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Math 515 Professor Lieberman November 8, 2004 HOMEWORK #11 SOLUTIONS Chapter 7 6. (a) Let the ball be B(x, δ) and let y ∈ B(x, δ). Set ε = δ − ρ(x, y). If z ∈ B(y, ε), then ρ(x, z) ≤ ρ(x, y) + ρ(y, z) < ε, so B(x, δ) is open. (b) Write E for {x : ρ(x, y) ≤ δ} and let z ∈ Ẽ. Then ρ(z, y) > δ. Therefore ε = ρ(z, y) − δ > 0 and, if w ∈ B(z, ε), we have ρ(y, w) ≥ ρ(y, z) − ρ(z, w) = ε + δ − ρ(z, w) > δ, so w ∈ Ẽ. Therefore Ẽ is open, so E is closed. (c) NO. Let X = {0, 1} and define ρ(x, y) = |x − y|. This is easily seen to be a metric space, but the closure of the ball {x : ρ(x, 0) < 1} is the set {0} while {x : ρ(x, 0) ≤ 1} = X. 7. All except L∞ . We go through the spaces in order. For Rn , the set of points with rational coordinates is dense and countable. For C, we construct a countable dense set of piecewise linear functions. (This avoids the Weierstrass approximation theorem entirely.) For each positive integer n, we write Ln for the set of functions f such that, for each integer m ∈ {1, . . . , n}, there is an integer k(m) such that f (m/n) =S k(m)/n, and f is linear between these points. Each Ln is clearly countable, so L = Ln is also countable. To see that L is dense, fix f ∈ C and ε > 0. Then f is uniformly continuous, so there is δ > 0 such that |f (x) − f (y)| < ε/2 if |x − y| < δ. Choose n ≥ max{2/ε, δ}. Then there is a function g ∈ Ln such that |f (x) − g(x)| ≤ ε/2 if x has the form m/n for some integer m. If x is between two such numbers, say m/n < x < (m + 1)/n, then g(x) ≤ max{g( m+1 m m+1 ε m ), g( ) < max{f ( ), f ( ) + < f (x) + ε, n n n n 2 and similarly, g(x) > f (x) − ε, so kf − gk < ε and hence L is dense in C. For L∞ , look at the set of functions {χ[0,x) : x ∈ [0, 1]}. It’s easy to check that the L∞ distance between any two of these functions is 1 and that this set is uncountable. Any dense set must contain at least one element within 1/3 of each χ[0,x) , so any dense set must be uncountable. For L1 , the set L from the discussion of C is dense and countable. 1 2 10. (a) Let i be the identity map. Then i is continuous if and only if given x ∈ X and ε > 0, there is a δ1 > 0 such that ρ(x, y) < δ1 ⇒ σ(x, y) = σ(i(x), i(y)) < ε. Similarly, i−1 is continuous if and only if given x ∈ X and ε > 0, there is a δ2 > 0 such that σ(x, y) < δ2 ⇒ ρ(x, y) = ρ(i−1 (x), i−1 (y)) < ε. The proof is completely by taking δ = min{δ1 , δ2 }. (b) We first observe that, for any n-tuple (ξ1 , . . . , ξn ) of nonnegative numbers, we have X 1/2 hX i2 1/2 X ξi2 ≤ = ξi , ξi X ξi ≤ n max{ξ1 , . . . , ξn }, X 1/2 max{ξ1 , . . . , ξn } ≤ ξi2 . Therefore ρ(x, y) ≤ ρ∗ (x, y) ≤ nρ+ (x, y) ≤ ρ(x, y). It follows that the implications of part (a) hold with δ = ε/n for any pair of metrics from the list ρ, ρ∗ , ρ+ , and hence these metrics are equivalent. (c) Take ( 1 if x 6= y, σ(x, y) = 0 if x = y. This is not equivalent to ρ because there is no positive number δ such that ρ(0, y) < δ implies that σ(0, y) < 1/2. (Here 0 is the origin in Rn .) 11. (a) First, we note that f (s) = s/(1 + s) is a strictly increasing function with f (0) = 0 and lims→∞ f (s) = 1. Properties i, ii, and iii are then immediate. To check iv, we write σ(x, y) = f (ρ(x, y)) ≤ f (ρ(x, z) + ρ(z, y)) ρ(x, z) ρ(z, y) + 1 + ρ(x, z) + ρ(z, y) 1 + ρ(x, z) + ρ(z, y) ρ(z, y) ρ(x, z) + = σ(x, z) + σ(z, y). ≤ 1 + ρ(x, z) 1 + ρ(z, y) = To see that σ and ρ are equivalent, let ε > 0 and x ∈ X be given. Set δ = min{ε, 1}/2. If ρ(x, y) < δ, then σ(x, y) ≤ ρ(x, y), so σ(x, y) < δ ≤ ε. On the other hand, if σ(x, y) < δ, then σ(x, y) < 12 , so ρ(x, y) ≤ f −1 (1/2) = 1. Therefore σ(x, y) ≥ ρ(x, y)/2 and hence ρ(x, y) ≤ 2σ(x, y) < ε. Finally, since f (s) ≤ 1 for all s, it follows that σ(x, y) ≤ 1 for all x and y in X.