2.5 Let E◦ Denote the Set of All Interior Points of a Set E. (A) Prove That E◦ Is Always Open

Home , Interior (topology)

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 ∈ 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 ∈ 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 ∈ E, then there exists δ > 0 such that N(p, δ) ⊂ E. This means that p is an interior point, so p ∈ 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 ∈ G, since G is open, there exists a neighborhood N of p such that p ∈ N ⊂ G ⊂ E. This implies that p is an interior point of E, so p ∈ E◦. Thus, G ⊂ E◦. (d) Suppose p ∈ (E◦)c. Then p∈ / 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 p∈ / E◦. It follows that either p ∈ Ec or there exists x ∈ N ∩ Ec, with x 6= p. Equivalently, either p ∈ Ec or p ∈ (Ec)0. Thus, p ∈ 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 ∈ X and q ∈ 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 ∈ 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 ∈ A, the neighborhood N 1 (p) = {p} ⊂ 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) = {p}, 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 ∈ R and y ∈ R, deﬁne Rudin’s Ex. 11

2 d1(x, y) = (x − y) , p d2(x, y) = |x − y|, 2 2 d3(x, y) = |x − y |,

d4(x, y) = |x − 2y|, |x − y| d (x, y) = . 5 1 + |x − y|

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 ∈ 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) = |x − y| > 0 if x 6= y; and d2(x, x) = 0 for all x ∈ R. p p d2(x, y) = |x − y| = |y − x| = d2(y, x). Since |x − y| ≤ |x − z| + |z − y|, we have

p|x − y| ≤ p|x − z| + |z − y| ≤ p|x − z| + p|z − y|.

This is the triangular inequality for d2. 2 2 d3 is not a metric, since d3(1, −2) = |1 − (−1) | = 0.

d4 is not a metric, since d4(1, 1) = |1 − 2| = 1 6= 0.

d5 is a metric. In fact, it is known that d(x, y) = |x − y| 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).