• 沒有找到結果。

N/A
N/A
Protected

Copied!
53
0
0

(1)

WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

2009

(2)

## Ch8: Euclidean Spaces

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

Definition (8.31)

Let E be a subset of a Euclidean SpaceRn. (i) The interior of E is the set

Eo:=[

{V : V ⊆ E and V is open in Rn}.

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

{B : B ⊇ E and B is closed in Rn}.

(3)

Let E ⊆Rn.Then (i) Eo ⊆ E ⊆ E,

(ii) if V is open and V ⊆ E then V ⊆ Eo,and (iii) if C is closed and C ⊇ E then C ⊇ E . Proof.

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

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

This proves (ii) and (iii).

(4)

Theorem (8.32) Let E ⊆Rn.Then

(i) Eo ⊆ E ⊆ E,

(ii) if V is open and V ⊆ E then V ⊆ Eo,and (iii) if C is closed and C ⊇ E then C ⊇ E . Proof.

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

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

This proves (ii) and (iii).

(5)

Let E ⊆Rn.Then (i) Eo ⊆ E ⊆ E,

(ii) if V is open and V ⊆ E then V ⊆ Eo,and (iii) if C is closed and C ⊇ E then C ⊇ E . Proof.

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

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

This proves (ii) and (iii).

(6)

Theorem (8.32) Let E ⊆Rn.Then

(i) Eo ⊆ E ⊆ E,

(ii) if V is open and V ⊆ E then V ⊆ Eo,and (iii) if C is closed and C ⊇ E then C ⊇ E . Proof.

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

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

This proves (ii) and (iii).

(7)

Let E ⊆Rn.Then (i) Eo ⊆ E ⊆ E,

(ii) if V is open and V ⊆ E then V ⊆ Eo,and (iii) if C is closed and C ⊇ E then C ⊇ E . Proof.

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

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

This proves (ii) and (iii).

(8)

Theorem (8.32) Let E ⊆Rn.Then

(i) Eo ⊆ E ⊆ E,

(ii) if V is open and V ⊆ E then V ⊆ Eo,and (iii) if C is closed and C ⊇ E then C ⊇ E . Proof.

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

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

This proves (ii) and (iii).

(9)

Let E ⊆Rn.Then (i) Eo ⊆ E ⊆ E,

(ii) if V is open and V ⊆ E then V ⊆ Eo,and (iii) if C is closed and C ⊇ E then C ⊇ E . Proof.

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

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

This proves (ii) and (iii).

(10)

Theorem (8.32) Let E ⊆Rn.Then

(i) Eo ⊆ E ⊆ E,

(ii) if V is open and V ⊆ E then V ⊆ Eo,and (iii) if C is closed and C ⊇ E then C ⊇ E . Proof.

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

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

This proves (ii) and (iii).

(11)

Let E ⊆Rn.Then (i) Eo ⊆ E ⊆ E,

(ii) if V is open and V ⊆ E then V ⊆ Eo,and (iii) if C is closed and C ⊇ E then C ⊇ E . Proof.

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

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

This proves (ii) and (iii).

(12)

Theorem (8.32) Let E ⊆Rn.Then

(i) Eo ⊆ E ⊆ E,

(ii) if V is open and V ⊆ E then V ⊆ Eo,and (iii) if C is closed and C ⊇ E then C ⊇ E . Proof.

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

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

This proves (ii) and (iii).

(13)

Let E ⊆Rn.Then (i) Eo ⊆ E ⊆ E,

(ii) if V is open and V ⊆ E then V ⊆ Eo,and (iii) if C is closed and C ⊇ E then C ⊇ E . Proof.

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

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

This proves (ii) and (iii).

(14)

Definition (8.34)

Let E ⊆Rn.The boundary of E is the set

∂E := {x ∈ Rn: for all r > 0, Br(x) ∩ E 6= ∅ and Br(x) ∩ Ec 6= ∅}.

(15)

Let E ⊆Rn.Then ∂E = E \Eo. Proof.

