## Advanced Calculus (II)

W^{EN}-C^{HING}L^{IEN}

Department of Mathematics National Cheng Kung University

2009

## Ch8: Euclidean Spaces

### 8.4: Interior, Closure, and, Boundary

Definition (8.31)

Let E be a subset of a Euclidean Space**R**^{n}.
(i) The interior of E is the set

E^{o}:=[

**{V : V ⊆ E and V is open in R**^{n}}.

(ii) The closure of E is the set E :=\

**{B : B ⊇ E and B is closed in R**^{n}}.

Let E ⊆**R**^{n}.Then
(i) E^{o} ⊆ E ⊆ E,

(ii) if V is open and V ⊆ E then V ⊆ E^{o},and
(iii) if C is closed and C ⊇ E then C ⊇ E .
Proof.

Since every open set V in the union defining E^{o} is a
subset of E , it is clear that the union of these V ’s is a
subset of E . Thus E^{o} ⊆ E. A similar argument establishes
E ⊆ E . This proves (i).

By Definition 8.31, if V is an open subset of E , then
V ⊆ E^{o} and if C is a closed set containing E , then E ⊆ C.

This proves (ii) and (iii).

Theorem (8.32)
Let E ⊆**R**^{n}.Then

(i) E^{o} ⊆ E ⊆ E,

(ii) if V is open and V ⊆ E then V ⊆ E^{o},and
(iii) if C is closed and C ⊇ E then C ⊇ E .
Proof.

Since every open set V in the union defining E^{o} is a
subset of E , it is clear that the union of these V ’s is a
subset of E . Thus E^{o} ⊆ E. A similar argument establishes
E ⊆ E . This proves (i).

By Definition 8.31, if V is an open subset of E , then
V ⊆ E^{o} and if C is a closed set containing E , then E ⊆ C.

This proves (ii) and (iii).

Let E ⊆**R**^{n}.Then
(i) E^{o} ⊆ E ⊆ E,

(ii) if V is open and V ⊆ E then V ⊆ E^{o},and
(iii) if C is closed and C ⊇ E then C ⊇ E .
Proof.

Since every open set V in the union defining E^{o} is a
subset of E ,it is clear that the union of these V ’s is a
subset of E . Thus E^{o} ⊆ E. A similar argument establishes
E ⊆ E . This proves (i).

By Definition 8.31, if V is an open subset of E , then
V ⊆ E^{o} and if C is a closed set containing E , then E ⊆ C.

This proves (ii) and (iii).

Theorem (8.32)
Let E ⊆**R**^{n}.Then

(i) E^{o} ⊆ E ⊆ E,

(ii) if V is open and V ⊆ E then V ⊆ E^{o},and
(iii) if C is closed and C ⊇ E then C ⊇ E .
Proof.

Since every open set V in the union defining E^{o} is a
subset of E , it is clear that the union of these V ’s is a
subset of E .Thus E^{o} ⊆ E. A similar argument establishes
E ⊆ E . This proves (i).

^{o} and if C is a closed set containing E , then E ⊆ C.

This proves (ii) and (iii).

Let E ⊆**R**^{n}.Then
(i) E^{o} ⊆ E ⊆ E,

(ii) if V is open and V ⊆ E then V ⊆ E^{o},and
(iii) if C is closed and C ⊇ E then C ⊇ E .
Proof.

Since every open set V in the union defining E^{o} is a
subset of E ,it is clear that the union of these V ’s is a
subset of E . Thus E^{o} ⊆ E.A similar argument establishes
E ⊆ E . This proves (i).

^{o} and if C is a closed set containing E , then E ⊆ C.

This proves (ii) and (iii).

Theorem (8.32)
Let E ⊆**R**^{n}.Then

(i) E^{o} ⊆ E ⊆ E,

(ii) if V is open and V ⊆ E then V ⊆ E^{o},and
(iii) if C is closed and C ⊇ E then C ⊇ E .
Proof.

Since every open set V in the union defining E^{o} is a
subset of E , it is clear that the union of these V ’s is a
subset of E .Thus E^{o} ⊆ E. A similar argument establishes
E ⊆ E . This proves (i).

^{o} and if C is a closed set containing E , then E ⊆ C.

This proves (ii) and (iii).

Let E ⊆**R**^{n}.Then
(i) E^{o} ⊆ E ⊆ E,

(ii) if V is open and V ⊆ E then V ⊆ E^{o},and
(iii) if C is closed and C ⊇ E then C ⊇ E .
Proof.

Since every open set V in the union defining E^{o} is a
subset of E , it is clear that the union of these V ’s is a
subset of E . Thus E^{o} ⊆ E.A similar argument establishes
E ⊆ E . This proves (i).

By Definition 8.31, if V is an open subset of E ,then
V ⊆ E^{o} and if C is a closed set containing E , then E ⊆ C.

This proves (ii) and (iii).

Theorem (8.32)
Let E ⊆**R**^{n}.Then

(i) E^{o} ⊆ E ⊆ E,

(ii) if V is open and V ⊆ E then V ⊆ E^{o},and
(iii) if C is closed and C ⊇ E then C ⊇ E .
Proof.

Since every open set V in the union defining E^{o} is a
subset of E , it is clear that the union of these V ’s is a
subset of E . Thus E^{o} ⊆ E. A similar argument establishes
E ⊆ E . This proves (i).

By Definition 8.31, if V is an open subset of E , then
V ⊆ E^{o} and if C is a closed set containing E ,then E ⊆ C.

