SOLUTION OF HW1
MINGFENG ZHAO
September 10, 2012
1. [2 Points, Exercises 5, on Page 27] Describe the following sets as open, closed, or neither, and find
S 0 , (S c )0 , and (S 0 )c .
a. S = (−1, 2)
S
[3, ∞).
S
b. S = (−∞, 1) (2, ∞).
S
c. S = [−3, −2] [7, 8].
d. S = {x : x = integer}.
S
Proof. a. Since S = (−1, 2) [3, ∞), then
(1) S is neither open nor closed.
S
(2) S 0 = (−1, 2) (3, ∞).
S
(3) S c = (−∞, −1] [2, 3).
S
(4) (S c )0 = (−∞, −1) (2, 3).
S
(5) (S 0 )c = (−∞, −1] [2, 3].
b. Since S = (−∞, 1)
S
(2, ∞), then
(1) S is open.
S
(2) S 0 = (−∞, 1) (2, ∞).
(3) S c = [1, 2].
(4) (S c )0 = (1, 2).
(5) (S 0 )c = [1, 2].
S
c. Since S = [−3, −2] [7, 8], then
1
2
MINGFENG ZHAO
(1) S is closed.
(2) S 0 = (−3, −2)
S
(7, 8).
S
S
(3) S c = (−∞, −3) (−2, 7) (8, ∞).
S
S
(4) (S c )0 = (−∞, −3) (−2, 7) (8, ∞).
S
S
(5) (S 0 )c = (−∞, −3] [−2, 7] [8, ∞).
d. Since S = {x : x = integer}, then
(1) S is closed.
(2) S 0 = ∅.
(3) S c = R\Z.
(4) (S c )0 = R\Z.
(5) (S 0 )c = R.
2. [2 Points, Exercises 11, on Page 28] Find the set of limit points of S, ∂S, S, the set of isolated
points of S, and the exterior of S.
a. S = (−∞, −2)
S
(2, 3)
S
{4}
S
(7, ∞).
b. S = {all integers}.
c. S =
S
d. S =
{(n, n + 1) : n = integer}.
1
x : x = , n = 1, 2, 3, · · · .
n
S
S
S
Proof. a. Since S = (−∞, −2) (2, 3) {4} (7, ∞), then
(1) The set of limit points of S is (−∞, −2]
S
S
[2, 3] [7, ∞).
(2) ∂S = {−2, 2, 3, 4, 7}.
(3) S = (−∞, −2]
S
[2, 3]
S
{4}
S
[7, ∞).
(4) The set of isolated points of S is {4}.
S
S
(5) The exterior of S is (−2, 2) (3, 4) (4, 7).
SOLUTION OF HW1
3
b. Since S = {all integers}, then
(1) The set of limit points of S is ∅.
(2) ∂S = Z.
(3) S = Z.
(4) The set of isolated points of S is Z.
(5) The exterior of S is R\Z.
c. Since S =
S
{(n, n + 1) : n = integer}, then
(1) The set of limit points of S is R.
(2) ∂S = Z.
(3) S = R.
(4) The set of isolated points of S is ∅.
(5) The exterior of S is ∅.
d. Since S =
1
x : x = , n = 1, 2, 3, · · · , then
n
(1) The set of limit points of S is {0}.
[
1
: n = 1, 2, 3, · · ·
{0}.
(2) ∂S =
n
[
1
(3) S =
: n = 1, 2, 3, · · ·
{0}.
n
1
(4) The set of isolated points of S is
: n = 1, 2, 3, · · · .
n
! [
∞ [ [
1
1
(5) The exterior of S is (−∞, 0)
,
(1, ∞).
n+1 n
n=1
3. [3 Points, Exercises 17, on Page 28] Prove: If the nonempty subset S of R is closed and bounded,
then inf S and sup S are both in S.
Proof. Since S is nonempty and bounded, then there exist α, β ∈ R such that
α = sup S
and β = inf S.
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MINGFENG ZHAO
Claim I: α ∈ S.
If the Claim I is not true, that is, α ∈
/ S, then α ∈ S c . Since S is closed, then S c is open. Hence
there exists some 0 > 0 such that (α − 0 , α + 0 ) ⊂ S, that is,
(1)
(α − 0 , α + 0 )
\
S = ∅.
By the definition of α = sup S, then for the above 0 , there exists some x0 ∈ S such that α − 0 <
x0 ≤ α, in particular, we get
x0 ∈ (α − 0 , α + 0 )
\
S.
Which contradicts with (1). Hence, we must have α ∈ S.
Claim II: β ∈ S.
