10/18/2005
CSCI 2670
4.6
Assume
B
is countable and that
f:N
→
B
is a functional correspondence.
Let
f
i
(j)
denote the
i
th
symbol in
f(j)
.
Consider the string
s
whose
i
th
element differs from
f
i
(i)
– i.e. if
f
i
(i)
= 0, then the
i
th
symbol in
s
is 1 and vice versa.
Then
s
cannot be
in the image of
f
since it differs from every string in the image of
f
by at least one
symbol.
Furthermore,
s
is in
B
since it is an infinite sequence over {0,1}.
Therefore, no functional correspondence can exist between
N
and
B
– i.e.,
B
is
uncountable.
4.7
We begin by lexicographically arranging the elements of T according to the sum
of the elements {(1,1,1),(1,1,2),(1,2,1),(2,1,1),(1,1,3),(1,2,2),(1,3,1),(2,1,2),
(2,2,1),(3,1,1),…}.
For each natural number
i
, let
f(i)
be the
i
th
element in the list.
Since the list never repeats,
f
is clearly one-to-one.
Also, for every element
t
in T,
the sum of the components of
t
is finite.
Therefore, there is some natural number
n
such that