This proves (ii) and (iii).

Let E ⊆**R**^{n}.Then
(i) E^{o} ⊆ E ⊆ E,

(ii) if V is open and V ⊆ E then V ⊆ E^{o},and
(iii) if C is closed and C ⊇ E then C ⊇ E .
Proof.

^{o} is a
subset of E , it is clear that the union of these V ’s is a
subset of E . Thus E^{o} ⊆ E. A similar argument establishes
E ⊆ E . This proves (i).

By Definition 8.31, if V is an open subset of E ,then
V ⊆ E^{o} and if C is a closed set containing E , then E ⊆ C.

This proves (ii) and (iii).

Theorem (8.32)
Let E ⊆**R**^{n}.Then

(i) E^{o} ⊆ E ⊆ E,

(ii) if V is open and V ⊆ E then V ⊆ E^{o},and
(iii) if C is closed and C ⊇ E then C ⊇ E .
Proof.

^{o} is a
subset of E , it is clear that the union of these V ’s is a
subset of E . Thus E^{o} ⊆ E. A similar argument establishes
E ⊆ E . This proves (i).

By Definition 8.31, if V is an open subset of E , then
V ⊆ E^{o} and if C is a closed set containing E ,then E ⊆ C.

This proves (ii) and (iii).

Let E ⊆**R**^{n}.Then
(i) E^{o} ⊆ E ⊆ E,

(ii) if V is open and V ⊆ E then V ⊆ E^{o},and
(iii) if C is closed and C ⊇ E then C ⊇ E .
Proof.

^{o} is a
subset of E , it is clear that the union of these V ’s is a
subset of E . Thus E^{o} ⊆ E. A similar argument establishes
E ⊆ E . This proves (i).

^{o} and if C is a closed set containing E , then E ⊆ C.

This proves (ii) and (iii).

Definition (8.34)

Let E ⊆**R**^{n}.The boundary of E is the set

∂E := {x ∈ R^{n}: for all r > 0, Br(x) ∩ E 6= ∅ and
Br(x) ∩ E^{c} 6= ∅}.

Let E ⊆**R**^{n}.Then ∂E = E \E^{o}.
Proof.

By Definition 8.34, it suffices to show that

(10) **x ∈ E if and only if B**_{r}(x) ∩ E 6= ∅ for all r > 0, and
(11) **x /**∈ E^{o} if and only if Br(x) ∩ E^{c} 6= ∅ for all r > 0.

We will provide the details for (10) and leave the proof of
(11) as an exercise. Suppose that**x ∈ E but B**_{r}_{0}(x) ∩ E = ∅
for some r0>0. Then (Br_{0}(x))^{c} is a closed set that

contains E ; hence, by theorem 8.32iii, E ⊆ (Br0(x))^{c}.It
follows that E ∩ Br0(x) = ∅, e.g., x /∈ E, a contradiction.

Conversely, suppose that**x /**∈ E. Since (E)^{c} is open,
there is an r0 >0 such that Br0(x) ⊆ (E )^{c}.In particular,

∅ = B_{r}_{0}(x) ∩ E ⊇ B_{r}_{0}(x) ∩ E for some r_{0} >0.

Theorem (8.36)

Let E ⊆**R**^{n}.Then ∂E = E \E^{o}.
Proof.

By Definition 8.34, it suffices to show that

(10) **x ∈ E if and only if B**_{r}(x) ∩ E 6= ∅ for all r > 0,and
(11) **x /**∈ E^{o} if and only if Br(x) ∩ E^{c} 6= ∅ for all r > 0.

We will provide the details for (10) and leave the proof of
(11) as an exercise.Suppose that**x ∈ E but B**_{r}_{0}(x) ∩ E = ∅
for some r0>0. Then (Br_{0}(x))^{c} is a closed set that

contains E ; hence, by theorem 8.32iii, E ⊆ (Br0(x))^{c}.It
follows that E ∩ Br0(x) = ∅, e.g., x /∈ E, a contradiction.

Conversely, suppose that**x /**∈ E. Since (E)^{c} is open,
there is an r0 >0 such that Br0(x) ⊆ (E )^{c}.In particular,

∅ = B_{r}_{0}(x) ∩ E ⊇ B_{r}_{0}(x) ∩ E for some r_{0} >0.

Let E ⊆**R**^{n}.Then ∂E = E \E^{o}.
Proof.

By Definition 8.34, it suffices to show that

(10) **x ∈ E if and only if B**_{r}(x) ∩ E 6= ∅ for all r > 0, and
(11) **x /**∈ E^{o} if and only if Br(x) ∩ E^{c} 6= ∅ for all r > 0.

We will provide the details for (10) and leave the proof of
(11) as an exercise. Suppose that**x ∈ E but B**_{r}_{0}(x) ∩ E = ∅
for some r0>0.Then (Br_{0}(x))^{c} is a closed set that

contains E ; hence, by theorem 8.32iii, E ⊆ (Br0(x))^{c}.It
follows that E ∩ Br0(x) = ∅, e.g., x /∈ E, a contradiction.

Conversely, suppose that**x /**∈ E. Since (E)^{c} is open,
there is an r0 >0 such that Br0(x) ⊆ (E )^{c}.In particular,

∅ = B_{r}_{0}(x) ∩ E ⊇ B_{r}_{0}(x) ∩ E for some r_{0} >0.

Theorem (8.36)

Let E ⊆**R**^{n}.Then ∂E = E \E^{o}.
Proof.

