S
5: S
{
C
x
:
x
∈
M
}
M
t
N
We now claim that the collection
{
C
x
:
x
∈
M
}
forms a parti
tion of
M
t
N
(recall that this means that any two given
C
x
s
are either identical or disjoint—see Definition A.3.2.9). If
C
x
1
is disjoint from
C
x
2
we are done, so instead suppose that there
is some element
x
0
that is in both
C
x
1
and
C
x
2
. First, let us do
the case in which
x
0
∈
M
. From the definition of
C
x
, we then
must have that
[
g
◦
f
]
k
(
x
1
)
=
x
0
=
[
g
◦
f
]
l
(
x
2
)
(A.5.1.11)
for some
k
,
l
∈
Z
. Without loss of generality, suppose that
k
≤
l
. Then, applying
f

1
◦
g

1
to both sides of this equation
k
times,
e
we find that
x
1
=
[
g
◦
f
]
l

k
(
x
2
)
.
(A.5.1.12)
In other words,
x
1
∈
C
x
2
. Not only this, but
f
(
x
1
) ∈
C
x
2
as well
because
f
(
x
1
)
=
f
(
[
g
◦
f
]
l

k
(
x
2
)
)
. Similarly,
g

1
(
x
1
) ∈
C
x
2
,
and so on. It follows that
C
x
1
⊆
C
x
2
. Switching
1
↔
2
and
applying the same arguments gives us
C
x
2
⊆
C
x
1
, and hence
C
x
1
=
C
x
2
. Thus, indeed,
{
C
x
:
x
∈
M
}
forms a partition of
M
t
N
. In particular, it follows that
C
x
=
C
x
0
for all
x
0
∈
C
x
.
(A.5.1.13)
S
6: D
X
1
,
X
2
,
Y
1
,
Y
2
Now define
A
B
x
∈
M
s.t.
C
x
∩
N
⊆
f
(
M
)
C
x
(A.5.1.14)
A.5 Cardinality, countability, and the naturals
565
as well as
X
1
B
M
∩
A
,
Y
1
B
N
∩
A
,
(A.5.1.15a)
X
2
B
M
∩
A
c
,
Y
2
B
N
∩
A
c
.
(A.5.1.15b)
Note that, as
{
C
x
:
x
∈
M
}
is a partition of
M
t
N
, we have
that
A
c
=
x
∈
M
s.t.
C
x
∩
N
*
f
(
M
)
C
x
.
(A.5.1.16)
S
7: S
f

X
1
:
X
1
→
Y
1
We claim that
f

X
1
:
X
1
→
Y
1
is a bijection. First of all, note
the it
x
∈
X
1
, then in fact
f
(
x
) ∈
Y
1
, so that this statement
indeed does make sense. Of course, it is injective because
f
is.
To show surjectivity, let
y
∈
Y
1
B
N
∩
A
. From the definition
of
A
(A.5.1.14)
, we see that
y
∈
C
x
∩
N
for some
C
x
with
C
x
∩
N
⊆
f
(
M
)
, so that
y
=
f
(
x
0
)
for some
x
0
∈
M
. We still
need to show that
x
0
∈
X
1
. However, we have that
x
0
=
f

1
(
y
)
,
and so as
y
∈
C
x
, we have that
x
0
=
f

1
(
y
) ∈
C
x
as well. We
already had that
C
x
∩
N
⊆
f
(
M
)
, so that indeed
x
0
∈
A
, and
hence
x
0
∈
X
1
. Thus,
f

X
1
:
X
1
→
Y
1
is a bijection.
S
8: S
g

Y
2
:
Y
2
→
X
2
We now show that
g

Y
2
:
Y
2
→
X
2
is a bijection. Once again,
all we must show is surjectivity, so let
x
∈
X
2
=
M
∩
A
c
. From
the definition of
A
(A.5.1.14)
, it thus cannot be the case that
C
x
∩
N
is contained in
f
(
M
)
, so that there is some
y
∈
C
x
∩
N
such that
y
<
f
(
M
)
. By virtue of
(A.5.1.13)
, we have that
C
x
=
C
y
, and in particular
x
∈
C
y
. From the definition of
C
y
,
it follows that either (i)
x
=
y
, (ii)
x
is in the image of
f

1
, or
(iii)
x
is in the image of
g
(the other possibilities are excluded
because
x
∈
M
). Of course it cannot be the case that
x
=
y
because
x
∈
M
and
y
∈
N
. Likewise, it cannot be the case
that
x
is in the image of
f

1
because
x
∈
A
c
. Thus, we must
A Basic set theory
566
have that
x
=
g
(
y
0
)
for some
y
0
∈
N
. Once again, we still must
show that
y
0
∈
Y
2
. However, we have that
y
0
=
g

1
(
x
)
, so that
y
0
∈
C
x
. Furthermore, as
C
x
∩
N
is not contained in
f
(
M
)
,
from
(A.5.1.16)
it follows that
C
x
⊆
A
c
. Thus,
y
0
∈
C
x
⊆
A
c
,
and so
y
0
∈
Y
2
. Thus,
g

Y
2
:
Y
2
→
X
2
is a bijection.
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 Fall '08
 GUREVITCH
 Linear Algebra, Algebra, Transformations