# Math4225 Homework 1

2.1.1 d(2, 3) need to satisfy the triangle inequalities: 2 + d(2, 3) ≥ 3, 3 + d(2, 3) ≥ 2, 2 + 3 ≥ d(2, 3). Therefore the allowable value is 5 ≥ d(2, 3) ≥ 1.

2.1.3 Let ai = |xi - yi|. Then

d2(x, y) = [a12 + … + an2]1/2 ≤ [a12 + … + an2 + 2a1a2 + 2a1a3 + … + 2an-1an]1/2 = a1 + … + an = d1(x, y).

On the other hand, by Schwartz inequality, we have

d1(x, y) = a1 + … + an = ⟨ (a1, …, an), (1, …, 1) ⟩ ≤ [a12 + … + an2]1/2 [12 + … + 12]1/2 = √n d2(x, y).

2.1.4 Basically we need to show limp→∞ [a1p + … + anp]1/p = max {a1, …, an} for non-negative a1, …, an. If all the numbers are zero, then the limits is true. If some number is nonzero, then we may assume max {a1, …, an} = a1 > 0, without loss of generality. Then [a1p + … + anp]1/p = a1[1 + (a2/a1)p + … + (an/a1)p]1/p. By 1 ≤ 1 + (a2/a1)p + … + (an/a1)pn, we get a1 ≤ [a1p + … + anp]1/pa1n1/p. By limp→∞ n1/p = 1 and the sandwich rule, we get limp→∞ [a1p + … + anp]1/p = a1.

2.1.6 The first inequality means d(x, y) - d(y, z) ≤ d(x, z) and d(y, z) - d(x, y) ≤ d(x, z). The inequality d(x, y) - d(y, z) ≤ d(x, z) follows from

d(y, z) + d(x, z)
= d(x, z) + d(z, y) (by symmetry)
d(x, y). (by triangle inequality)

The inequality d(y, z) - d(x, y) ≤ d(x, z) can be proved similarly.

The second inequality means d(x, y) - d(z, w) ≤ d(x, z) + d(y, w) and d(z, w) - d(x, y) ≤ d(x, z) + d(y, w). The inequality d(x, y) - d(z, w) ≤ d(x, z) + d(y, w) follows from

d(z, w) + d(x, z) + d(y, w)
= d(x, w) + d(y, w) (by triangle inequality)
= d(x, w) + d(w, y) (by symmetry)
d(x, y). (by triangle inequality)

The inequality d(z, w) - d(x, y) ≤ d(x, z) + d(y, w) can be proved similarly.

2.1.8 The subtle point here is the order of elements in the triangle inequality. First we have

d(x, y) + d(x, y)
d(y, y) (by (b))
= 0. (by (a))

Combined with (a), we get the positivity property. Second we have

d(y, x)
= d(y, x) + d(y, y) (by (a))
d(x, y). (by (b))

By switching the roles of x and y, we also get d(x, y) ≥ d(y, x). This proves the symmetry property. Once the symmetry is proved, the property (b) becomes the usual triangle inequality.

2.1.11 min{d, 1} is a metric: It clearly satisfies positivity and symmetry. The triangle in equality follows from min{a + b, 1} ≤ min{a, 1} + min{b, 1}.

d/(1 + d) is a metric: It clearly satisfies positivity and symmetry. The triangle inequality follows from

• t/(1 + t) is a strictly increasing function for t ≥ 0.
• a/(1 + a) + b/(b + 1) ≥ (a + b)/(1+ a + b) for a, b ≥ 0.

d is a metric: It clearly satisfies positivity and symmetry. The triangle inequality follows from a + bc ⇒ √a + √b ≥ √c.

d2 is not a metric: It satisfies positivity and symmetry. However, it does not satisfy the triangle inequality. For example, on the set {1, 2, 3} we have the metric d(1,2) = d(1,3) = 1, d(2,3) = 2. But d2(1,2) = d2(1,3) = 1, d2(2,3) = 4 fails the triangle inequality.

2.1.12 max{d1, d2} is a metric: It clearly satisfies positivity and symmetry. The triangle inequality follows from a1 + b1c1, a2 + b2c2 ⇒ max{a1, a2} + max{b1, b2} ≥ max{c1, c2}.

min{d1, d2} is not a metric: Although it satisfies positivity and symmetry. It fails the triangle inequality: Consider the following two metrics on three points.

