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Math 350
Advanced Calculus I
Fall 2009
Homework Solutions
Section 8
CARDINALITY
Text Problems
8.10
Prove: For all sets
S
, if
S
is denumerable, then
S
is equinumerous with a proper subset of itself.
Proof:
Suppose
S
is denumerable. We need to show there exists a proper subset
T
of
S
such that
S
∼
T
. So we need to ±rst come up with a subset
T
and then show that
S
is equinumerous with
this subset
T
.
Since
S
is denumerable, then
S
is equinumerous with
N
so we can list the elements in
S
S
=
{
s
1
, s
2
, s
3
, . . .
}
,
where all elements in the listing are distinct. Let
T
be the set of all elements of
S
with an even
subscripts in this listing
T
=
{
s
2
, s
4
, s
6
, . . .
}
.
Then
T
is a proper subset of
S
. We now need to show that
S
is equinumerous with
T
. To show
this, we need to show there exists a bijection
g
:
S
→
T
. De±ne
g
:
S
→
T
by
g
(
s
n
) =
s
2
n
for each
s
n
∈
S
(or for each
n
∈
N
).
Injective.
To show that
g
is injective, we need to show
(
∀
s
n
∈
S
) (
∀
s
m
∈
S
) [
g
(
s
m
) =
g
(
s
n
)
→
s
m
=
s
n
]
.
Suppose
s
m
and
s
n
are arbitrary elements in
S
. Suppose
g
(
s
m
) =
g
(
s
n
)
.
We need to show
s
m
=
s
n
.
Since
g
(
s
m
) =
s
2
m
and
g
(
s
n
) =
s
2
n
, then
g
(
s
m
) =
g
(
s
n
) implies
s
2
m
=
s
2
n
. Since the elements in
the listing for
S
are distinct, then this implies 2
m
= 2
n
so
m
=
n
. But then
s
m
=
s
n
, as required.
Therefore
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 Spring '08
 QIAN
 Calculus, Sets

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