Math 104 Interiors, Closures, and Boundaries Solutions

Exercises: 1. For the following subsets, determine (without proof) their interiors, closures, and boundaries

(a) S = [0, 1] in R with the usual metric. (b) S = (0, 1) in R with the usual metric. (c) S = Z in R with the usual metric. (d) S = Q in R with the usual metric. (e) S = R in R with the usual metric. (f) S = [0, 1) × [0, 1) in R2 with the 2-dimensional Euclidean metric. 2. Let S ⊂ E. (a) Show that S◦ is the union of all open subsets U ⊂ S. (b) Show that S◦ is open. (c) Show that ∂S is closed. 3. Let T ⊂ S ⊂ E. (a) Show that T ⊂ S. (b) Show that T ◦ ⊂ S◦. 4. Let S ⊂ E. (a) Show that S is open if and only if S = S◦. (b) Show that S is closed if and only if S = S. 5. Let S ⊂ E. (a) Show that S = ((Sc)◦)c.  c (b) Show that S◦ = (Sc) .

(c) Show that ∂S = S ∩ Sc. (d) Show ∂S = ∂ (Sc). 6. Let S ⊂ E (a) Show S = {x ∈ E : B(x, r) ∩ S 6= ∅ ∀r > 0}. (b) Show ∂S = {x ∈ E : B(x, r) ∩ S 6= ∅ and B(x, r) ∩ Sc 6= ∅ ∀r > 0}. 7. For S ⊂ E show that the following are equivalent: (i) S is dense in E. (ii)( Sc)◦ = ∅. (iii) S = E. 8. For S ⊂ E, show that E is the disjoint union of S◦, ∂S, and (Sc)◦. 9. Let S ⊂ E. (a) Show that S is closed if and only if ∂S ⊂ S. (b) Show that S is open if and only if ∂S ∩ S = ∅. 10. Let A, B ⊂ E. (a) Show that A ∪ B = A ∪ B.

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(b) Show that (A ∩ B)◦ = A◦ ∩ B◦. (c) For E = R with the usual metric, give examples of subsets A, B ⊂ R such that A ∩ B 6= A ∩ B and (A ∪ B)◦ 6= A◦ ∪ B◦. 11. Let S ⊂ E be a connected set. Suppose T ⊂ E satisfies S ⊂ T ⊂ S. Show that T is also connected. ———————————————————————————————————————————– Solutions: 1. (a) S◦ = (0, 1), S = [0, 1], and ∂S = {0, 1}. (b) S◦ = (0, 1), S = [0, 1], and ∂S = {0, 1}. (c) S◦ = ∅, S = Z, and ∂S = Z. (d) S◦ = ∅, S = R, and ∂S = R. (e) S◦ = R, S = R, and ∂S = ∅. (f) S◦ = (0, 1) ⊗ (0, 1), S = [0, 1] × [0, 1], and

 2 ∂S = (x, y) ∈ R : either x ∈ {0, 1} or y ∈ {0, 1} (or both) .

2. (a) Denote by T the union of all open subsets of S. Let x ∈ S◦. Since x is an interior point, there exists r > 0 such that B(x, r) ⊂ S, but then B(x, r) is an open subset of S. Hence x ∈ T , and since x ∈ S◦ was arbitrary we have S◦ ⊂ T . Conversely, if x ∈ T then x ∈ U for some open subset U ⊂ S. Since U is open, there exists r > 0 such that B(x, r) ⊂ U ⊂ S; that is, x is an interior point of S and therefore x ∈ S◦. Since x ∈ T was arbitrary, we have T ⊂ S◦, which yields ◦ T = S .  ◦ (b) By part (a), S is a union of open sets and is therefore open.  (c) We have ∂S = S \ S◦ = S ∩ (S◦)c. We know S is closed, and by part (b) (S◦)c is closed as the complement of an open set. Thus ∂S is closed as an intersection of closed sets.  3. (a) Since T ⊂ S ⊂ S, we have that S is a closed set containing T . Thus T ⊂ S.  (b) If x ∈ T ◦, then there exists r > 0 such that B(x, r) ⊂ T ⊂ S. Hence x is also an interior point of ◦ ◦ ◦ S and so x ∈ S . Consequently T ⊂ S .  4. (a)( ⇒) : If S is open, then U = S is an open subset of S. Hence S ⊂ S◦. The reverse inclusion always holds, so we have S = S◦. ◦ (⇐) : By Exercise 3.(b), S = S is open.  (b)( ⇒) : If S is closed, then S is a closed set containing S. Hence S ⊂ S. The reverse inclusion always holds, so we have S = S. (⇐): S = S is closed as the intersection of closed sets.  5. (a) Denote T := ((Sc)◦)c. We first note that (Sc)◦ is open and is contained in Sc. Consequently, its complement, T , is a closed set containing S. Hence S ⊂ T . On the other hand, if V is a closed set containing S, then we have that V c is an open subset of Sc. Hence V c ⊂ (Sc)◦, which implies V ⊃ T . Since V was an arbitrary closed set containing S, we have T ⊂ S, and therefore T = S. (b) Denote T = Sc. Then using part (a) we have

 c c (Sc) = T
= (((T c)◦)c)c = (T c)◦ = S◦.

