120
Solutions to Exercises
has a smallest element, say
y,
by the induction hypothesis.
If
y
>
x,
then
x
is the smallest element
in 8, while, if
x
>
y,
then
y
is the smallest element in 8.
In
either case, 8 has a smallest element,
as required. By the Principle of Mathematical Induction, the result is true for all
n
;:::: 1.
(b) The real numbers is an infinite set without a smallest element: For any
r
E R,
r

1 is always
smaller than
r.
On the other hand, the natural numbers does have a smallest element.
(c) The WellOrdering Principle!
18. [BB] For
n
=
1,
n~=1
Bi
= Blo
and so
A
U
(n~=1
Bi)
=
AU BI
=
n~=1
(A UBi)
and the result
holds for
n
=
1. Now assume that
A
U
(n~=1
=
n~=1
for
k
;::::
1. Then, given a set
A
and
k
+
1 sets Bl,
B2,
... ,
Bk+1.
we have
A
U
(n~~l
=
AU
[(
n~=1
n
Bk+I]
=
[A
U
(n~=1
]
n
(A
U
Bk+1
)
since
A
U
(B
n
C)
=
U
B)
n
U C)
=
[n~=1
(A UBi)]
n
U
Bk+I)
(by the induction hypothesis)
=
n~~11
(
A
U
) ,
giving the result for
k
+
1. By the Principle of Mathematical Induction, we conclude that
A
U
(
n~=1
=
n~=1
(
AU Bi
),
for all
n
;:::: 1.
19. For
n
=
U~=I
=
BI
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory

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