Ex4-10

Question:
10. (a) If

f

{\displaystyle \displaystyle{f}}

 is deined and continuous on E, show that

{
a
<
f
<
b
}

{\displaystyle \{a < f < b\}}

 is relatively open and that

{
a
≤
f
≤
b
}

{\displaystyle \{a \le f \le b\}}

and

{
# f
a
}

{\displaystyle \{ f = a\}}

 are relatively closed.
(b) Let

f

{\displaystyle \displaystyle{f}}

 be a finite function on

R
n

{\displaystyle \mathbb{R^n}}

. Show that

f

{\displaystyle \displaystyle{f}}

 is continuous on

R
n

{\displaystyle \mathbb{R^n}}

 if and only if

f

−
1

(
G
)

{\displaystyle f^{-1}(G)}

 is open for every open

G

{\displaystyle G}

 in

R
1

{\displaystyle \mathbb{R^1}}

, or if and only if

f

−
1

(
F
)

{\displaystyle f^{-1}(F)}

is closed for every closed

F

{\displaystyle F}

 in

R
1

{\displaystyle \mathbb{R^1}}

.

Answer:

(a) We first prove two Lemma.

**Lemma 1: The function

f
:
E
→

R

{\displaystyle f: E \to \R}

 is continuous on

E
⊆

R

n

{\displaystyle E \subseteq \R^n}

 if and only if

f

−
1

(
G
)

{\displaystyle f^{-1}(G)}

 is relatively open in

E

{\displaystyle E}

 for every open set

G
⊆

R

{\displaystyle G \subseteq \R}

.**

Proof: Suppose

f
:
E
→

R

{\displaystyle f: E \to \R}

 is continuous on

E

{\displaystyle E}

. Given an open set

G
⊆

R

{\displaystyle G \subseteq \R}

. Let

x

0

∈

f

−
1

(
G
)

{\displaystyle \mathbf{x}\_0 \in f^{-1}(G)}

. Then

f
(

x

0

)
∈
G

{\displaystyle f(\mathbf{x}\_0) \in G}

. Since

G

{\displaystyle G}

 is open in

R

{\displaystyle \R}

, there exits

ε
>
0

{\displaystyle \varepsilon > 0}

 such that

(
f
(

x

0

)
−
ε
,
f
(

x

0

)
ε
)
⊆
G

{\displaystyle (f(\mathbf{x}\_0) - \varepsilon, f(\mathbf{x}\_0) + \varepsilon) \subseteq G}

. Since

f

{\displaystyle f}

 is continuous at

x

0

{\displaystyle \mathbf{x}\_0}

, there exists

δ
>
0

{\displaystyle \delta > 0}

 such that if

x

∈
E

{\displaystyle \mathbf{x} \in E}

 and

‖

x

−

x

0

‖
<
δ

{\displaystyle \|\mathbf{x} - \mathbf{x}\_0\| < \delta}

 then

f
(

x

)
−
f
(

x

0

)

<
ε

{\displaystyle |f(\mathbf{x}) - f(\mathbf{x}\_0)| < \varepsilon}

. Thus, there is a ball

B
δ

(

x

0

)

{\displaystyle B\_\delta(\mathbf{x}\_0)}

 such that

f
(
E
∩

B
δ

(

x

0

)
)
⊆
(
f
(

x

0

)
−
ε
,
f
(

x

0

)
ε
)
⊆
G
.

{\displaystyle f(E \cap B\_\delta(\mathbf{x}\_0)) \subseteq (f(\mathbf{x}\_0) - \varepsilon, f(\mathbf{x}\_0) + \varepsilon) \subseteq G.}

Then, we have

E
∩

B
δ

(

x

0

)
⊆

f

−
1

(
G
)

{\displaystyle E \cap B\_\delta(\mathbf{x}\_0) \subseteq f^{-1}(G)}

. Hence,

f

−
1

(
G
)

{\displaystyle f^{-1}(G)}

 is relatively open in

E

{\displaystyle E}

.

