(b)
S
=
(
[0
,
3)
∩
(1
,
4)
)
∪ {
1
,
5
}
S
= (1
,
3)
∪ {
1
,
5
}
:
int
S
= (1
,
3)
,
bd
S
=
{
1
,
1
,
3
,
5
}
,
S
0
= [1
,
3],
isolated points:
{
1
,
5
}
.
(c)
S
=
{
r
∈
Q
:
0
< r <
√
2
}
int
S
=
∅
,
bd
S
= [0
,
√
2]
,
S
0
= [0
,
√
2],
no isolated points.
(d)
S
=
n
n
2
+ 1
:
n
∈
N
S
=
{
1
/
2
,
2
/
5
,
3
/
10
,
4
/
17
,
5
/
26
, . . .
}
.
int
S
=
∅
,
bd
S
=
S
∪ {
0
}
,
S
0
=
{
0
}
,
isolated points:
{
1
/
2
,
2
/
5
,
3
/
10
,
4
/
17
,
5
/
26
, . . .
}
=
S
.
(e)
S
=
{
x
∈
R
: 0
<

x

3
 ≤
2
}
S
= [1
,
3)
∪
(3
,
5] :
int
S
= (1
,
3)
∪
(3
,
5)
,
bd
S
=
{
1
,
3
,
5
}
,
S
0
= [1
,
5],
no isolated points.
4
(f)
S
=
1
n
:
n
∈
N
∪
[1
,
∞
)
S
=
1
,
1
2
,
1
3
,
1
4
,
1
5
,
· · ·
∪
[1
,
∞
)
int
S
= (1
,
∞
)
,
bd
S
=
{
0
} ∪
1
,
1
2
,
1
3
,
1
4
,
1
5
,
· · ·
,
S
0
= [1
,
∞
)
∪ {
0
}
,
isolated points:
1
2
,
1
3
,
1
4
,
1
5
,
· · ·
.
5
3. Prove that if
α
= sup
S
and
α /
∈
S
,
then
(a)
α
is an accumulation point of
S
.
Let
N
*
(
α,
) be a deleted neighborhood of
α
. Since
α
= sup
S
, there is
an element
s
∈
S
such that
α

< s
≤
α
but, since
α /
∈
S,
α

< s < α.
Thus, every deleted neighborhood of
α
contains a point of
S
which
implies that
m
is an accumulation point of
S
.
(b)
S
is an infinite set.
Let
s
1
be any element of
S
.
By Theorem 3 (lecture notes), there exists
s
2
∈
S
such that
s
1
< s
2
< α.
Now assume we have selected
s
1
, s
2
, . . . , s
n
∈
S
such that
s
1
< s
2
< . . . < s
n
< α.
By Theorem 3, there is an element
s
n
+1
∈
S
such that
s
n
< s
n
+1
< α
.
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 Fall '08
 Staff
 Math, Sets, accumulation point, Closed set, General topology