Solutions homework 9.
Page 60 Problem 2.
:
(i) Let
x
≤
y
with
x, y
∈
I
∩
J
. Then
I
s an interval implies that [
x, y
]
⊂
I
and similarly
[
x, y
]
⊂
J
. Hence [
x, y
]
⊂
I
∩
J
. It follows therefore from Theorem 4.1.4 that
I
∩
J
is an interval.
(ii) False, take e.g.
I
= [0
,
1) and
J
= [1
,
2].
(iii) True. Let
I
= (
a
1
, b
1
) and
J
= (
a
2
, b
2
). Let
a
= min
{
a
1
, a
2
}
. Then
a
= inf(
I
∪
J
),
Similarly if
b
= max
{
b
1
, b
2
}
, then
b
= sup(
I
∪
J
). If
I
∪
J
is an interval, it is equal
to (
a, b
). Assume
a
=
a
1
. If
a
2
=
a
1
, then obviously
I
∩
J
6
=
∅
. If
a
2
> b
1
, then
I
∩
J
= (
b
1
, a
2
) is non-empty. If
a
2
≤
b
1
, then
a
2
∈
(
a, b
)
\
(
I
∪
J
), which is impossible.
(iv) True. Let
I
= [
a
1
, b
1
] and
J
= [
a
2
, b
2
]. Let
a
= min
{
a
1
, a
2
}
. Then
a
= inf(
I
∪
J
),
Similarly if
b
= max
{
b
1
, b
2
}
, then
b
= sup(
I
∪
J
). If
I
∪
J
is an interval, it is equal
to [
a, b
].
Assume
a
=
a
1
.
If
a
2
=
a
1
, then obviously
I
∩
J
6
=
∅
.
If
a
2
≥
b
1
, then
I
∩
J
= [
b
1
, a
2
] is non-empty.
If
a
2
< b
1
, then
a
2
+
b
1
2
∈
(
a, b
)
\
(
I
∪
J
), which is
impossible.