Conversely, suppose

f

−
1

(
G
)

{\displaystyle f^{-1}(G)}

 is relatively open in

E

{\displaystyle E}

 for every open set

G
⊆

R

{\displaystyle G \subseteq \R}

. Let

x

0

∈
E

{\displaystyle \mathbf{x}\_0 \in E}

. Then for every

ε
>
0

{\displaystyle \varepsilon > 0}

,

f

−
1

(
(
f
(

x

0

)
)
−
ε
,
f
(

x

0

)
ε
)
)

{\displaystyle f^{-1}((f(\mathbf{x}\_0)) - \varepsilon, f(\mathbf{x}\_0) + \varepsilon))}

 is relatively open in

E

{\displaystyle E}

. Thus, there exists

δ
>
0

{\displaystyle \delta > 0}

 such that

B
δ

(

x

0

)
∩
E
⊆

f

−
1

(
(
f
(

x

0

)
)
−
ε
,
f
(

x

0

)
ε
)
)
.

{\displaystyle B\_\delta(\mathbf{x}\_0) \cap E \subseteq f^{-1}((f(\mathbf{x}\_0)) - \varepsilon, f(\mathbf{x}\_0) + \varepsilon)).}

Then we have

f
(

B
δ

(

x

0

)
∩
E
)
⊆
(
f
(

x

0

)
−
ε
,
f
(

x

0

)
ε
)

{\displaystyle f(B\_\delta(\mathbf{x}\_0) \cap E) \subseteq (f(\mathbf{x}\_0) - \varepsilon, f(\mathbf{x}\_0) + \varepsilon)}

. This shows that

f
(

x

)
−
f
(

x

0

)

<
ε

{\displaystyle |f(\mathbf{x}) - f(\mathbf{x}\_0)| < \varepsilon}

 whenever

‖

x

−

x

0

‖
<
δ

{\displaystyle \|\mathbf{x} - \mathbf{x}\_0\| < \delta}

 and

x

∈
E

{\displaystyle \mathbf{x} \in E}

. Hence,

f

{\displaystyle f}

 is continuous at

x

0

{\displaystyle \mathbf{x}\_0}

.

**Lemma 2: The function

f
:
E
→

R

{\displaystyle f: E \to \R}

 is continuous on

E
⊆

R

n

{\displaystyle E \subseteq \R^n}

 if and only if

f

−
1

(
F
)

{\displaystyle f^{-1}(F)}

 is relatively closed in

E

{\displaystyle E}

 for every closed set

F
⊆

R

{\displaystyle F \subseteq \R}

.**

Proof: We first note that

S
⊆
E

{\displaystyle S \subseteq E}

 is relatively closed in

E

{\displaystyle E}

 if and only if

E
∖
S

{\displaystyle E \setminus S}

 is relatively open in

E

{\displaystyle E}

.
Suppose

f
:
E
→

R

{\displaystyle f: E \to \R}

 is continuous on

E

{\displaystyle E}

. Given

F
⊆

R

{\displaystyle F \subseteq \R}

 be a closed set. Then

R

∖
F

{\displaystyle \R \setminus F}

 is open. By Lemma 1,

f

−
1

(

R

∖
F
)

{\displaystyle f^{-1}(\R \setminus F)}

 is relatively open in

E

{\displaystyle E}

. Since

f

−
1

(

R

∖
F
# )
E
∖

f

−
1

(
F
)

{\displaystyle f^{-1}(\R \setminus F) = E \setminus f^{-1}(F)}

, we have

f

−
1

(
F
)

{\displaystyle f^{-1}(F)}

 is relatively closed in

E

{\displaystyle E}

.

Conversely, suppose

f

−
1

(
F
)

{\displaystyle f^{-1}(F)}

 is relatively closed in

E

{\displaystyle E}

 for every closed set

F
⊆

R

{\displaystyle F \subseteq \R}

. Then

R

∖
F

{\displaystyle \R \setminus F}

 is open. Thus

f

−
1

(

R

∖
F
# )
E
∖

f

−
1

(
F
)

{\displaystyle f^{-1}(\R \setminus F) = E \setminus f^{-1}(F)}

 is relatively open in

E

{\displaystyle E}

 for every open set

(

R

∖
F
)
⊆

R

{\displaystyle (\R \setminus F) \subseteq \R}

. Hence, by Lemma 1

f
:
E
→

R

{\displaystyle f: E \to \R}

 is continuous on

E

{\displaystyle E}

.

