Advanced Calculus (II)
WEN-CHINGLIEN
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 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}.
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).
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).
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).
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).
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).
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).
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).
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).
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).
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).
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).
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= ∅}.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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).
(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).
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).
(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).
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).
(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).
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).
(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).
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).
(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).
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).
(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).
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).
(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).
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.
