§8.3 # 4 (a) A union of open sets is open. So if V =S α

§8.3 # 4

(a) A union of open sets is open. So if V = S

α∈AB_α with B_α open, then V is open. For the other direction, we assume that V is open, then for x ∈ V , there exists B_rx(x) ⊂ V for some r_x > 0. Since B_rx(x) is a subset of V for all x ∈ V , we have S

x∈V B_rx(x) ⊆ V . On the other hand, for any x ∈ V , we must have x ∈ B_rx(x).

Thus, V ∈S

x∈V B_rx(x).

(b) Using the complement argument, we can see that if V is closed then there is a collection of B_α^c for α ∈ A such that V =T

α∈AB_α^c.

§8.3 # 5

E is closed iff E^c is open. If a /∈ E then x ∈ E^c. So there exists ε > 0 such that B_ε(x) ⊂ E^c. Therefore, kx − ak ≥ ε > 0 for all x ∈ E. Hence inf_x∈Ekx − ak > 0.

§8.3 # 8

(a) If C is closed in Rⁿ and C = C ∩ E (C is a subset of E), then C is relatively closed in E. On the other hand, if C is relatively open in E, then there exists a closed set B such that C = B ∩ E. Hence C^c = B^c ∪ E^c. For both B^c and E^c are open, we know that C^c is open. So C is closed.

(b) C is relatively closed iff there exists a closed set B such that C = B ∩ E.

Thus, we have E \ C = E ∩ B^c. So E \ C is relatively open. The proof of the other direction is similar.

§8.3 # 9

Assume that E is not connected. That is, there exist open sets U and V such that U ∩ E 6= ∅, V ∩ E 6= ∅, U ∩ V = ∅, and E ⊆ U ∪ V . From E = S

α∈AE_α we have U ∩ (S

α∈AE_α) =S

α∈A(U ∩ E_α) 6= ∅. Therefore U ∩ E_α1 6= ∅ for some α₁ ∈ A.

Likewise, we can show that V ∩ E_α2 6= ∅ for some α₂ ∈ A. If α₁ = α₂ = α, then E_α is not connected. Then we have a contradiction. The problem is that α₁ is not necessarily equal to α₂. However, in fact, we want to show that either U ∩ E_α 6= ∅ or V ∩ E_α 6= ∅ for all α ∈ A. Assume not, namely, U ∩ E_α = ∅ for some α ∈ A and V ∩ E_α⁰ = ∅ for some α⁰ ∈ A. Thus, U ∩T

α∈AE_α = ∅ and V ∩T

α∈AE_α = ∅. This implies that E ⊆ U ∪ V is not true sinceT

α∈AE_α 6= ∅. So we must have U ∩ E_α 6= ∅ and V ∩ E_α 6= ∅ for some α ∈ A. Additionally, E_α ⊆ E ⊆ U ∪ V . Therefore, E_α is not connected. This is a contradiction.

§8.3 # 10

(i) Any connected set in R is an interval (single point is included). If ∅ 6= E = T

α∈AE_α is not connected, then there exists at least one point c in E such that (a, c) ⊂ E and (c, b) ⊂ E. In other words, (a, c) ⊂ E_α and (c, b) ⊂ E_α for all α.

Hence E_α is not connected. This is a contradiction.

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(ii) Let E₁ = {(x, y) : x² + y² = 1, y ≥ 0} and E₂ = {(x, y) : y = 0}. Then E₁ ∩ E₂ = {(−1, 0), (1, 0)}. Both E₁ and E₂ are connected, but E₁ ∩ E₂ is not connected.

§8.4 # 2

(a) E⁰ = {(x, y) : x²+ 4y² < 1}, E = E, and ∂E = {(x, y) : x²+ 4y² = 1}.

(b) E⁰ = ∅, E = E, and ∂E = E.

(c) E⁰ = {(x, y) : y > x², y < 1}, E = {(x, y) : y ≥ x², 0 ≤ y ≤ 1}, ∂E = {(x, y) : y = x², −1 ≤ x ≤ 1} ∪ {(x, y) : −1 < x < 1, y = 1}.

(d) E⁰ = E, E = {(x, y) : x²− y² ≤ 1, −1 ≤ y ≤ 1}, and ∂E = {(x, y) : x²− y² = 1, −1 ≤ y ≤ 1} ∪ {(x, y) : −√

2 < x <√

2, y = 1} ∪ {(x, y) : −√

2 < x <√

2, y = −1}

§8.4 # 3

A ⊆ B ⊆ B ⇒ A ⊆ B; A⁰ ⊆ A ⊆ B ⇒ A⁰ ⊆ B⁰.

§8.4 # 7

Suppose that A is not connected. Then there exist two open sets U and V such that U ∩ A 6= ∅, V ∩ A 6= ∅, U ∩ V = ∅, and A ⊆ U ∪ V . It is clear that E ⊆ U ∪ V . Since A ⊆ E, we have U ∩ E 6= ∅ and V ∩ E 6= ∅. Then both U ∩ E and V ∩ E are non-empty. For if U ∩ E = ∅ then E ⊂ U^c. Since U^c is closed, we get E ⊆ U^c, i.e., U ∩ E = ∅. This is again a contradiction. Similar proof works for V ∩ E. Now because U ∩ E 6= ∅ and V ∩ E 6= ∅, E is not connected. This is a contradiction. So A must be connected.

§8.4 # 8 Note that we use the canonical metric in Rⁿ here.

(a) ∅ and Rⁿ.

(b) We assume that E is connected. We know that ∅ and E are relatively clopen sets. Assume that E contains another relatively clopen set, say U . Then E \ U is relatively open. So E = U ∪ U^c and U ∩ U^c = ∅. Thus E is not connected. This is a contradiction.

On the other hand, if E has only two relatively clopen sets, i.e. ∅ and E, and E is not connected. Hence, there exist relatively open sets U and V such that E = U ∪ V and U ∩ V = ∅. So V and U are relatively clopen. This is a contradiction.

(c) If ∂E = ∅, then E⁰ = E. Hence E is clopen. Hence E and E^c are open and Rⁿ = E ∪ E^c. Also, E 6= ∅, E^c 6= ∅, and E ∩ E^c = ∅. So Rⁿ is not connected. This is a contradiction since Rⁿ is connected.

§8.4 # 10 Answers can be found in the proof of Theorem 10.40.

§8.4 # 11

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(a) U is relatively open iff ∃ open set Ω in Rⁿsuch that U = E ∩ Ω. Since U ⊂ E⁰, we have U = E⁰∩ Ω, i.e. U is open in Rⁿ. Thus, U ∩ ∂U = ∅.

(b) If x ∈ U ∩ ∂E, then x ∈ U and B_r(x) ∩ E 6= ∅, B_r(x) ∩ E^c 6= ∅ for all r > 0.

From U ⊂ E, we know that E^c ⊂ U^c. So B_r(x) ∩ U^c 6= ∅. That x ∈ U implies B_r(x) ∩ U 6= ∅ for all r > 0. Therefore, we have U ∩ ∂E = U ∩ ∂U .

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