Now, we have

{
a
<
f
<
b
# }
{

x

∈
E
∣
a
<
f
(

x

)
<
b
}
∩
E

{\displaystyle \{a < f < b\} = \{\mathbf{x} \in E \mid a < f(\mathbf{x}) < b\} \cap E }

=
{

x

∈
E
∣
f
(

x

)
<
b
}
∩
{

x

∈
E
∣
f
(

x

)
>
a
}

{\displaystyle = \{\mathbf{x} \in E \mid f(\mathbf{x}) < b\} \cap \{\mathbf{x} \in E \mid f(\mathbf{x}) > a\} }

=

f

−
1

(
(
−
∞
,
b
)
)
∩

f

−
1

(
(
a
,
∞
)
)

{\displaystyle = f^{-1}((-\infty, b)) \cap f^{-1}((a,\infty))}

Note that

(
a
,
∞
)

{\displaystyle (a, \infty)}

 and

(
−
∞
,
b
)

{\displaystyle (-\infty, b)}

 are open in

R

{\displaystyle \R}

. Since

f

{\displaystyle f}

 is continuous on

E

{\displaystyle E}

, we have

f

−
1

(
(
−
∞
,
b
)
)

{\displaystyle f^{-1}((-\infty, b))}

 and

f

−
1

(
(
a
,
∞
)
)

{\displaystyle f^{-1}((a,\infty))}

 are relatively open in

E

{\displaystyle E}

 by Lemma 1. Thus,

{
a
<
f
<
b
}

{\displaystyle \{a < f < b\}}

 is relatively open.

Similarly, we have

{
a
≤
f
≤
b
# }

f

−
1

(
(
−
∞
,
b
]
)
∩

f

−
1

(
[
a
,
∞
)
)

{\displaystyle \{a \le f \le b\} = f^{-1}((-\infty, b]) \cap f^{-1}([a,\infty))}

. Note that

[
a
,
∞
)

{\displaystyle [a, \infty)}

 and

(
−
∞
,
b
]

{\displaystyle (-\infty, b]}

 are closed in

R

{\displaystyle \R}

. Since

f

{\displaystyle f}

 is continuous on

E

{\displaystyle E}

, we have

f

−
1

(
(
−
∞
,
b
]
)

{\displaystyle f^{-1}((-\infty, b])}

 and

f

−
1

(
[
a
,
∞
)
)

{\displaystyle f^{-1}([a,\infty))}

 are relatively closed by Lemma 2. Thus,

{
a
≤
f
≤
b
}

{\displaystyle \{a \le f \le b\}}

 is relatively closed in

E

{\displaystyle E}

.

Moreover, note that

{
# }

f

−
1

(
{
a
}
)

{\displaystyle \{f = a\} = f^{-1}(\{a\})}

. Since

{
a
}

{\displaystyle \{a\} }

 is closed in

R

{\displaystyle \R}

, we have

{
# f
a
}

{\displaystyle \{ f = a\}}

 is relatively closed in

E

{\displaystyle E}

.

(b)

***(Statement on Open Sets)***

Assume that

x
∈

{\displaystyle x \in }

f

−
1

(
G
)

{\displaystyle f^{-1}(G)}

, then

f
(
x
)
∈
G

{\displaystyle f(x) \in G }

.

Firstly, we prove that if

f

{\displaystyle \displaystyle{f}}

 is continuous on

R
n

{\displaystyle \mathbb{R^n}}

 then

f

−
1

(
G
)

{\displaystyle f^{-1}(G)}

 is open for every open

G

{\displaystyle G}

 in

R
1

{\displaystyle \mathbb{R^1}}

.

Because G is an open set, there exist a

ε
>
0

{\displaystyle \varepsilon > 0}

,

(
f
(
x
)
−
ε
,
f
(
x
)
ε
)
⊂
G

{\displaystyle (f(x)-\varepsilon,f(x)+\varepsilon) \subset G }

And follow definition of continuous function.

For each

ε
>
0

{\displaystyle \varepsilon > 0}

, there exist a

δ
>
0

{\displaystyle \delta > 0}

 such that,

f
(

x
1

)
−
f
(

x
0

)

<
ε
,

{\displaystyle |f(x\_1)-f(x\_0)| < \varepsilon, }

 whenever

x
1

−

x
0

<
δ

{\displaystyle |x\_1-x\_0|<\delta }

.

f
(

x
1

)
−
f
(
x
)

<
ε
,

{\displaystyle |f(x\_1)-f(x)| < \varepsilon, }

 then

f
(

x
1

)
⊂
G

{\displaystyle f(x\_1) \subset G }

 for

x
1

−
x

<
δ

{\displaystyle |x\_1-x|<\delta }

.

x
1

∈

{\displaystyle x\_1 \in }

f

−
1

(
G
)

{\displaystyle f^{-1}(G)}

 for

x
1

−
x

<
δ

{\displaystyle |x\_1-x|<\delta }

,

(
x
−
δ
,
x
δ
)
⊂

{\displaystyle (x-\delta,x+\delta) \subset }

f

−
1

(
G
)

{\displaystyle f^{-1}(G)}

,

Then we can get that

f

−
1

(
G
)

{\displaystyle f^{-1}(G)}

 is open set.