By Definition 8.34, it suffices to show that

(10) x ∈ E if and only if Br(x) ∩ E 6= ∅ for all r > 0, and (11) x /∈ Eo if and only if Br(x) ∩ Ec 6= ∅ for all r > 0.

We will provide the details for (10) and leave the proof of (11) as an exercise. Suppose thatx ∈ E but Br0(x) ∩ E = ∅ for some r0>0. Then (Br0(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 thatx /∈ E. Since (E)c is open, there is an r0 >0 such that Br0(x) ⊆ (E )c.In particular,

∅ = Br0(x) ∩ E ⊇ Br0(x) ∩ E for some r0 >0.

(16)

Theorem (8.36)

Let E ⊆Rn.Then ∂E = E \Eo. Proof.

By Definition 8.34, it suffices to show that

(10) x ∈ E if and only if Br(x) ∩ E 6= ∅ for all r > 0,and (11) x /∈ Eo if and only if Br(x) ∩ Ec 6= ∅ for all r > 0.

We will provide the details for (10) and leave the proof of (11) as an exercise.Suppose thatx ∈ E but Br0(x) ∩ E = ∅ for some r0>0. Then (Br0(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 thatx /∈ E. Since (E)c is open, there is an r0 >0 such that Br0(x) ⊆ (E )c.In particular,

∅ = Br0(x) ∩ E ⊇ Br0(x) ∩ E for some r0 >0.

(17)

Let E ⊆Rn.Then ∂E = E \Eo. Proof.

By Definition 8.34, it suffices to show that

(10) x ∈ E if and only if Br(x) ∩ E 6= ∅ for all r > 0, and (11) x /∈ Eo if and only if Br(x) ∩ Ec 6= ∅ for all r > 0.

We will provide the details for (10) and leave the proof of (11) as an exercise. Suppose thatx ∈ E but Br0(x) ∩ E = ∅ for some r0>0.Then (Br0(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 thatx /∈ E. Since (E)c is open, there is an r0 >0 such that Br0(x) ⊆ (E )c.In particular,

∅ = Br0(x) ∩ E ⊇ Br0(x) ∩ E for some r0 >0.

(18)

Theorem (8.36)

Let E ⊆Rn.Then ∂E = E \Eo. Proof.

By Definition 8.34, it suffices to show that

(10) x ∈ E if and only if Br(x) ∩ E 6= ∅ for all r > 0, and (11) x /∈ Eo if and only if Br(x) ∩ Ec 6= ∅ for all r > 0.

We will provide the details for (10) and leave the proof of (11) as an exercise.Suppose thatx ∈ E but Br0(x) ∩ E = ∅ for some r0>0. Then (Br0(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 thatx /∈ E. Since (E)c is open, there is an r0 >0 such that Br0(x) ⊆ (E )c.In particular,

∅ = Br0(x) ∩ E ⊇ Br0(x) ∩ E for some r0 >0.

(19)

Let E ⊆Rn.Then ∂E = E \Eo. Proof.

By Definition 8.34, it suffices to show that

(10) x ∈ E if and only if Br(x) ∩ E 6= ∅ for all r > 0, and (11) x /∈ Eo if and only if Br(x) ∩ Ec 6= ∅ for all r > 0.

We will provide the details for (10) and leave the proof of (11) as an exercise. Suppose thatx ∈ E but Br0(x) ∩ E = ∅ for some r0>0.Then (Br0(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 thatx /∈ E. Since (E)c is open, there is an r0 >0 such that Br0(x) ⊆ (E )c.In particular,

∅ = Br0(x) ∩ E ⊇ Br0(x) ∩ E for some r0 >0.

(20)

Theorem (8.36)

Let E ⊆Rn.Then ∂E = E \Eo. Proof.

By Definition 8.34, it suffices to show that

(10) x ∈ E if and only if Br(x) ∩ E 6= ∅ for all r > 0, and (11) x /∈ Eo if and only if Br(x) ∩ Ec 6= ∅ for all r > 0.