By Definition 8.34, it suffices to show that

(10) **x ∈ E if and only if B**_{r}(x) ∩ E 6= ∅ for all r > 0, and
(11) **x /**∈ E^{o} if and only if Br(x) ∩ E^{c} 6= ∅ for all r > 0.

We will provide the details for (10) and leave the proof of
(11) as an exercise.Suppose that**x ∈ E but B**_{r}_{0}(x) ∩ E = ∅
for some r0>0. Then (Br_{0}(x))^{c} is a closed set that

contains E ;hence, by theorem 8.32iii, E ⊆ (Br0(x))^{c}.It
follows that E ∩ Br0(x) = ∅, e.g., x /∈ E, a contradiction.

**x /**∈ E. Since (E)^{c} is open,
there is an r0 >0 such that Br0(x) ⊆ (E )^{c}.In particular,

∅ = B_{r}_{0}(x) ∩ E ⊇ B_{r}_{0}(x) ∩ E for some r_{0} >0.

Let E ⊆**R**^{n}.Then ∂E = E \E^{o}.
Proof.

By Definition 8.34, it suffices to show that

**x ∈ E if and only if B**_{r}(x) ∩ E 6= ∅ for all r > 0, and
(11) **x /**∈ E^{o} if and only if Br(x) ∩ E^{c} 6= ∅ for all r > 0.

We will provide the details for (10) and leave the proof of
(11) as an exercise. Suppose that**x ∈ E but B**_{r}_{0}(x) ∩ E = ∅
for some r0>0.Then (Br_{0}(x))^{c} is a closed set that

^{c}.It
follows that E ∩ Br0(x) = ∅, e.g., x /∈ E, a contradiction.

**x /**∈ E. Since (E)^{c} is open,
there is an r0 >0 such that Br0(x) ⊆ (E )^{c}.In particular,

∅ = B_{r}_{0}(x) ∩ E ⊇ B_{r}_{0}(x) ∩ E for some r_{0} >0.

Theorem (8.36)

Let E ⊆**R**^{n}.Then ∂E = E \E^{o}.
Proof.

By Definition 8.34, it suffices to show that

**x ∈ E if and only if B**_{r}(x) ∩ E 6= ∅ for all r > 0, and
(11) **x /**∈ E^{o} if and only if Br(x) ∩ E^{c} 6= ∅ for all r > 0.

We will provide the details for (10) and leave the proof of
(11) as an exercise. Suppose that**x ∈ E but B**_{r}_{0}(x) ∩ E = ∅
for some r0>0. Then (Br_{0}(x))^{c} is a closed set that

contains E ;hence, by theorem 8.32iii, E ⊆ (Br0(x))^{c}.It
follows that E ∩ Br0(x) = ∅,e.g.,**x /**∈ E, a contradiction.

**x /**∈ E. Since (E)^{c} is open,
there is an r0 >0 such that Br0(x) ⊆ (E )^{c}.In particular,

∅ = B_{r}_{0}(x) ∩ E ⊇ B_{r}_{0}(x) ∩ E for some r_{0} >0.

Let E ⊆**R**^{n}.Then ∂E = E \E^{o}.
Proof.

By Definition 8.34, it suffices to show that

**x ∈ E if and only if B**_{r}(x) ∩ E 6= ∅ for all r > 0, and
(11) **x /**∈ E^{o} if and only if Br(x) ∩ E^{c} 6= ∅ for all r > 0.

We will provide the details for (10) and leave the proof of
(11) as an exercise. Suppose that**x ∈ E but B**_{r}_{0}(x) ∩ E = ∅
for some r0>0. Then (Br_{0}(x))^{c} is a closed set that

^{c}.It
follows that E ∩ Br0(x) = ∅, e.g., x /∈ E, a contradiction.

**x /**∈ E. Since (E)^{c} is open,
there is an r0 >0 such that Br0(x) ⊆ (E )^{c}.In particular,

∅ = B_{r}_{0}(x) ∩ E ⊇ B_{r}_{0}(x) ∩ E for some r_{0} >0.

Theorem (8.36)

Let E ⊆**R**^{n}.Then ∂E = E \E^{o}.
Proof.

By Definition 8.34, it suffices to show that

**x ∈ E if and only if B**_{r}(x) ∩ E 6= ∅ for all r > 0, and
(11) **x /**∈ E^{o} if and only if Br(x) ∩ E^{c} 6= ∅ for all r > 0.

**x ∈ E but B**_{r}_{0}(x) ∩ E = ∅
for some r0>0. Then (Br_{0}(x))^{c} is a closed set that

contains E ; hence, by theorem 8.32iii, E ⊆ (Br0(x))^{c}.It
follows that E ∩ Br0(x) = ∅,e.g.,**x /**∈ E, a contradiction.

Conversely, suppose that**x /**∈ E. Since (E )^{c} is open,
there is an r0 >0 such that Br0(x) ⊆ (E )^{c}.In particular,

∅ = B_{r}_{0}(x) ∩ E ⊇ B_{r}_{0}(x) ∩ E for some r_{0} >0.

Let E ⊆**R**^{n}.Then ∂E = E \E^{o}.
Proof.

By Definition 8.34, it suffices to show that

**x ∈ E if and only if B**_{r}(x) ∩ E 6= ∅ for all r > 0, and
(11) **x /**∈ E^{o} if and only if Br(x) ∩ E^{c} 6= ∅ for all r > 0.