 (c) By definition we have ∂S = S \ S◦ = S ∩ (S◦)c.

By part (b) it follows that ∂S = S ∩ Sc. 

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c (d) From part (c), we have ∂S = S ∩ Sc = Sc ∩ S = ∂(S ).  6. (a) Denote T = {x ∈ E : B(x, r) ∩ S 6= ∅ ∀r > 0}. Observe that x ∈ T c iff ∃r > 0 such that B(x, r) ∩ S = ∅ iff x ∈ (Sc)◦ iff (by Exercise 5.(a)) c c c x ∈ S . Thus T = S which implies T = S.  (b) By part (a),

{x ∈ E : B(x, r) ∩ S 6= ∅ and B(x, r) ∩ Sc 6= ∅ ∀r > 0} = S ∩ Sc,

and this latter set equals ∂S by Exercise 5.(c).  7.[( i) ⇒ (ii)] : Assume S is dense in E. Then for every x ∈ E and every r > 0 we have B(x, r) ∩ S 6= ∅. In particular, this is true for x ∈ Sc and means that B(x, r) 6⊂ Sc for any r > 0. Hence Sc has no interior points which means (Sc)◦ = ∅. [(ii) ⇒ (iii)] : Assume (Sc)◦ = ∅. Then by Exercise 5.(a) we have S = (∅)c = E. [(iii) ⇒ (i)] : Assume S = E. Let x ∈ E = S. Then by Exercise 6.(a), B(x, r) ∩ S 6= ∅ for all r > 0. Since this holds for all x ∈ E, we have that S is dense.  c 8. We know E is the disjoint union of S and S . By definition of the boundary we see that S is the ◦ c c ◦ disjoint union of S and ∂S, and by Exercise 5.(a) we see that S = (S ) .  9. (a) If S is closed then S = S by Exercise 4.(b), but then ∂S ⊂ S = S. Conversely, if ∂S ⊂ S then ◦ S = ∂S ∪ S ⊂ S ⊂ S. Thus S = S, which implies S is closed.  (b) If S is open then S = S◦ by Exercise 4.(a), and hence ∂S ∩ S = ∂S ∩ S◦ = ∅. Conversely, if ∂S ∩ S = ∅, then S ⊂ ∂Sc = S◦ ∪ (Sc)◦ by Exercise 8. Since (S◦)◦ ⊂ Sc, we know that S ⊂ S◦ ◦ and the reverse inclusion always holds. Thus S = S which implies S is open by Exercise 4.(a). 10. (a) Note that A ∪ B is closed as the finite union of closed sets, and it contains A ∪ B. Hence A ∪ B ⊂ A ∪ B. On the other hand, if V is a closed set containing A ∪ B, then it is also a closed set containing A. Hence A ⊂ V . Similarly, B ⊂ V , which means A ∪ B ⊂ V . Since V was an arbitrary closed set containing A ∪ B, we have A ∪ B ⊂ A ∪ B, which gives equality.  (b) We can proof this direclty in a similar fashion to part (a), or we can appeal to part (a) along with Exercise 5.(b):

 c c c c c (A ∩ B)◦ = (A ∩ B)c = Ac ∪ Bc
= Ac ∪ Bc
= Ac
∩ Bc
= A◦ ∩ B◦.

 (c) Consider A = [0, 1) and B = [1, 2]. Then A ∩ B = ∅, which is closed so A ∩ B = ∅. However, A = [0, 1] and B = [1, 2] so that A ∩ B = {1}= 6 ∅. Also, A ∪ B = [0, 2] which has (A ∪ B)◦ = (0, 2), while A◦ = (0, 1) and B◦ = (1, 2) so that A◦ ∪ B◦ = (0, 1) ∪ (1, 2) 6= (0, 2). 11. Suppose T = A ∪ B for disjoint non-empty subsets A, B ⊂ T which are open relative to T . Since S ⊂ T , we have S = [A ∩ S] ∪ [B ∩ S], and A ∩ S and B ∩ S are open relative to S. As S is connected, we know either A ∩ S or B ∩ S is empty. Without loss of generality, assume B ∩ S = ∅ and hence A ∩ S = S. This means S ⊂ A, and consequently Ac ⊂ Sc. Thus

B ⊂ Ac ∩ T ⊂ Sc.

On the other hand, B ⊂ T ⊂ S. Combining these two inclusions yields B ⊂ S ∩ Sc. Since S ∩ Sc = S \ S ⊂ ∂S, we have B ⊂ ∂S. Now, since B is open relative to T , for each x ∈ B there exists r > 0 such that B(x, r) ∩ T ⊂ B. On the other hand, by Exercise 6.(b), x ∈ B ⊂ ∂S means this ball must intersect S and hence A, a contradiction. Thus T is connected. 

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