1.13.
Let
x
∈
A
. Then
x
= 2
k

1 for some
k
∈
Z
. Let
k
1
=
k

1. We then have
2
k
1
+ 1
∈
B
, but also 2
k
1
+ 1 = 2(
k

1) + 1 = 2
k

1 =
x
, so
x
∈
B
. Hence
A
⊂
B
.
Conversely, let
x
∈
B
so that
x
= 2
m
+ 1 for some
m
∈
Z
and let
m
1
= 2
m

1.
Then 2
m
1

1
∈
A
and
x
= 2
m
+1 = 2(
m
+1)

1 = 2
m
1

1 so
x
∈
A
. Thus
B
⊂
A
.
It follows that
A
=
B
.
1.14.
We claim that [
a,b
]
∪
[
c,d
] = [
a,d
]

(
b,c
). We ﬁrst show [
a,b
]
∪
[
c,d
]
⊂
[
a,d
]

(
b,c
).
Let
x
∈
[
a,b
]
∪
[
c,d
]. Then
a
≤
x
≤
b
or
c
≤
x
≤
d
. In the ﬁrst case,
x
≤
d
, since
x
≤
b < d
, and
x
6
> b
, since
x
≤
b
, so
x
∈
[
a,d
]

(
b,c
). In the latter case,
x
≥
a
, since
x
≥
c > a
, and
x
6
< c
, since
x
≥
c
, so
x
∈
[
a,d
]

(
b,c
). Thus [
a,b
]
∪
[
c,d
]
⊂
[
a,d
]

(
b,c
)
as desired.
Conversely we show [
a,d
]

(
b,c
)
⊂
[
a,b
]
∪
[
c,d
]. Let
x
∈
[
a,d
]

(
b,c
). Then
a
≤
x
≤
d
and
x /
∈
(
b,c
). Thus
x
≤
b
or
x
≥
c
. In the former case,