**x ∈ E but B**_{r}_{0}(x) ∩ E = ∅
for some r0>0. Then (Br_{0}(x))^{c} is a closed set that

^{c}.It
follows that E ∩ Br0(x) = ∅, e.g., x /∈ E, a contradiction.

**x /**∈ E. Since (E)^{c} is open,
there is an r0 >0 such that Br0(x) ⊆ (E )^{c}.In particular,

∅ = B_{r}_{0}(x) ∩ E ⊇ B_{r}_{0}(x) ∩ E for some r_{0} >0.

Theorem (8.36)

Let E ⊆**R**^{n}.Then ∂E = E \E^{o}.
Proof.

By Definition 8.34, it suffices to show that

**x ∈ E if and only if B**_{r}(x) ∩ E 6= ∅ for all r > 0, and
(11) **x /**∈ E^{o} if and only if Br(x) ∩ E^{c} 6= ∅ for all r > 0.

**x ∈ E but B**_{r}_{0}(x) ∩ E = ∅
for some r0>0. Then (Br_{0}(x))^{c} is a closed set that

^{c}.It
follows that E ∩ Br0(x) = ∅, e.g., x /∈ E, a contradiction.

Conversely, suppose that**x /**∈ E. Since (E )^{c} is open,
there is an r0 >0 such that Br0(x) ⊆ (E )^{c}.In particular,

∅ = B_{r}_{0}(x) ∩ E ⊇ B_{r}_{0}(x) ∩ E for some r_{0} >0.

Let E ⊆**R**^{n}.Then ∂E = E \E^{o}.
Proof.

By Definition 8.34, it suffices to show that

**x ∈ E if and only if B**_{r}(x) ∩ E 6= ∅ for all r > 0, and
(11) **x /**∈ E^{o} if and only if Br(x) ∩ E^{c} 6= ∅ for all r > 0.

**x ∈ E but B**_{r}_{0}(x) ∩ E = ∅
for some r0>0. Then (Br_{0}(x))^{c} is a closed set that

^{c}.It
follows that E ∩ Br0(x) = ∅, e.g., x /∈ E, a contradiction.

**x /**∈ E. Since (E)^{c} is open,
there is an r0 >0 such that Br0(x) ⊆ (E )^{c}.In particular,

∅ = B_{r}_{0}(x) ∩ E ⊇ B_{r}_{0}(x) ∩ E for some r_{0} >0.

Theorem (8.36)

Let E ⊆**R**^{n}.Then ∂E = E \E^{o}.
Proof.

By Definition 8.34, it suffices to show that

**x ∈ E if and only if B**_{r}(x) ∩ E 6= ∅ for all r > 0, and
(11) **x /**∈ E^{o} if and only if Br(x) ∩ E^{c} 6= ∅ for all r > 0.

**x ∈ E but B**_{r}_{0}(x) ∩ E = ∅
for some r0>0. Then (Br_{0}(x))^{c} is a closed set that

^{c}.It
follows that E ∩ Br0(x) = ∅, e.g., x /∈ E, a contradiction.

**x /**∈ E. Since (E)^{c} is open,
there is an r0 >0 such that Br0(x) ⊆ (E )^{c}.In particular,

∅ = B_{r}_{0}(x) ∩ E ⊇ B_{r}_{0}(x) ∩ E for some r_{0} >0.

Let E ⊆**R**^{n}.Then ∂E = E \E^{o}.
Proof.

By Definition 8.34, it suffices to show that

**x ∈ E if and only if B**_{r}(x) ∩ E 6= ∅ for all r > 0, and
(11) **x /**∈ E^{o} if and only if Br(x) ∩ E^{c} 6= ∅ for all r > 0.

**x ∈ E but B**_{r}_{0}(x) ∩ E = ∅
for some r0>0. Then (Br_{0}(x))^{c} is a closed set that

^{c}.It
follows that E ∩ Br0(x) = ∅, e.g., x /∈ E, a contradiction.

**x /**∈ E. Since (E)^{c} is open,
there is an r0 >0 such that Br0(x) ⊆ (E )^{c}.In particular,

∅ = B_{r}_{0}(x) ∩ E ⊇ B_{r}_{0}(x) ∩ E for some r_{0} >0.

Theorem (8.37)
Let A, B ⊆**R**^{n} Then

(i) (A ∪ B)^{o}⊇ A^{o}∪ B^{o}, (A ∩ B)^{o} =A^{o}∩ B^{o},
(ii) A ∪ B = A ∪ B, A ∩ B ⊆ A ∩ B,

(iii) ∂(A ∪ B) ⊆ ∂A ∪ ∂B, and ∂(A ∩ B) ⊆ ∂A ∪ ∂B.

Proof.

(i) Since the union of two open sets is open, A^{o}∪ B^{o}is an
open subset of A ∪ B. Hence, by Theorem 8.32ii,

A^{o}∪ B^{o} ⊆ (A ∪ B)^{o}.

Similarly, (A ∩ B)^{o} ⊇ A^{o}∩ B^{o}. On the other hand, if
V ⊂ A ∩ B, then V ⊂ A and V ⊂ B. Thus

(A ∩ B)^{o} ⊆ A^{o}∩ B^{o}.

Let A, B ⊆**R**^{n} Then

(i) (A ∪ B)^{o}⊇ A^{o}∪ B^{o}, (A ∩ B)^{o} =A^{o}∩ B^{o},
(ii) A ∪ B = A ∪ B, A ∩ B ⊆ A ∩ B,

(iii) ∂(A ∪ B) ⊆ ∂A ∪ ∂B, and ∂(A ∩ B) ⊆ ∂A ∪ ∂B.