Then the corresponding min{d1, d2} is

which does not satisfy the triangle inequality.

d1 + d2 is a metric: It clearly satisfies positivity and symmetry. The triangle inequality follows from a1 + b1c1, a2 + b2c2 ⇒ (a1 + a2) + (b1 + b2) ≥ (c1 + c2).

|d1 - d2| is not a metric: It may not satisfy positivity. An example is given by the case d1 = d2.

2.2.2 The following picture illustrates the relation between Euclidean ball and Taxicab ball. It suggests us that δ = ε/√n.

Specifically, let ai = |yi - xi|. Then we need to show

dEuclidean(x, y) = [a12 + … + an2]1/2 < δ ⇒ dtaxicab(x, y) = a1 + … + an < ε,

and δ is the biggest that makes the above happen. The implication above is in fact the Cauchy-Schwarz inequality

a1 + … + an = ⟨ (a1, …, an), (1, …, 1) ⟩ ≤ [a12 + … + an2]1/2 [12 + … + 12]1/2 ≤ δ √n = ε.

If we have δ > ε/√n, then the implication fails for a1 = … = an taken to be any number between δ/√n and ε/n.

2.2.5 The numbers in B2-adic(0, 2) are of the form x = (m/n)2a, where m, n are odd integers, a is an integer, and d(x, 0) = 2-a < 2. Thus 2a > 1/2, and this means x is of the form (any integer)/(odd integer).

The numbers in B2-adic(0, 2) are of the form x = (m/n)2a, where m, n are odd integers, a is an integer, and d(x, 0) = 2-a < 1. Thus 2a > 1, and this means x is of the form (even integer)/(odd integer). The difference B2-adic(0, 2) - B2-adic(0, 1) then consists of numbers of the form (odd integer)/(odd integer).

The numbers in B2-adic(0, 3) are of theform x = (m/n)2a, where m, n are odd integers, a is an integer, and d(x, 0) = 2-a < 3. The numbers in B2-adic(0, 4) are similar, with 2-a < 4 instead. Since for integers a, the conditions 2-a < 3 and 2-a < 4 are equivalent, we see that the two sets are the same.

2.2.6 A subset A of N is in B(E, 0.1) if and only if the minimum of the difference between A and E is > 0.1-1 = 10. In other words, A is obtained from E by modifying the part of E consisting of numbers > 10. Thus B(E, 0.1) consists of subsets of the form A = {2, 4, 6, 8, 10}∪B, where B consists of numbers > 10.

The condition d(A, C) < maxA-1 means that the minimum of the difference between A and C is > (maxA-1)-1 = maxA. In particular, this means that C is obtained by adding numbers > maxA to A (there is no number > maxA to delete from A). This is equivalent to AC.

2.2.9 The following picture suggests us to take ε = d(x, y)/2.

Thus we claim there is no intersection between B(x, ε) and B(y, ε). Suppose not and z is in both balls. Then

zB(x, ε) ⇒ d(x, z)< d(x, y)/2.
zB(y, ε) ⇒ d(z, y)< d(x, y)/2.

This implies d(x, z) + d(z, y)< d(x, y), which contradicts the triangle inequality.

2.3.1 In a finite metric space, ε = min{d(x, y): xy} is a positive number, because the set of pairs is finite. Then aU and d(x, a) < ε implies x = aU.

2.3.2 U1, U2 are open.

U3, U4 are not open.

2.3.3 (0,1] is open in the last four, but not in the first two.

2.3.4 We need to consider two cases for a subset U of X.

If 0 ∉ U, then U is always open. In fact, since any aU is nonzero, we have a = n-1. Then by taking ε = (3n)-1, we have (a - ε , a + ε) ∩ X = {a} which is still contained in U. Therefore U is open.

If 0 ∈ U, then by applying the definition of open subset to 0 in U, we see that there is big n, such that U contains all n-1, (n + 1)-1, (n + 2)-1, (n + 3)-1, … . In addition to these, there are only finitely many more points 1, 2-1, 3-1, …, (n - 1)-1 to consider. Adding any finitely many of these to U will still make U open (the rigorous argument is the same as the case U does not contain 0). In other words, we have

U = X - {finitely many nonzero numbers}.