We will provide the details for (10) and leave the proof of (11) as an exercise. Suppose thatx ∈ E but Br0(x) ∩ E = ∅ for some r0>0. Then (Br0(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 thatx /∈ E. Since (E)c is open, there is an r0 >0 such that Br0(x) ⊆ (E )c.In particular,

∅ = Br0(x) ∩ E ⊇ Br0(x) ∩ E for some r0 >0.

(21)

Let E ⊆Rn.Then ∂E = E \Eo. Proof.

By Definition 8.34, it suffices to show that

(10) x ∈ E if and only if Br(x) ∩ E 6= ∅ for all r > 0, and (11) x /∈ Eo if and only if Br(x) ∩ Ec 6= ∅ for all r > 0.

We will provide the details for (10) and leave the proof of (11) as an exercise. Suppose thatx ∈ E but Br0(x) ∩ E = ∅ for some r0>0. Then (Br0(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 thatx /∈ E. Since (E)c is open, there is an r0 >0 such that Br0(x) ⊆ (E )c.In particular,

∅ = Br0(x) ∩ E ⊇ Br0(x) ∩ E for some r0 >0.

(22)

Theorem (8.36)

Let E ⊆Rn.Then ∂E = E \Eo. Proof.

By Definition 8.34, it suffices to show that

(10) x ∈ E if and only if Br(x) ∩ E 6= ∅ for all r > 0, and (11) x /∈ Eo if and only if Br(x) ∩ Ec 6= ∅ for all r > 0.

We will provide the details for (10) and leave the proof of (11) as an exercise. Suppose thatx ∈ E but Br0(x) ∩ E = ∅ for some r0>0. Then (Br0(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 thatx /∈ E. Since (E )c is open, there is an r0 >0 such that Br0(x) ⊆ (E )c.In particular,

∅ = Br0(x) ∩ E ⊇ Br0(x) ∩ E for some r0 >0.

(23)

Let E ⊆Rn.Then ∂E = E \Eo. Proof.

By Definition 8.34, it suffices to show that

(10) x ∈ E if and only if Br(x) ∩ E 6= ∅ for all r > 0, and (11) x /∈ Eo if and only if Br(x) ∩ Ec 6= ∅ for all r > 0.

We will provide the details for (10) and leave the proof of (11) as an exercise. Suppose thatx ∈ E but Br0(x) ∩ E = ∅ for some r0>0. Then (Br0(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 thatx /∈ E. Since (E)c is open, there is an r0 >0 such that Br0(x) ⊆ (E )c.In particular,

∅ = Br0(x) ∩ E ⊇ Br0(x) ∩ E for some r0 >0.

(24)

Theorem (8.36)

Let E ⊆Rn.Then ∂E = E \Eo. Proof.

By Definition 8.34, it suffices to show that

(10) x ∈ E if and only if Br(x) ∩ E 6= ∅ for all r > 0, and (11) x /∈ Eo if and only if Br(x) ∩ Ec 6= ∅ for all r > 0.

We will provide the details for (10) and leave the proof of (11) as an exercise. Suppose thatx ∈ E but Br0(x) ∩ E = ∅ for some r0>0. Then (Br0(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 thatx /∈ E. Since (E )c is open, there is an r0 >0 such that Br0(x) ⊆ (E )c.In particular,

∅ = Br0(x) ∩ E ⊇ Br0(x) ∩ E for some r0 >0.

(25)

Let E ⊆Rn.Then ∂E = E \Eo. Proof.

By Definition 8.34, it suffices to show that

(10) x ∈ E if and only if Br(x) ∩ E 6= ∅ for all r > 0, and (11) x /∈ Eo if and only if Br(x) ∩ Ec 6= ∅ for all r > 0.

We will provide the details for (10) and leave the proof of (11) as an exercise. Suppose thatx ∈ E but Br0(x) ∩ E = ∅ for some r0>0. Then (Br0(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 thatx /∈ E. Since (E)c is open, there is an r0 >0 such that Br0(x) ⊆ (E )c.In particular,

∅ = Br0(x) ∩ E ⊇ Br0(x) ∩ E for some r0 >0.