Proof.

(i) Since the union of two open sets is open,A^{o}∪ B^{o}is an
open subset of A ∪ B. Hence, by Theorem 8.32ii,

A^{o}∪ B^{o} ⊆ (A ∪ B)^{o}.

Similarly, (A ∩ B)^{o} ⊇ A^{o}∩ B^{o}. On the other hand, if
V ⊂ A ∩ B, then V ⊂ A and V ⊂ B. Thus

(A ∩ B)^{o} ⊆ A^{o}∩ B^{o}.

Theorem (8.37)
Let A, B ⊆**R**^{n} Then

(i) (A ∪ B)^{o}⊇ A^{o}∪ B^{o}, (A ∩ B)^{o} =A^{o}∩ B^{o},
(ii) A ∪ B = A ∪ B, A ∩ B ⊆ A ∩ B,

(iii) ∂(A ∪ B) ⊆ ∂A ∪ ∂B, and ∂(A ∩ B) ⊆ ∂A ∪ ∂B.

Proof.

(i) Since the union of two open sets is open, A^{o}∪ B^{o}is an
open subset of A ∪ B.Hence, by Theorem 8.32ii,

A^{o}∪ B^{o} ⊆ (A ∪ B)^{o}.

Similarly, (A ∩ B)^{o} ⊇ A^{o}∩ B^{o}. On the other hand, if
V ⊂ A ∩ B, then V ⊂ A and V ⊂ B. Thus

(A ∩ B)^{o} ⊆ A^{o}∩ B^{o}.

Let A, B ⊆**R**^{n} Then

^{o}⊇ A^{o}∪ B^{o}, (A ∩ B)^{o} =A^{o}∩ B^{o},
(ii) A ∪ B = A ∪ B, A ∩ B ⊆ A ∩ B,

(iii) ∂(A ∪ B) ⊆ ∂A ∪ ∂B, and ∂(A ∩ B) ⊆ ∂A ∪ ∂B.

Proof.

(i) Since the union of two open sets is open,A^{o}∪ B^{o}is an
open subset of A ∪ B. Hence, by Theorem 8.32ii,

A^{o}∪ B^{o} ⊆ (A ∪ B)^{o}.

^{o} ⊇ A^{o}∩ B^{o}. On the other hand, if
V ⊂ A ∩ B, then V ⊂ A and V ⊂ B. Thus

(A ∩ B)^{o} ⊆ A^{o}∩ B^{o}.

Theorem (8.37)
Let A, B ⊆**R**^{n} Then

^{o}⊇ A^{o}∪ B^{o}, (A ∩ B)^{o} =A^{o}∩ B^{o},
(ii) A ∪ B = A ∪ B, A ∩ B ⊆ A ∩ B,

(iii) ∂(A ∪ B) ⊆ ∂A ∪ ∂B, and ∂(A ∩ B) ⊆ ∂A ∪ ∂B.

Proof.

(i) Since the union of two open sets is open, A^{o}∪ B^{o}is an
open subset of A ∪ B.Hence, by Theorem 8.32ii,

A^{o}∪ B^{o} ⊆ (A ∪ B)^{o}.

Similarly, (A ∩ B)^{o} ⊇ A^{o}∩ B^{o}.On the other hand, if
V ⊂ A ∩ B, then V ⊂ A and V ⊂ B. Thus

(A ∩ B)^{o} ⊆ A^{o}∩ B^{o}.

Let A, B ⊆**R**^{n} Then

^{o}⊇ A^{o}∪ B^{o}, (A ∩ B)^{o} =A^{o}∩ B^{o},
(ii) A ∪ B = A ∪ B, A ∩ B ⊆ A ∩ B,

(iii) ∂(A ∪ B) ⊆ ∂A ∪ ∂B, and ∂(A ∩ B) ⊆ ∂A ∪ ∂B.

Proof.

(i) Since the union of two open sets is open, A^{o}∪ B^{o}is an
open subset of A ∪ B. Hence, by Theorem 8.32ii,

A^{o}∪ B^{o} ⊆ (A ∪ B)^{o}.

^{o} ⊇ A^{o}∩ B^{o}. On the other hand, if
V ⊂ A ∩ B, then V ⊂ A and V ⊂ B. Thus

(A ∩ B)^{o} ⊆ A^{o}∩ B^{o}.

Theorem (8.37)
Let A, B ⊆**R**^{n} Then

^{o}⊇ A^{o}∪ B^{o}, (A ∩ B)^{o} =A^{o}∩ B^{o},
(ii) A ∪ B = A ∪ B, A ∩ B ⊆ A ∩ B,

(iii) ∂(A ∪ B) ⊆ ∂A ∪ ∂B, and ∂(A ∩ B) ⊆ ∂A ∪ ∂B.

Proof.

(i) Since the union of two open sets is open, A^{o}∪ B^{o}is an
open subset of A ∪ B. Hence, by Theorem 8.32ii,

A^{o}∪ B^{o} ⊆ (A ∪ B)^{o}.

Similarly, (A ∩ B)^{o} ⊇ A^{o}∩ B^{o}.On the other hand, if
V ⊂ A ∩ B, then V ⊂ A and V ⊂ B. Thus

(A ∩ B)^{o} ⊆ A^{o}∩ B^{o}.

