This preview shows page 1. Sign up to view the full content.
Unformatted text preview: ion
X A Þ X B Þ X C X
holds. This equation tells us that every member of X must belong to at least one of the sets X A
and X B and X C which means that every member of X must fail to belong to at least one of the
sets A and B and C. Thus no member of X can belong to A B C. Since the sets A and B and C
are subsets of X we also know that no object x lying outside the set X can belong to A B C.
Therefore A B C .
8. Given sets A, B and C, determine which of the following identities are true. Hint: The truth values are given here. Explain carefully why they are correct.
a. A B CA B A C True b. A Þ B C AÞB A Þ C False c. A Þ B C AÞB A C False d. A Þ B C AÞB A C False e. A B CA B f. A B CA B A C False g. A B CA B A C False h. A B CA BÞA C False i. A B CA BÞA C True j. A B CA B k. A A BÞA B True C False A Þ C False l. p A Þ B p A Þ p B False
m. p A B pA n. A BÞC A o. A B p. A B p B True
BÞA C True CA B A C True CA B A C True 9. Is it true that if A and B are sets and A A B then the sets A and B are disjoint from each other? Hint: Yes
10. Given that A is a set with ten members, B is a set with seven members and that the set A 39 B has four members, how many members does the set A Þ B have?
11. Give an example of a set A that contains at least three members and that satisfies the condition A pA. Hint: Define
A
12. For which sets A do we have A , , pA? Solution: This happens for all sets. Every set is a subset of itself.
Exercises on Families of Sets
For each of the following families of subsets of R, describe the sets Þ and : a. 11
n is a positive integer .
n, n
1, 1 and we see at once that Þ 1, 1 . We can
If n is any positive integer then 1 , 1
nn
11
also see that
0 because if x is any nonzero number then the condition x
n , n will
1
fail whenever n x  which holds when n 1/x . b. 1. 1 1 ,3 1
n is a positive integer .
n
n
Using methods similar to part a we can see that Þ 1, 3 and . c. is the family of all those subsets of R that have no more than two members.
We have Þ R and
. d. is the family of all those subsets A of R for which the set R
We have Þ R and
. A has no more than two members. 2. Write out careful proofs for the three DeMorgan laws for families of sets that were not proved above.
We show just one more proof, the proof that if is a nonempty family of sets then
S S S A A . Suppose that is a nonempty family of sets. Given any x S
we know that x S and that
x fails to be in . Thus if A is any member of then we know that x S and x A, and we
conclude that x S A for every A
. We have therefore shown that
S A A . Now suppose that x S A A
. Using the fact that
we choose a member B of .
Since x S B we know that x S. Finally, if A is any member of then it follows from the
condition x S A that x A and we conclude that x . We have therefore shown that S A A S . 3. A family of sets is said to be nested if for any two members A and B of we have either A B or B A.
Given...
View
Full
Document
This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.
 Fall '08
 STAFF
 Math, Calculus

Click to edit the document details