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Unformatted text preview: e set A
u is disjoint from
AB
u , Exercise 3 guarantees that
card A B card A B u ÞA u mn m m n 1 6. Given that S is a finite set and card S n, prove that the power set p S of S is a finite set and that
card p S 2 n . Solution: We use mathematical induction. For each nonnegative integer n we take p n to be the
assertion that whenever S is a finite set satisfying card S n then p S is a finite set and card p S 2 n .
If S then p S
and so
card p S 1 2 0 2 card S .
Therefore the assertion p 0 is true.
To see that the assertion p 1 is true, suppose that card S 1. We write S in the form S u . Since
pS , u
we see that
card p S 2 2 1 2 card S .
Now suppose that n is any positive integer for which the assertion p n is true and suppose that
card S n 1. Choose a member u of the set S. Since card S
u n we know that
card p S
We now define 55 u 2n. W AÞ u A pS u and we observe that W ß p S
u . (To see this, define f A A Þ u for every A
W is disjoint from p S
u and card W n we deduce from Exercise 3 that
card p S card p S
u Þ W 2 n 2 n 2 n 1 . pS u .) Since 7. Prove that if S is a nonempty finite set of real numbers then S has a largest member.
We use mathematical induction. For each positive integer n we take p n to be the assertion that
whenever S is a finite set satisfying card S n then S has a largest member. The assertion p 1 is
obvious.
Now suppose that n is any positive integer for which the assertion p n is true and suppose that S is a
finite set satisfying card S n 1. We choose a member u of the set S. Since card S
u n we
know that the set S
u has a largest member that we shall call v. In the event that u v we see
that v is the largest member of S and in the event that v u we see that u is the largest member of
S. In either event the set S must have a largest ,member.
8. Prove that none of the sets the set Z , Z, Q, and R are finite.
All of these sets are nonempty and none of them has a largest member.
9. Prove that a subset S of Z is finite if and only if it is possible to find an integer n such that the inequality
m n holds for every member m of S.
We already know that if n is any positive integer then every subset of the set 1, 2, , n is finite. On
the other hand, if S is a finite subset of Z then either S is empty, in which case m 1 for every
m S, or S has a largest member n, in which case m n for every m S.
10. Prove that if m and n are positive integers then the number of possible functions from the set 1, , n into
the set 1, , m is m n . Solution: We want to know that if is the set of all functions from 1, 2, , n into 1, 2, , m then
the set is finite and card
m n . We suppose that m is a positive integer and for each positive integer n
we define n to be the set of all functions from 1, 2, , n into 1, 2, , m . For each positive integer n
we take p n to be the assertion that card n m n .
1, 2, , m to be the
To see that the assertion p 1 is true we observe that...
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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

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