Let A, B ⊆**R**^{n} Then

^{o}⊇ A^{o}∪ B^{o}, (A ∩ B)^{o} =A^{o}∩ B^{o},
(ii) A ∪ B = A ∪ B, A ∩ B ⊆ A ∩ B,

(iii) ∂(A ∪ B) ⊆ ∂A ∪ ∂B, and ∂(A ∩ B) ⊆ ∂A ∪ ∂B.

Proof.

^{o}∪ B^{o}is an
open subset of A ∪ B. Hence, by Theorem 8.32ii,

A^{o}∪ B^{o} ⊆ (A ∪ B)^{o}.

^{o} ⊇ A^{o}∩ B^{o}. On the other hand, if
V ⊂ A ∩ B, then V ⊂ A and V ⊂ B. Thus

(A ∩ B)^{o} ⊆ A^{o}∩ B^{o}.

Theorem (8.37)
Let A, B ⊆**R**^{n} Then

^{o}⊇ A^{o}∪ B^{o}, (A ∩ B)^{o} =A^{o}∩ B^{o},
(ii) A ∪ B = A ∪ B, A ∩ B ⊆ A ∩ B,

(iii) ∂(A ∪ B) ⊆ ∂A ∪ ∂B, and ∂(A ∩ B) ⊆ ∂A ∪ ∂B.

Proof.

^{o}∪ B^{o}is an
open subset of A ∪ B. Hence, by Theorem 8.32ii,

A^{o}∪ B^{o} ⊆ (A ∪ B)^{o}.

^{o} ⊇ A^{o}∩ B^{o}. On the other hand, if
V ⊂ A ∩ B, then V ⊂ A and V ⊂ B. Thus

(A ∩ B)^{o} ⊆ A^{o}∩ B^{o}.

Let A, B ⊆**R**^{n} Then

^{o}⊇ A^{o}∪ B^{o}, (A ∩ B)^{o} =A^{o}∩ B^{o},
(ii) A ∪ B = A ∪ B, A ∩ B ⊆ A ∩ B,

(iii) ∂(A ∪ B) ⊆ ∂A ∪ ∂B, and ∂(A ∩ B) ⊆ ∂A ∪ ∂B.

Proof.

^{o}∪ B^{o}is an
open subset of A ∪ B. Hence, by Theorem 8.32ii,

A^{o}∪ B^{o} ⊆ (A ∪ B)^{o}.

^{o} ⊇ A^{o}∩ B^{o}. On the other hand, if
V ⊂ A ∩ B, then V ⊂ A and V ⊂ B. Thus

(A ∩ B)^{o} ⊆ A^{o}∩ B^{o}.

Proof.

(ii) Since A ∪ B is closed and contains A ∪ B,it is clear
that, by Theorem 8.32iii, A ∪ B ⊆ A ∪ B. Similarly,
A ∩ B ⊆ A ∩ B.To prove the reverse inequlity for union,
suppose that**x /**∈ A ∪ B. Then, by, Definition 8.31, there is
a closed set E that contains A ∪ B such that**x /**∈ E. Since
E contains both A and B, it follows that**x /∈ A and x /**∈ B.

This proves part(ii).

(iii) Let**x ∈ ∂(A ∪ B); i.e., suppose that B**r(x) intersects
A ∪ B and (A ∪ B)^{c} for all r > 0, Since (A ∪ B)^{c} =A^{c}∩ B^{c},
it follows that Br(x) interscets both A^{c} and B^{c} for all r > 0.

Thus Br(x) intersects A and A^{c} for all r > 0, or Br(x)
intersects B and B^{c} for all r > 0; i.e.,**x ∈ ∂A ∪ ∂B. This**
proves the first set inequality in part (iii). A similar
argument establishes the second inequality in part (iii).

(ii) Since A ∪ B is closed and contains A ∪ B, it is clear
that, by Theorem 8.32iii, A ∪ B ⊆ A ∪ B. Similarly,
A ∩ B ⊆ A ∩ B. To prove the reverse inequlity for union,
suppose that**x /**∈ A ∪ B. Then, by, Definition 8.31, there is
a closed set E that contains A ∪ B such that**x /**∈ E. Since
E contains both A and B, it follows that**x /∈ A and x /**∈ B.

This proves part(ii).

(iii) Let**x ∈ ∂(A ∪ B); i.e., suppose that B**r(x) intersects
A ∪ B and (A ∪ B)^{c} for all r > 0, Since (A ∪ B)^{c} =A^{c}∩ B^{c},
it follows that Br(x) interscets both A^{c} and B^{c} for all r > 0.

Thus Br(x) intersects A and A^{c} for all r > 0, or Br(x)
intersects B and B^{c} for all r > 0; i.e.,**x ∈ ∂A ∪ ∂B. This**
proves the first set inequality in part (iii). A similar
argument establishes the second inequality in part (iii).

Proof.

(ii) Since A ∪ B is closed and contains A ∪ B, it is clear
that, by Theorem 8.32iii, A ∪ B ⊆ A ∪ B. Similarly,
A ∩ B ⊆ A ∩ B.To prove the reverse inequlity for union,
suppose that**x /**∈ A ∪ B. Then, by, Definition 8.31, there is
a closed set E that contains A ∪ B such that**x /**∈ E. Since
E contains both A and B, it follows that**x /∈ A and x /**∈ B.

This proves part(ii).

(iii) Let**x ∈ ∂(A ∪ B); i.e., suppose that B**r(x) intersects
A ∪ B and (A ∪ B)^{c} for all r > 0, Since (A ∪ B)^{c} =A^{c}∩ B^{c},
it follows that Br(x) interscets both A^{c} and B^{c} for all r > 0.

