2.5 Let E◦ denote the set of all interior points of a set E. Rudin' Ex. 9 (a) Prove that E◦ is always open. (b) Prove that E is open if and only if E◦ = E. (c) If G ⊂ E and G is open, prove that G ⊂ E◦. (d) Prove that the complement of E◦ is the closure of the complement of E. (e) Do E and E always have the same interiors? (f) Do E and E◦ always have the same closures? Proof (a) Suppose p 2 E◦. There there exists δ > 0 such that N(p; δ) ⊂ E. By Theorem 2.19, N(p; δ) is open. So, for any point x 2 N(p; δ), there exists > 0 such that N(x; ) ⊂ N(p; δ) ⊂ E. This means that every point in N(p; δ) is an interior point of E. Hence N(p; δ) ⊂ E◦. This implies E◦ is open. (b) If E = E◦, by (a), E is open. Conversely, suppose E is open. If p 2 E, then there exists δ > 0 such that N(p; δ) ⊂ E. This means that p is an interior point, so p 2 E◦. Hence E ⊂ E◦. On the other hand, interior points are E are necessarily in E, since any neighborhood of a point contains that point. Hence E◦ ⊂ E. We conclude that E = E◦. (c) If p 2 G, since G is open, there exists a neighborhood N of p such that p 2 N ⊂ G ⊂ E. This implies that p is an interior point of E, so p 2 E◦. Thus, G ⊂ E◦. (d) Suppose p 2 (E◦)c. Then p2 = E◦, that is, p is not an interior point of E. Hence, for any neighborhood N of p, necessarily N \ Ec 6= ;, otherwise N ⊂ E that contradicts to p2 = E◦. It follows that either p 2 Ec or there exists x 2 N \ Ec, with x 6= p. Equivalently, either p 2 Ec or p 2 (Ec)0. Thus, p 2 Ec [ (Ec)0 = Ec. This concludes that (E◦)c ⊂ Ec. The other inclusion Ec ⊂ (E◦)c can be obtained exactly by reversing the above arguments. (e) In general, E◦ 6= (E)◦. For example, take E = (−1; 0) [ (0; 1) ⊂ R. Then Ec = (−1; 0) [ (0; 1), and (E)◦ = [−1; 1]. (f) In general, E 6= E◦. For example, take E = Q. Then E = R, and E◦ = ;. 2.6 Let X be an infinite set. For p 2 X and q 2 X, define Rudin's Ex. 10 1; if p 6= q, d(p; q) = 0; if p = q. Prove that this is a metric. Which subsets of the resulting metric space are open? Which are closed? Which are compact? Proof To show that d is a metric, we only need to show the triangular inequality, since the other conditions hold obviously. Let p; q; r 2 X. If p = q, then d(p; q) = 0 ≤ d(p; r) + d(r; q). If p 6= q, then d(p; q) = 1, and d(p; r) + d(r; q) ≥ 1 since p; q; r cannot all be the same element. Hence again we have d(p; q) ≤ d(p; r) + d(r; q). Thus, d is a metric. 1 We claim that every subset A of X is open. In fact, if A = ;, then it is open. If A 6= ;, for any p 2 A, the neighborhood N 1 (p) = fpg ⊂ A. Again, A is open. 2 For any set A, from A = X − (X − A), since X − A is open, so A is closed. We claim that A is compact if and only if A is a finite set. It is clear from the definition that any finite set of X is compact. On the other hand, since N 1 (p) = fpg, 2 every point is isolated. Hence an infinite set of X has no any limit point, so cannot be compact, by Theorem 2.37. 2.7 For x 2 R and y 2 R, define Rudin's Ex. 11 2 d1(x; y) = (x − y) ; p d2(x; y) = jx − yj; 2 2 d3(x; y) = jx − y j; d4(x; y) = jx − 2yj; jx − yj d (x; y) = : 5 1 + jx − yj Determine, for each of these, whether it is a metric or not. Proof For each of these we must determine whether d(x; y) > 0 for x 6= y; d(x; x) = 0 for any x 2 R; d(x; y) = d(y; x); and d(x; y) ≤ d(x; z) + d(z; y). d1 is not a metric because it does not satisfy the triangle inequality. For example, d1(0; 2) = 4 > d1(0; 1) + d1(1; 2) = 1 + 1 = 2. p d2 is a metric. d2(x; y) = jx − yj > 0 if x 6= y; and d2(x; x) = 0 for all x 2 R. p p d2(x; y) = jx − yj = jy − xj = d2(y; x). Since jx − yj ≤ jx − zj + jz − yj, we have pjx − yj ≤ pjx − zj + jz − yj ≤ pjx − zj + pjz − yj: This is the triangular inequality for d2. 2 2 d3 is not a metric, since d3(1; −2) = j1 − (−1) j = 0. d4 is not a metric, since d4(1; 1) = j1 − 2j = 1 6= 0. d5 is a metric. In fact, it is known that d(x; y) = jx − yj is a metric of R. We shall show that whenever d is a metric, then d(x; y) D(x; y) = 1 + d(x; y) is a metric. We only check the triangular inequality for D since the other conditions are easy to be verified. 2 In fact, since d(x; y) ≤ d(x; z) + d(z; y), we have d(x; y) d(x; z) + d(z; y) d(x; y) d(x; z) + d(z; y) = + − 1 + d(x; y) 1 + d(x; z) + d(z; y) 1 + d(x; y) 1 + d(x; z) + d(z; y) d(x; z) d(z; y) ≤ + 1 + d(x; z) + d(z; y) 1 + d(x; z) + d(z; y) d(x; y) − [d(x; z) + d(z; y)] + [1 + d(x; y)][1 + d(x; z) + d(z; y)] d(x; z) d(z; y) ≤ + ; 1 + d(x; z) 1 + d(z; y) that is, D(x; y) ≤ D(x; z) + D(z; y). 3.