M
453
S
A
1
S
8, 2004
Exercise 4 from Section 3, page 28
(a) Reﬂexivity, symmetry and transitivity for
∼
follow from the corresponding properties
for equality, so
∼
is an equivalence relation. For instance, to check symmetry, we
proceed as (
a
0
∼
a
1
)
⇒
(
f
(
a
0
)
=
f
(
a
1
))
⇒
(
f
(
a
1
)
=
f
(
a
0
))
⇒
(
a
1
∼
a
0
).
(b) Let
A
*
=
{
[
a
]

a
∈
A
}
be the equivalence classes under
∼
. The map
f
:
A
→
B
induces a map
f
*
:
A
*
→
B
given by
f
*
([
a
])
=
f
(
a
). We will check that
f
*
is a
bijection.
Given
b
∈
B
, there is an
a
∈
A
such that
f
(
a
)
=
b
since
f
is surjective. Hence
f
*
([
a
])
=
b
and so
f
*
is surjective. For injectivity, suppose
f
*
([
a
0
])
=
f
*
([
a
1
]). By
the deﬁnition of
f
*
we get
f
(
a
0
)
=
f
(
a
1
), or
a
0
∼
a
1
. Hence [
a
0
]
=
[
a
1
] and
f
*
is
injective.
±
Exercise 5 from Section 3, page 28
(a) Again we have to check the three conditions for an equivalence relation. For reﬂex
ivity, note that
x

x
=
0, so (
x
,
x
)
∈
S
0
. If
y

x
is an integer, then
x

y
=

(
y

x
)
is too, and symmetry follows. Finally, if
y

x
and
z

y
are integers, then so is
z

x
=
(
z

y
)
+
(
y

x
). Hence (
x
,
z
)
∈
S
0
and we have transitivity.
If (
x
,
y
)
∈
S
, then
y
=
x
+
1, or
y

x
=
1, an integer. Hence (
x
,
y
)
∈
S
0
. This shows
that
S
0
⊃
S
.
Given
x
∈
R
, the equivalence class [
x
] consists of all
y
∈
R
such that
y

x
∈
Z
. In
other words,
y
is of the form
x
+
n
where
n
∈
Z
. Denoting the set
{
x
+
n

n
∈
Z
}
by
x
+
Z
, we have [
x
]
=
x
+
Z
. Furthermore, given any
x
∈
R
, there is an
x
0
∈
[0
,
1) and
an integer
n
such that
x
=
x
0
+
n
. Hence [
x
]
=
[
x
0
] for some
x
0
∈
[0
,
1). Also, no two
distinct elements of [0
,
1) are
S
0
related, so
R
/
S
0
=
{
x
+
Z

x
∈
[0
,
1)
}
.
±
Exercise 5 from Section 5, page 39
Before looking at the speciﬁc subsets in this problem, let us consider an arbitrary subset
A
=
{
x

x
satisﬁes some condition
P
}
. Suppose
P
does not mix the indices, i.e. it is of
the form “for all
i
,
x
i
satisﬁes some condition
C
(
i
) that depends only on
i
”. This means
that the
i
th coordinate
x
i
can only take values from the subsets
A
i
of