Thus Br(x) intersects A and A^{c} for all r > 0, or Br(x)
intersects B and B^{c} for all r > 0; i.e.,**x ∈ ∂A ∪ ∂B. This**
proves the first set inequality in part (iii). A similar
argument establishes the second inequality in part (iii).

(ii) Since A ∪ B is closed and contains A ∪ B, it is clear
that, by Theorem 8.32iii, A ∪ B ⊆ A ∪ B. Similarly,
A ∩ B ⊆ A ∩ B. To prove the reverse inequlity for union,
suppose that**x /**∈ A ∪ B. Then, by, Definition 8.31, there is
a closed set E that contains A ∪ B such that**x /**∈ E.Since
E contains both A and B, it follows that**x /∈ A and x /**∈ B.

This proves part(ii).

**x ∈ ∂(A ∪ B); i.e., suppose that B**r(x) intersects
A ∪ B and (A ∪ B)^{c} for all r > 0, Since (A ∪ B)^{c} =A^{c}∩ B^{c},
it follows that Br(x) interscets both A^{c} and B^{c} for all r > 0.

^{c} for all r > 0, or Br(x)
intersects B and B^{c} for all r > 0; i.e.,**x ∈ ∂A ∪ ∂B. This**
proves the first set inequality in part (iii). A similar
argument establishes the second inequality in part (iii).

Proof.

(ii) Since A ∪ B is closed and contains A ∪ B, it is clear
that, by Theorem 8.32iii, A ∪ B ⊆ A ∪ B. Similarly,
A ∩ B ⊆ A ∩ B. To prove the reverse inequlity for union,
suppose that**x /**∈ A ∪ B. Then, by, Definition 8.31, there is
a closed set E that contains A ∪ B such that**x /**∈ E. Since
E contains both A and B,it follows that**x /∈ A and x /**∈ B.

This proves part(ii).

**x ∈ ∂(A ∪ B); i.e., suppose that B**r(x) intersects
A ∪ B and (A ∪ B)^{c} for all r > 0, Since (A ∪ B)^{c} =A^{c}∩ B^{c},
it follows that Br(x) interscets both A^{c} and B^{c} for all r > 0.

^{c} for all r > 0, or Br(x)
intersects B and B^{c} for all r > 0; i.e.,**x ∈ ∂A ∪ ∂B. This**
proves the first set inequality in part (iii). A similar
argument establishes the second inequality in part (iii).

(ii) Since A ∪ B is closed and contains A ∪ B, it is clear
that, by Theorem 8.32iii, A ∪ B ⊆ A ∪ B. Similarly,
A ∩ B ⊆ A ∩ B. To prove the reverse inequlity for union,
suppose that**x /**∈ A ∪ B. Then, by, Definition 8.31, there is
a closed set E that contains A ∪ B such that**x /**∈ E.Since
E contains both A and B, it follows that**x /∈ A and x /**∈ B.

This proves part(ii).

**x ∈ ∂(A ∪ B); i.e., suppose that B**r(x) intersects
A ∪ B and (A ∪ B)^{c} for all r > 0, Since (A ∪ B)^{c} =A^{c}∩ B^{c},
it follows that Br(x) interscets both A^{c} and B^{c} for all r > 0.

^{c} for all r > 0, or Br(x)
intersects B and B^{c} for all r > 0; i.e.,**x ∈ ∂A ∪ ∂B. This**
proves the first set inequality in part (iii). A similar
argument establishes the second inequality in part (iii).

Proof.

(ii) Since A ∪ B is closed and contains A ∪ B, it is clear
that, by Theorem 8.32iii, A ∪ B ⊆ A ∪ B. Similarly,
A ∩ B ⊆ A ∩ B. To prove the reverse inequlity for union,
suppose that**x /**∈ A ∪ B. Then, by, Definition 8.31, there is
a closed set E that contains A ∪ B such that**x /**∈ E. Since
E contains both A and B,it follows that**x /∈ A and x /**∈ B.

This proves part(ii).

(iii) Let**x ∈ ∂(A ∪ B);** i.e., suppose that Br(x) intersects
A ∪ B and (A ∪ B)^{c} for all r > 0, Since (A ∪ B)^{c} =A^{c}∩ B^{c},
it follows that Br(x) interscets both A^{c} and B^{c} for all r > 0.

^{c} for all r > 0, or Br(x)
intersects B and B^{c} for all r > 0; i.e.,**x ∈ ∂A ∪ ∂B. This**
proves the first set inequality in part (iii). A similar
argument establishes the second inequality in part (iii).

(ii) Since A ∪ B is closed and contains A ∪ B, it is clear
that, by Theorem 8.32iii, A ∪ B ⊆ A ∪ B. Similarly,
A ∩ B ⊆ A ∩ B. To prove the reverse inequlity for union,
suppose that**x /**∈ A ∪ B. Then, by, Definition 8.31, there is
a closed set E that contains A ∪ B such that**x /**∈ E. Since
E contains both A and B, it follows that**x /∈ A and x /**∈ B.

This proves part(ii).

(iii) Let**x ∈ ∂(A ∪ B); i.e., suppose that B**r(x) intersects
A ∪ B and (A ∪ B)^{c} for all r > 0,Since (A ∪ B)^{c} =A^{c}∩ B^{c},
it follows that Br(x) interscets both A^{c} and B^{c} for all r > 0.