(26)

Theorem (8.36)

Let E ⊆Rn.Then ∂E = E \Eo. Proof.

By Definition 8.34, it suffices to show that

(10) x ∈ E if and only if Br(x) ∩ E 6= ∅ for all r > 0, and (11) x /∈ Eo if and only if Br(x) ∩ Ec 6= ∅ for all r > 0.

We will provide the details for (10) and leave the proof of (11) as an exercise. Suppose thatx ∈ E but Br0(x) ∩ E = ∅ for some r0>0. Then (Br0(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 thatx /∈ E. Since (E)c is open, there is an r0 >0 such that Br0(x) ⊆ (E )c.In particular,

∅ = Br0(x) ∩ E ⊇ Br0(x) ∩ E for some r0 >0.

(27)

Let E ⊆Rn.Then ∂E = E \Eo. Proof.

By Definition 8.34, it suffices to show that

(10) x ∈ E if and only if Br(x) ∩ E 6= ∅ for all r > 0, and (11) x /∈ Eo if and only if Br(x) ∩ Ec 6= ∅ for all r > 0.

We will provide the details for (10) and leave the proof of (11) as an exercise. Suppose thatx ∈ E but Br0(x) ∩ E = ∅ for some r0>0. Then (Br0(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 thatx /∈ E. Since (E)c is open, there is an r0 >0 such that Br0(x) ⊆ (E )c.In particular,

∅ = Br0(x) ∩ E ⊇ Br0(x) ∩ E for some r0 >0.

(28)

Theorem (8.37) Let A, B ⊆Rn Then

(i) (A ∪ B)o⊇ Ao∪ Bo, (A ∩ B)o =Ao∩ Bo, (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, Ao∪ Bois an open subset of A ∪ B. Hence, by Theorem 8.32ii,

Ao∪ Bo ⊆ (A ∪ B)o.

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

(A ∩ B)o ⊆ Ao∩ Bo.

(29)

Let A, B ⊆Rn Then

(i) (A ∪ B)o⊇ Ao∪ Bo, (A ∩ B)o =Ao∩ Bo, (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,Ao∪ Bois an open subset of A ∪ B. Hence, by Theorem 8.32ii,

Ao∪ Bo ⊆ (A ∪ B)o.

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

(A ∩ B)o ⊆ Ao∩ Bo.

(30)

Theorem (8.37) Let A, B ⊆Rn Then

(i) (A ∪ B)o⊇ Ao∪ Bo, (A ∩ B)o =Ao∩ Bo, (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, Ao∪ Bois an open subset of A ∪ B.Hence, by Theorem 8.32ii,

Ao∪ Bo ⊆ (A ∪ B)o.

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

(A ∩ B)o ⊆ Ao∩ Bo.

(31)

Let A, B ⊆Rn Then

(i) (A ∪ B)o⊇ Ao∪ Bo, (A ∩ B)o =Ao∩ Bo, (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,Ao∪ Bois an open subset of A ∪ B. Hence, by Theorem 8.32ii,

Ao∪ Bo ⊆ (A ∪ B)o.

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

(A ∩ B)o ⊆ Ao∩ Bo.

(32)

Theorem (8.37) Let A, B ⊆Rn Then

(i) (A ∪ B)o⊇ Ao∪ Bo, (A ∩ B)o =Ao∩ Bo, (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, Ao∪ Bois an open subset of A ∪ B.Hence, by Theorem 8.32ii,

Ao∪ Bo ⊆ (A ∪ B)o.

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

(A ∩ B)o ⊆ Ao∩ Bo.

(33)

Let A, B ⊆Rn Then

(i) (A ∪ B)o⊇ Ao∪ Bo, (A ∩ B)o =Ao∩ Bo, (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, Ao∪ Bois an open subset of A ∪ B. Hence, by Theorem 8.32ii,

Ao∪ Bo ⊆ (A ∪ B)o.

