intersect C)

Answer 10

Exercise 2.4.4, pg. 48

Question 11

C union (the intersection over alpha in the lambda of the B-alphas) = the intersection over alpha in the lambda of (the B-alphas union C)

Answer 11

Theorem 2.4.5 pg 49

Question 12

C intersect (the union over alpha in the lambda of the B-alphas) = the union over alpha in the lambda of (the B-alphas intersect C)

Answer 12

Theorem 2.4.5 pg 49

Question 13

C union (the union over alpha in the lambda of the B-alphas) = (the union over alpha in the lambda of the B-alphas union C)

Answer 13

Theorem 2.4.6 pg 49

Question 14

C intersect (the intersection over alpha in the lambda of the B-alphas) = (the intersection over alpha in the lambda of the B-alphas intersect C)

Answer 14

Theorem 2.4.6 pg. 49

Question 15

Definition of set complement

Answer 15

The set containing all the elements in U not in S, 2.3.1 pg 43

Question 16

The complement of (A union B) = The complement of A intersect the complement of B

Answer 16

Problem 2.4.8 pg. 49

Question 17

The complement of (A intersect B)=The complement of A union the complement of B

Answer 17

Problem 2.4.8 pg 49

Question 18

Definition of set difference

Answer 18

A\B is the set containing those elements of A that are not elements of B; is the same thing as A intersect the complement of B; 2.4.10 pg 50

Question 19

C(A union B)=(C\A) intersect (C\B)

Answer 19

Theorem 2.4.11.1 pg 50

Question 20

C(A intersect B)=(C\A) union (C\B)

Answer 20

Theorem 2.4.11.2 pg 50

Question 21

B(B\A)= A intersect B

Answer 21

Theorem 2.4.11.3 pg 50

Question 22

(A\B) union (B\A)=(A union B)(A intersect B)

Answer 22

Theorem 2.4.11.4 pg 50

Question 23

Definition of symmetric difference

Answer 23

(A union B)(A intersect B) or A delta B pg 50

Question 24

Definition of a power set

Answer 24

The power set of A is the set of all subset of A.

Question 25

A is a subset of B iff P(A) is a subset of P(B)

Answer 25

Theorem 2.5.4 pg 51

Question 26

P(A intersect B)=P(A) intersect P(B)

Answer 26

Theorem 2.5.5.1 pg 51

Question 27

P(A) union P(B) is a subset of P(A union B)

Answer 27

Theorem 2.5.5.2 pg 51

Question 28

Adding one element to a set doubles the size of the power set

Answer 28

Problem 2.5.7 pg 51x

Question 29

The Axiom of Induction

Answer 29

Let S be a subset of the natural numbers such that 1 is an element of S and, if k is an element of S, then k+1 is an element of S for some arbitrary k. Then S is the natural numbers.

Question 30

Principle of Mathematical Induction

Answer 30

Suppose that P(1) is true and, if P(k) is true, then P(k+1) is true. Then P(n) is true for all n in the natural numbers. 3.1.2 pg 60

Question 31

If S is a set with n elements, then the power set of S has 2^n elements

Answer 31

3.2.1 pg 61

Question 32

Principle of Complete Induction

Answer 32

Suppose that P(1) is true and, if P(j) is true for all j≤k, then P(k+1) is true. Then P(n) is true for all n in the natural numbers. 3.3.1 pg. 62