^{c} for all r > 0, or Br(x)
intersects B and B^{c} for all r > 0; i.e.,**x ∈ ∂A ∪ ∂B. This**
proves the first set inequality in part (iii). A similar
argument establishes the second inequality in part (iii).

Proof.

(ii) Since A ∪ B is closed and contains A ∪ B, it is clear
that, by Theorem 8.32iii, A ∪ B ⊆ A ∪ B. Similarly,
A ∩ B ⊆ A ∩ B. To prove the reverse inequlity for union,
suppose that**x /**∈ A ∪ B. Then, by, Definition 8.31, there is
a closed set E that contains A ∪ B such that**x /**∈ E. Since
E contains both A and B, it follows that**x /∈ A and x /**∈ B.

This proves part(ii).

(iii) Let**x ∈ ∂(A ∪ B);** i.e., suppose that Br(x) intersects
A ∪ B and (A ∪ B)^{c} for all r > 0, Since (A ∪ B)^{c} =A^{c}∩ B^{c},
it follows that Br(x) interscets both A^{c} and B^{c} for all r > 0.

^{c} for all r > 0, or Br(x)
intersects B and B^{c} for all r > 0; i.e.,**x ∈ ∂A ∪ ∂B. This**
proves the first set inequality in part (iii). A similar
argument establishes the second inequality in part (iii).

**x /**∈ A ∪ B. Then, by, Definition 8.31, there is
a closed set E that contains A ∪ B such that**x /**∈ E. Since
E contains both A and B, it follows that**x /∈ A and x /**∈ B.

This proves part(ii).

(iii) Let**x ∈ ∂(A ∪ B); i.e., suppose that B**r(x) intersects
A ∪ B and (A ∪ B)^{c} for all r > 0,Since (A ∪ B)^{c} =A^{c}∩ B^{c},
it follows that Br(x) interscets both A^{c} and B^{c} for all r > 0.

^{c} for all r > 0, or Br(x)
intersects B and B^{c} for all r > 0; i.e.,**x ∈ ∂A ∪ ∂B. This**
proves the first set inequality in part (iii). A similar
argument establishes the second inequality in part (iii).

Proof.

**x /**∈ A ∪ B. Then, by, Definition 8.31, there is
a closed set E that contains A ∪ B such that**x /**∈ E. Since
E contains both A and B, it follows that**x /∈ A and x /**∈ B.

This proves part(ii).

**x ∈ ∂(A ∪ B); i.e., suppose that B**r(x) intersects
A ∪ B and (A ∪ B)^{c} for all r > 0, Since (A ∪ B)^{c} =A^{c}∩ B^{c},
it follows that Br(x) interscets both A^{c} and B^{c} for all r > 0.

^{c} for all r > 0, or Br(x)
intersects B and B^{c} for all r > 0; i.e.,**x ∈ ∂A ∪ ∂B. This**
proves the first set inequality in part (iii). A similar
argument establishes the second inequality in part (iii).

**x /**∈ A ∪ B. Then, by, Definition 8.31, there is
a closed set E that contains A ∪ B such that**x /**∈ E. Since
E contains both A and B, it follows that**x /∈ A and x /**∈ B.

This proves part(ii).

**x ∈ ∂(A ∪ B); i.e., suppose that B**r(x) intersects
A ∪ B and (A ∪ B)^{c} for all r > 0, Since (A ∪ B)^{c} =A^{c}∩ B^{c},
it follows that Br(x) interscets both A^{c} and B^{c} for all r > 0.

^{c} for all r > 0, or Br(x)
intersects B and B^{c} for all r > 0; i.e.,**x ∈ ∂A ∪ ∂B. This**
proves the first set inequality in part (iii). A similar
argument establishes the second inequality in part (iii).

Proof.

**x /**∈ A ∪ B. Then, by, Definition 8.31, there is
a closed set E that contains A ∪ B such that**x /**∈ E. Since
E contains both A and B, it follows that**x /∈ A and x /**∈ B.

This proves part(ii).

**x ∈ ∂(A ∪ B); i.e., suppose that B**r(x) intersects
A ∪ B and (A ∪ B)^{c} for all r > 0, Since (A ∪ B)^{c} =A^{c}∩ B^{c},
it follows that Br(x) interscets both A^{c} and B^{c} for all r > 0.

^{c} for all r > 0, or Br(x)
intersects B and B^{c} for all r > 0; i.e.,**x ∈ ∂A ∪ ∂B. This**
proves the first set inequality in part (iii). A similar
argument establishes the second inequality in part (iii).

**x /**∈ A ∪ B. Then, by, Definition 8.31, there is
a closed set E that contains A ∪ B such that**x /**∈ E. Since
E contains both A and B, it follows that**x /∈ A and x /**∈ B.

This proves part(ii).

**x ∈ ∂(A ∪ B); i.e., suppose that B**r(x) intersects
A ∪ B and (A ∪ B)^{c} for all r > 0, Since (A ∪ B)^{c} =A^{c}∩ B^{c},
it follows that Br(x) interscets both A^{c} and B^{c} for all r > 0.

^{c} for all r > 0, or Br(x)
intersects B and B^{c} for all r > 0; i.e.,**x ∈ ∂A ∪ ∂B. This**
proves the first set inequality in part (iii). A similar
argument establishes the second inequality in part (iii).

Theorem (8.38)

Let E ⊆**R**^{n}. If there exist nonempty, relatively open sets
U, V which separate E , then there is a pair of open sets
A, B such that A ∩ E 6= ∅, B ∩ E 6= ∅, A ∩ B = ∅, and
E ⊆ A ∪ B.