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

(A ∩ B)o ⊆ Ao∩ Bo.

(34)

Theorem (8.37) Let A, B ⊆Rn Then

(i) (A ∪ B)o⊇ Ao∪ Bo, (A ∩ B)o =Ao∩ Bo, (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, Ao∪ Bois an open subset of A ∪ B. Hence, by Theorem 8.32ii,

Ao∪ Bo ⊆ (A ∪ B)o.

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

(A ∩ B)o ⊆ Ao∩ Bo.

(35)

Let A, B ⊆Rn Then

(i) (A ∪ B)o⊇ Ao∪ Bo, (A ∩ B)o =Ao∩ Bo, (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, Ao∪ Bois an open subset of A ∪ B. Hence, by Theorem 8.32ii,

Ao∪ Bo ⊆ (A ∪ B)o.

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

(A ∩ B)o ⊆ Ao∩ Bo.

(36)

Theorem (8.37) Let A, B ⊆Rn Then

(i) (A ∪ B)o⊇ Ao∪ Bo, (A ∩ B)o =Ao∩ Bo, (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, Ao∪ Bois an open subset of A ∪ B. Hence, by Theorem 8.32ii,

Ao∪ Bo ⊆ (A ∪ B)o.

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

(A ∩ B)o ⊆ Ao∩ Bo.

(37)

Let A, B ⊆Rn Then

(i) (A ∪ B)o⊇ Ao∪ Bo, (A ∩ B)o =Ao∩ Bo, (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, Ao∪ Bois an open subset of A ∪ B. Hence, by Theorem 8.32ii,

Ao∪ Bo ⊆ (A ∪ B)o.

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

(A ∩ B)o ⊆ Ao∩ Bo.

(38)

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

This proves part(ii).

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

Thus Br(x) intersects A and Ac for all r > 0, or Br(x) intersects B and Bc 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).

(39)

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

This proves part(ii).

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

Thus Br(x) intersects A and Ac for all r > 0, or Br(x) intersects B and Bc 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).

(40)

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

This proves part(ii).

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

Thus Br(x) intersects A and Ac for all r > 0, or Br(x) intersects B and Bc 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).

(41)

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

This proves part(ii).

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

Thus Br(x) intersects A and Ac for all r > 0, or Br(x) intersects B and Bc 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).

(42)

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

This proves part(ii).

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

Thus Br(x) intersects A and Ac for all r > 0, or Br(x) intersects B and Bc 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).

(43)

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

This proves part(ii).

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

Thus Br(x) intersects A and Ac for all r > 0, or Br(x) intersects B and Bc 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).

(44)

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

This proves part(ii).

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

Thus Br(x) intersects A and Ac for all r > 0, or Br(x) intersects B and Bc 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).

(45)

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

This proves part(ii).

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

Thus Br(x) intersects A and Ac for all r > 0, or Br(x) intersects B and Bc 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).

(46)

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

This proves part(ii).

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

Thus Br(x) intersects A and Ac for all r > 0, or Br(x) intersects B and Bc 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).

(47)

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

This proves part(ii).

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

Thus Br(x) intersects A and Ac for all r > 0, or Br(x) intersects B and Bc 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).

(48)

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

This proves part(ii).

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

Thus Br(x) intersects A and Ac for all r > 0, or Br(x) intersects B and Bc 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).

(49)

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

This proves part(ii).

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

Thus Br(x) intersects A and Ac for all r > 0, or Br(x) intersects B and Bc 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).

(50)

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

This proves part(ii).

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

Thus Br(x) intersects A and Ac for all r > 0, or Br(x) intersects B and Bc 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).

(51)

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

This proves part(ii).

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

Thus Br(x) intersects A and Ac for all r > 0, or Br(x) intersects B and Bc 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).

(52)

Theorem (8.38)

Let E ⊆Rn. 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.

(53)

## Thank you.

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung