This preview shows page 1. Sign up to view the full content.
Unformatted text preview: he event that m 1 S we know that S
1, 2, , m and so card S
m m 1. Suppose now that
m 1 S. Using the fact that that S is a proper subset of the set 1, 2, , m 1 we choose an integer
k
1, 2, , m
S. Since
S ß SÞ k
m1
and since
card S Þ k
m1
m1
we see that card S m 1. 2. Prove that if A is a finite set and B is a proper subset of A then we do not have A ß B.
Using the fact that A is a nonempty finite set we choose a positive integer n such that
A ß 1, 2, , n . Choose a oneone function f from A onto 1, 2, , n . Since B is a proper subset of
A the set f B must be a proper subset of 1, 2, , n . If we had A ß B then we would have
f B ß B ß A ß 1, 2, , n
which is impossible by Exercise 1. Therefore we can’t have A ß B.
3. Prove that if A and B are finite sets and A B then card A Þ B card A card B . Solution: We use mathematical induction. Suppose that A is a finite set and define m card A . For
each nonnegative integer n we take p n to be the assertion that if B is any finite set disjoint from A and
satisfying card B n then the set A Þ B is finite and card A Þ B m n. 54 The assertion p 0 is obvious. We shall now prove that p 1 is true. Suppose that B is any set disjoint from A
and satisfying card B 1. We can express B in the form u . Choose a oneone function f from A onto
1, 2, , m and extend the function f to A Þ B by defining f u m 1. Since this extension of f is a
oneone function from A Þ B onto 1, 2, , m, m 1 , we have card A Þ B m 1.
Now suppose that n is any positive integer for which the assertion p n is true and that B is a set disjoint
from A and satisfying card B n 1. Choose a member u of the set B. Since the assertion p 1 is true and
u is disjoint from the finite set B
u we have
card B card B
u 1
which tells us that card B
u n. Therefore, since the assertion p n is true we have
card A Þ B
u
mn
and since p 1 is true and since u is disjoint from A Þ B
u we have
card A Þ B card AÞ B u Þu mn1 4. Given that A and B are finite sets and that card A m and card B n, prove that A Þ B is finite and that
card A Þ B
m n.
The point is that A Þ B A Þ B A and that since B A is a subset of B we have
card B A
card B .
5. Given that A and B are finite sets and that card A m and card B n, prove that A
card A B mn. B is finite and that Solution: We use mathematical induction. Suppose that A is a finite set and that card A m. For
each nonnegative integer n we take p n to be the assertion that whenever B is a finite set and card B n
then the set A B is finite and card A B mn.
The assertion p 0 is obvious because, if B is empty then A B is empty.
The assertion p 1 is obvious because, if B u then A B ß A.
Now suppose that n is any positive integer for which the assertion p n is true and suppose that
card B n 1. Choose a member u of the set B. Since card B
u n we have
card A B
u
mn
and since the assertion p 1 gives us card A
u m and since th...
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