Math 3333
Homework 3
NAME (Print):
1. TRUE–FALSE. If your answer is “True,” justify by quoting a definition or
theorem, or by giving a proof. If your answer is “False,” give a counter-example.
\
(a) Let G be a collection of closed sets. Then
A is closed.
A∈G
!c
TRUE:
\
A
[
=
A∈G
Ac . Since A is closed for all A ∈ G, Ac is
A∈G
open for all A. Since the union of any collection of open sets is open,
!c
[
\
Ac =
A
A∈G
is open. Therefore,
\
A∈G
A is closed.
A∈G
[
(b) Let F be a collection of closed sets. Then
B is closed.
B∈G
FALSE:
Let B1 =
1
,2
3 3
, B2 =
1
,3
4 4
, B3 =
∞
[
n=1
which is open.
1
1
,4
5 5
, . . . , Bn =
Bn = (0, 1)
1
n
, n−1
, . . ..
n
(c) If p is an isolated point of S, then p is a boundary point of S.
TRUE: Since p is an isolated point of S, p ∈ S and there is a neighborhood N (p) such that the deleted neighborhood N ∗ (p) ∩ S = ∅. That
is, N ∗ (p) ⊆ S c . Now let U (p) be any neighborhood of p. Then, either
U (p) ⊆ N (p) or N (p) ⊆ U (p). In either case, U (p) ∩ S 6= ∅ (e.g.,
p ∈ U (p) ∩ S) and U (p) ∩ S c 6= ∅ since N ∗ (p) ⊆ S c .
(d) If x is an interior point of S, then x is an accumulation point of S.
TRUE: Let p be an interior point of S, and let N (p, ) be any neighborhood of p. Since p is an interior point of S, there is a neighborhood
N (p, δ) of p such that N (p, δ) ⊆ S. Let γ = min {, δ}. Then
N (p, γ) ⊆ N (p, δ) ⊆ S; every point in N (p, γ) is in S. Also, since
N (p, γ) ⊆ N (p, ), N (p, ) contains points of S other than p.
(e) If x is a boundary point of S, then x is an accumulation point of S.
FALSE: Let S = [0, 1] ∪ {2}. Then 2 is a boundary point of S but 2
is not an accumulation point of S. (2 is an isolated point of S.)
(f) If x ∈ S and x is not an isolated point of S, then x is an accumulation
point of S.
TRUE: If x ∈ S and x is an isolated point of S, then there is a deleted
neighborhood N ∗ (x) such that N ∗ (x) ∩ S = ∅.
If x not an isolated point of S, then for every deleted neighborhood
N ∗ (x), N ∗ (x) ∩ S 6= ∅. This is the definition of accumulation point of S,
i.e., x must be an accumulation point.
2
(g) If α = sup S, then α is an accumulation point of S.
FALSE:
You can use the example in (e): 2 = sup S but 2 is not an accumulation
point of S.
3
2. For each of the following sets, give the interior points, the boundary points, the
accumulation points and the isolated points.
(a) S = (−1, 1) ∪ [1, 2) ∪ [2, 5)
S = (−1, 5) : int S = (−1, 5),
isolated points.
bd S = {−1, 5},
S 0 = [−1, 5],
no
(b) S = [0, 3) ∩ (1, 4) ∪ {−1, 5}
S = (1, 3) ∪ {−1, 5} : int S = (1, 3),
isolated points: {−1, 5}.
(c) S = {r ∈ Q : 0 < r <
int S = ∅,
(d) S =
√
bd S = {−1, 1, 3, 5},
S 0 = [1, 3],
2}
√
bd S = [0, 2],
√
S 0 = [0, 2],
no isolated points.
n
: n∈N
n2 + 1
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] :
no isolated points.
int S = (1, 3) ∪ (3, 5),
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bd S = {1, 3, 5},
S 0 = [1, 5],
(f) S =
1
: n ∈ N ∪ [1, ∞)
n
S = 1, 12 , 31 , 14 , 15 , · · · ∪ [1, ∞)
bd S = {0} ∪ 1, 21 , 13 , 14 , 15 , · · · ,
S 0 = [1, ∞) ∪ {0}, isolated points: 21 , 31 , 14 , 15 , · · · .
int S = (1, ∞),
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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 s1 be any element of S. By Theorem 3 (lecture notes), there exists
s2 ∈ S such that
s1 < s2 < α.
Now assume we have selected s1 , s2 , . . . , sn ∈ S such that
s1 < s2 < . . . < sn < α.
By Theorem 3, there is an element sn+1 ∈ S such that sn < sn+1 < α.
That is, we have
s1 < s2 < . . . < sn < sn+1 < α.
It follows by mathematical induction that there exists and infinite set of
elements {s1 , s2 , s3 , . . .} ∈ S (which are all distinct and all less than α).
4. Let I = (a, b) be an open interval. Prove that I is an open set.
Choose any point p ∈ I. Then a < p < b. Let = min {p − a, b − p} and
consider the neighborhood N (p, ) of p. Choose any point x ∈ N (p, ). Then
p−x<p−a
⇒x>a
and
x−p<b−p
⇒ x < b.
Therefore x ∈ I which implies that N (p, ) ⊆ I and I is an open set.
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5. Determine whether the given set is open, closed, or neither open nor closed.
(a) S = (−1, 2) ∪ [3, ∞)
Neither
Not open: 3 is not an interior point.
Not closed: −1 and 2 are boundary points of S which are not in S.
(b) S = (−∞, 1) ∪ (2, ∞) Open: every point of S is an interior point of S.
(c) S = [−3, 2] ∪ [7, 8] Closed: The union of two closed sets is closed. Or,
S contains all its boundary points.
n−1
1 2 3
, · · · ∪ [2, 4]
(d) S = 0, , , , · · · ,
2 3 4
n
Neither
1 2 3
n−1
Not open: The points 0, , , , · · · ,
, · · · are not interior points.
2 3 4
n
Not closed: 1 is an accumulation point of S which is not in S.
1 1 1
1
(e) S = 1, , , , · · · , , · · · ∪ (1, 3]
2 3 4
n
Neither
1 1 1
1
Not open: The points 1, , , , · · · , , · · ·
are not interior points.
2 3 4
n
Not closed: 0 is an accumulation point of S which is not in S.
1 1 1
1
(f) S = [−1, 0] ∪ 1, , , , · · · , , · · ·
Closed: S contains all its
2 3 4
n
accumulation points; it contains all its boundary points.
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