Advanced Calculus I MAA4226 Fall 2012 Homework 4 Solutions I ...

Advanced Calculus I MAA4226 Fall 2015 Homework 4 Solutions Due Wednesday, September 30, 2015 Do problems 8,9,10,12,14 from Rudin, pages 43-44. 8. If E ⊂ R2 is open, then every point of E is a limit point of E. If E ⊂ R2 is closed, then it is not necessarily true that every point of E is a limit point of E. Proof: Suppose E ⊂ R2 is open and p ∈ E. Hence, there exists r > 0 such that Nr (p) ⊂ E. We now show p is a limit point of E. Let Nt (p) be given, where t > 0. Let s be the smaller of r, t. Clearly, Ns (p) ⊂ Nt (p) ∩ E. Since this set contains many elements other than p, it follows that p is a limit point of E. We now give a counterexample for the case in which E ⊂ R2 is closed. Let E = {0}. Since E is a finite set, it has no limit points (by the Corollary to Theorem 2.20). So 0 is not a limit point! QED. 9. Let E ◦ denote the set of all interior points of a subset E of a metric space X. (a) Proof that E ◦ is open: Let p ∈ E ◦ . We need to prove p is an interior point of E ◦ . Since p is interior to E, there exists r > 0 such that Nr (p) ⊂ E. I claim that Nr (p) ⊂ E ◦ . This will show that p is interior to E ◦ , completing the proof. So, let y ∈ Nr (p). We have proved that Nr (p) is open, so there exists a neighborhood Ns (y) ⊂ Nr (p) ⊂ E. It follows that y is interior to E, so y ∈ E ◦ . Since y ∈ Nr (p) was chosen arbitrarily, this proves that Nr (p) ⊂ E ◦ . QED. (b) Proof that E is open if and only if E = E ◦ : Suppose E is open. Then by definition, every point of E is interior to E, so E ⊂ E ◦ . But it is immediate from the definition that E ◦ ⊂ E, so E = E ◦ . On the other hand, suppose that E = E ◦ . Then by definition of E ◦ , every point of E is interior to E, so E is open. QED. (c) Proof that if G ⊂ E and G is open, then G ⊂ E ◦ : Let p ∈ G. Then since G is open, there exists r > 0 such that Nr (p) ⊂ G ⊂ E. Therefore p is interior to E, so p ∈ E ◦ . QED. (d) Proof that (E ◦ )c = E c : First we prove that (E ◦ )c ⊂ E c . Let p ∈ (E ◦ )c . We need to prove that p ∈ E c . We consider two separate cases. Case I: Suppose p ∈ E. Reason as follows: p ∈ (E ◦ )c ⇒ p 6∈ E ◦ ⇒ (∀r > 0)Nr (p) 6⊂ E ⇒ (∀r > 0)Nr (p) ∩ E c 6= ∅. Since p ∈ E, it follow that (∀r > 0)Nr (p) ∩ E c contains elements other than p. Therefore, p is a limit point of E c , so p ∈ E c . Case II: Suppose that p 6∈ E. Then p ∈ E c , so it is immediate that p ∈ E c . Now, we need to prove that E c ⊂ (E ◦ )c . Suppose p ∈ E c ; hence either p ∈ E c or p is a limit point of E c . In either of these cases, (∀r > 0)Nr (p) ∩ E c 6= ∅. Since every neighborhood of p intersects E c , no neighborhood of p is contained in E. Hence, p is not an interior point of E, so p ∈ (E ◦ )c . QED. (e) No! Let X = R, and E = (0, 1) ∪ (1, 2). Then E ◦ = E, while (E)◦ = (0, 2). (f) No! Let X = R, and E = (0, 1) ∩ Q. (Where Q denotes the rationals). Then E ◦ = ∅ while E = [0, 1]. 10. Properties (a) and (b) of a metric space are trivial to verify and I will omit them here. So we just need to verify the triangle inequality: for all p, q, r ∈ X, d(p, r) ≤ d(p, q) + d(q, r). This is

also quite easy: First consider the case p = r. In that case, d(p, r) = 0, so the inequality follows immediately since neither term on the right can be negative. Now consider the case in which p 6= r. In that case d(p, r) = 1. However, we also know that either p 6= q or q 6= r (otherwise p = r), so the right hand side of the inequality is at least 1 also, and we are done. Every subset of X is open with respect to this metric! To see this, first note that for all p ∈ X, N1/2 (p) = {p}. Hence, since we know all neighborhoods are open, we know all singleton sets are open! Since every set can be expressed as a union of singleton sets, it follows that every set is open. It follows immediately that every subset is also closed, since complements of open sets are closed. The compact subsets of X are precisely the finite subsets: We already know that all finite sets are compact. To see that an infinite subset of X cannot be compact, let E ⊂ X be an infinite set. Then the collection of singleton sets {{p}}p∈E is clearly an open cover of E that has no finite subcover! 12. Let {Uα }α∈A be a cover of K by open sets in R. One of these, say Uα0 , contains 0. Since Uα0 is open, there exists r > 0 such that Nr (0) ⊂ Uα0 . Now, choose a positive integer N such that 1/N < r. Then, for all natural numbers n ≥ N , 1/n ∈ Uα0 . For each j = 1, . . . , n − 1, choose αj ∈ A such that 1/j ∈ Uαj , which is possible since {Uα }α∈A covers K. Then clearly, {Uαj }j=0,1,...,n−1 is a finite subcover of K. This proves K is compact. 14. Consider the collection of open sets {( n1 , 1 − n1 )}n=3,4,5,... . It’s easy to see that this is an open cover of (0, 1). However, the union of any finite subcollection of these sets is equal to a single one of these sets (corresponding to the largest value of n in the subcollection), and hence is not a cover of (0, 1).