If the Claim II is not true, that is, β ∈
/ S, then β ∈ S c . Since S is closed, then S c is open. Hence
there exists some 0 > 0 such that (β − 0 , β + 0 ) ⊂ S, that is,
(2)
(β − 0 , β + 0 )
\
S = ∅.
By the definition of β = inf S, then for the above 0 , there exists some x0 ∈ S such that β + 0 >
x0 ≥ β, in particular, we get
x0 ∈ (β − 0 , β + 0 )
\
S.
Which contradicts with (2). Hence, we must have β ∈ S.
By the Claim I and the Claim II, then we know that inf S and sup S are both in S.
4. [3 Points, Exercises 3, on Page 531] a. Suppose that (A, ρ) is a metric space, and define
ρ1 (u, v) =
ρ(u, v)
,
1 + ρ(u, v)
∀u, v ∈ A.
Show that (A, ρ1 ) is a metric space.
b. Show that infinitely many metrics can be defined on any set A with more than one member.
SOLUTION OF HW1
5
Proof. a. Non-negativity: For any u, v ∈ A, since ρ is a metric on A, then ρ(u, v) ≥ 0, in particular,
we have
ρ1 (u, v) =
ρ(u, v)
≥ 0.
1 + ρ(u, v)
Positive Definite: For any u ∈ A, since ρ is a metric on A, then ρ(u, u) = 0, in particular, we have
ρ1 (u, u) =
ρ(u, u)
= 0.
1 + ρ(u, u)
If ρ1 (u, v) = 0 for some u, v ∈ A, that is,
ρ1 (u, v) =
ρ(u, v)
= 0.
1 + ρ(u, v)
Which implies that ρ(u, v) = 0, hence u = v. Hence, we know that ρ1 (u, v) = 0 if and only if u = v.
Symmetric: For any u, v ∈ A, since ρ is a metric on A, then ρ(u, v) = ρ(v, u). Hence we get
ρ1 (u, v) =
ρ(v, u)
ρ(u, v)
=
= ρ1 (v, u).
1 + ρ(u, v)
1 + ρ(v, u)
Triangle Inequality: For any u, v, w ∈ A, since ρ is a metric on A, then
ρ(u, v) ≤ ρ(u, w) + ρ(w, v).
(3)
Now we consider the function f (t) =
f 0 (t) =
t
for t > −1, then
1+t
1+t−t
1
=
> 0,
(1 + t)2
(1 + t)2
∀t > −1.
Hence f is strictly increasing on (−1, ∞). By (3), then we have
f (ρ(u, v)) ≤ f (ρ(u, w) + ρ(w, v)).
That is, we get
ρ1 (u, v)
=
ρ(u, v)
1 + ρ(u, v)
= f (ρ(u, v))
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MINGFENG ZHAO
≤ f (ρ(u, w) + ρ(w, v))
≤
ρ(u, w) + ρ(w, v)
1 + ρ(u, w) + ρ(w, v)
=
ρ(w, v)
ρ(u, w)
+
1 + ρ(u, w) + ρ(w, v) 1 + ρ(u, w) + ρ(w, v)
≤
ρ(u, w)
ρ(w, v)
+
1 + ρ(u, w) 1 + ρ(w, v)
Since ρ(u, w) ≥ 0 and ρ(w, v) ≥ 0
= ρ1 (u, w) + ρ1 (w, v).
In summary, we can conclude that ρ1 is a metric on A.
b. Since A has more than one element, then we can define the discrete metric on A: for any u, v ∈ A,
we define
ρ1 (u, v) =




 1,
if u 6= v



 0,
if u = v.
Now for any n ∈ N, we define
ρn (u, v) = nρ(u, v),
∀u, v ∈ A.
Obviously, ρn is a metric on A for all n ∈ N. Since A has more than one element, then we can find
u, v ∈ A such that u 6= v. Hence ρ1 (u, v) 6= 0, which implies that
ρi (u, v) 6= ρj (u, v),
∀i 6= j.
Hence, {ρn }∞
n=1 is a sequence of different metrics on A. Therefore, we can conclude that infinitely
many metrics can be defined on any set A with more than one member.
Remark 1. For the part b, let ρ1 be the discrete metric on A, we also can use the following inductive
definition: for any a, c ∈ R+ and any b ≥ 0, let
ρca,b (u, v) =
cρ1 (u, v)
,
a + bρ1 (u, v)
∀u, v ∈ A.
SOLUTION OF HW1
Notice that if we consider the function g(t) =
g 0 (t)
7
ct
for all t ≥ 0, then
a + bt
=
c(a + bt) − bct
(a + bt)2
=
ac
(a + bt)2
> 0,
∀t ≥ 0.
That is, g is strictly increasing.
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu