B s 0 3 1 4 1 5 s 1 3 1 5 int s 1 3 bd s 1 1 3 5 s 1

(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 .