Secondly , we prove that if

f

−
1

(
G
)

{\displaystyle f^{-1}(G)}

 is open for every open

G

{\displaystyle G}

 in

R
1

{\displaystyle \mathbb{R^1}}

 then

f

{\displaystyle \displaystyle{f}}

 is continuous on

R
n

{\displaystyle \mathbb{R^n}}

.

Assume

ε
>
0
,
x
∈

R
1

.

{\displaystyle \varepsilon >0,x \in \mathbb{R^1}.
 }

# G
(
f
(
x
)
−
ε
,
f
(
x
)
ε
)

{\displaystyle G=(f(x)-\varepsilon,f(x)+\varepsilon) }

 is an open set in

R
1

{\displaystyle \mathbb{R^1}}

 and

f

−
1

(
G
)

{\displaystyle f^{-1}(G)}

 also an open set.

Hence, there exist

δ
<
0

{\displaystyle \delta<0 }

 such that,

(
x
−
δ
,
x
δ
)
⊂

{\displaystyle (x-\delta,x+\delta) \subset }

f

−
1

(
G
)

{\displaystyle f^{-1}(G)}

, And

x
1

∈
(
x
−
δ
,
x
δ
)
,
f
(

x
1

)
∈
G

{\displaystyle x\_1 \in (x-\delta,x+\delta) ,f(x\_1) \in G }

,

We can get that,

f
(

x
1

)
−
f
(
x
)

<
ε
,

{\displaystyle |f(x\_1)-f(x)| < \varepsilon, }

 whenever

x
1

−
x

<
δ

{\displaystyle |x\_1-x|<\delta }

.

Therefore,

f

{\displaystyle \displaystyle{f}}

 is continuous on

R
n

{\displaystyle \mathbb{R^n}}

.

***(Statement on Closed Sets)***

*Using the fact that a set is open if and only if its complement is closed, the statement on the closed sets immediately follows from that of open sets. Nevertheless, we can still give a direct proof as follows.*

Suppose

f

{\displaystyle f}

  is continuous on

R

n

{\displaystyle \R^n}

. Let

F
⊆

R

{\displaystyle F \subseteq \R}

 be a closed set. Suppose

{

x

k

}

{\displaystyle \{\mathbf{x}\_k\}}

 is a sequence in

f

−
1

(
F
)

{\displaystyle f^{-1}(F)}

 that converges to some

x

0

∈

R

n

{\displaystyle \mathbf{x}\_0 \in \R^n}

. Since

f

{\displaystyle f}

 is continuous at

x

0

{\displaystyle \mathbf{x}\_0}

, the sequence

{
f
(

x

k

)
}

{\displaystyle \{f(\mathbf{x}\_k)\}}

 is a sequence in

F

{\displaystyle F}

 that converges to the point

f
(

x

0

)
∈

R

{\displaystyle f(\mathbf{x}\_0) \in \R}

. But since

F

{\displaystyle F}

 is closed in

R

{\displaystyle \R}

, we have

f
(

x

0

)
∈
F

{\displaystyle f(\mathbf{x}\_0) \in F}

. Therefore,

x

0

∈

f

−
1

(
F
)

{\displaystyle \mathbf{x}\_0 \in f^{-1}(F)}

. This proves that

f

−
1

(
F
)

{\displaystyle f^{-1}(F)}

 is closed in

R

n

{\displaystyle \R^n}

.

Conversely, suppose that

f

{\displaystyle f}

 is not continuous at some

x

0

∈

R

n

{\displaystyle \mathbf{x}\_0 \in \R^n}

. We want to show that there is a closed set

F
⊆

R

{\displaystyle F \subseteq \R}

 such that

f

−
1

(
F
)

{\displaystyle f^{-1}(F)}

 is not closed in

R

n

{\displaystyle \R^n}

. There exists

ε
>
0

{\displaystyle \varepsilon > 0}

 such that for any

δ
>
0

{\displaystyle \delta > 0}

, there is a point

x

∈

B
δ

(

x

0

)

{\displaystyle \mathbf{x} \in B\_\delta(\mathbf{x}\_0)}

 such that

f
(

x

)
−
f
(

x

0

)

≥
ε
.

{\displaystyle |f(\mathbf{x}) - f(\mathbf{x}\_0)| \ge \varepsilon.}

For

k
∈

Z

{\displaystyle k\in\mathbb{Z}^+}

, let

x

k

∈

B

1
k

(

x

0

)

{\displaystyle \mathbf{x}\_k \in B\_{\frac{1}{k}}(\mathbf{x}\_0)}

 be such that

f
(

x

k

)
−
f
(

x

0

)

≥
ε
.

{\displaystyle |f(\mathbf{x}\_k) - f(\mathbf{x}\_0)| \ge \varepsilon.}

Since

‖

x

k

−

x

0

‖
<

1
k

{\displaystyle \|\mathbf{x}\_k - \mathbf{x}\_0\| < \dfrac{1}{k}}

 for all

k
∈

Z

{\displaystyle k\in\mathbb{Z}^+}

, the sequence

{

x

k

}

{\displaystyle \{\mathbf{x}\_k\}}

 in

R

n

{\displaystyle \R^n}

 converges to

x

0

{\displaystyle \mathbf{x}\_0}

. Let

# F
{
a
∈

R

∣

a
−
f
(

x

0

)

≥
ε
}
.

{\displaystyle F = \{a \in \R \mid |a - f(\mathbf{x}\_0)| \ge \varepsilon\}.}

Then

# F

(

B
ε

(
f
(

x

0

)

)

c

{\displaystyle F = \left(B\_\varepsilon(f(\mathbf{x}\_0)\right)^c}

, so that

F

{\displaystyle F}

 is closed in

R

{\displaystyle \R}

 and

f
(

x

k

)
∈
F

{\displaystyle f(\mathbf{x}\_k) \in F}

 for all

k
∈

Z

{\displaystyle k\in\mathbb{Z}^+}

. But

f
(

x

0

)
∉
F

{\displaystyle f(\mathbf{x}\_0) \notin F}

. Therefore, the set

f

−
1

(
F
)

{\displaystyle f^{-1}(F)}

 contains the sequence

{

x

k

}

{\displaystyle \{\mathbf{x}\_k\}}

 but does not contain its limit point

x

0

{\displaystyle \mathbf{x}\_0}

. This implies that

f

−
1

(
F
)

{\displaystyle f^{-1}(F)}

 is not closed.

*Proof on the Statements of Closed Sets using Sets Complement*

First suppose that

f
:

R

n

→

R

{\displaystyle f: \R^n \to \R}

 is continuous. Let

F
⊆

R

{\displaystyle F \subseteq \R}

 be a closed set. Then

R

∖
F

{\displaystyle \R \setminus F}

 is open. Thus by the result on open sets we have

f

−
1

(

R

∖
F
)

{\displaystyle f^{-1}(\R \setminus F)}

 is open in

R

n

{\displaystyle \R^n}

. Since

f

−
1

(

R

∖
F
# )

R

n

∖

f

−
1

(
F
)

{\displaystyle f^{-1}(\R \setminus F) = \R^n \setminus f^{-1}(F)}

, we find that

f

−
1

(
F
)

{\displaystyle f^{-1}(F)}

 is closed in

R

n

{\displaystyle \R^n}

.

Conversely, suppose that

f

−
1

(
F
)

{\displaystyle f^{-1}(F)}

 is closed in

R

n

{\displaystyle \R^n}

 for every closed subset

F
⊆

R

{\displaystyle F \subseteq \R}

. Let

G
⊆

R

{\displaystyle G \subseteq \R}

 be an open set. Then

R

∖
G

{\displaystyle \R \setminus G}

 is closed. Thus

R

n

∖

f

−
1

(
G
# )

f

−
1

(

R

∖
G
)

{\displaystyle \R^n \setminus f^{-1}(G) = f^{-1}(\R \setminus G)}

 is closed in

R

n

{\displaystyle \R^n}

. It follows that

f

−
1

(
G
)

{\displaystyle f^{-1}(G)}

 is open for every open set

G
⊆

R

{\displaystyle G \subseteq \R}

, which then implies that

f
:

R

n

→

R

{\displaystyle f: \R^n \to \R}

 is continuous.