In conclusion, there are two kinds of open subsets of X. Either 0 ∉ U, or U is X with finitely many nonzero numbers deleted.

2.3.5 U1 is L-open: Suppose f(0) > 1. Then take ε = f(0) - 1. For gBL(f, ε), we have |g(0) - f(0)| ≤ d(g, f) < f(0) - 1, which implies g(0) > 1.

U2 and U3 are not L-open: The constant function 1 is in U. For any ε, the constant function g = 1 - ε/2 ∈ BL(1, ε) but gU2. Moreover, g = 1 + εt/2 ∈ BL(1, ε) but gU3.

U4 is L-open: Similar proof to Example 2.9, by choosing ε = min{(t + 1) - f(t), f(t) -t2: t ∈ [0, 1]}.

U5 is not L-open: t2 is in U, but t2 - ε/2 is not in U.

U6 is L-open: If ∫f > 0 , then take ε = ∫f. For gBL(f, ε), we have |∫f - ∫g| ≤ ∫|f - g| ≤ d(f, g)(1 - 0) < ε = ∫f. This implies ∫g > 0.

U7 and U8 are not L-open: The constant function 0 ∈ U. For any ε, the constant function, g = - ε/2 ∈ BL(0, ε) but gU6. Moreover, g = εt/2 ∈ BL(0, ε) but gU7.

U1, U2 and U3 are not L1-open: The constant function 2 ∈ U. For any ε, we can always construct a function g such that g(0) < 0 and the area between the graphs of 2 and g is less than ε. This means BL1(2, ε) is not contained in U. The following picture illustrates the construction.

U4 and U5 are not L1-open: The construction of the counterexample is similar to the first three. We can find a function g, such that g(0) is as negative as we want (so that g(0) < a(0)), but the area between f and g is as small as we want.

U6 is L1-open: If ∫f > 0 , then take ε = ∫f. For gBL1(f, ε), we have |∫f - ∫g| < ε = ∫f. This implies ∫g > 0. This shows BL1(f, ε) ⊂ U.

U7 and U8 are not L1-open: The constant function 0 ∈ U. For any ε, the constant function, g = - ε/2 ∈ BL1(0, ε) but gU6. Moreover, g = εt/2 ∈ BL1(0, ε) but gU7.

2.3.6 U1 is not open: If U1 is open, then there is ε > 0, such that all numbers in B(0, ε) are integers. However, we can always find big n such that 2-n < ε. Then 2n/3 ∈ B(0, ε) but is not an integer.

U2 is not open: The reason is similar to the case of U1, since 2n/3 ∈ B(0, ε) but is not an integer, let alone even or odd number.

U3 is open: The subset is thos rational numbers satisfying d(x, 0) > 0. The subset is open by Exercise 2.3.8.

U4 is open: The subset is actually the ball B(0, 1).

U5 and U6 are not open: For any ε, we have 2n ∈ B(0, ε) for big n.

2.3.7 U1 is open: If d(A, B) < 0.1, then the difference between A and B are made up of some numbers > 10. In particular, if AU1, then B (and the ball B(A, 0.1)) is also in U1.

U2 is not open: Let AU2. No matter how small ε is, we can always find a number n satisfying ε > n-1. Then d( A∪{n}, A ) < ε, but A∪{n} ∉ U2.

U3 is not open: This is similar to the previous case. No matter how small ε is, we can always find an odd number n satisfying ε > n-1.

U4 is open: If d(A, B) < 0.01, then the difference between A and B are made up of some numbers > 100. In particular, if A does not contain 100, then B also does not contain 100.

U5 is not open: d(A, B) < ε if and only if the difference between the two subsets are > ε-1. In particular, for any infinite A, we can delete all the numbers in A that are > ε-1 and form B. Then B is finite and yet d(A, B) < ε.

U6 is not open: Given any A and ε, we can always add arbitrarily many numbers > ε-1 to get B so that d(A, B) < ε is satisfied. In particular, the number of elements can be increased arbitrarily.

2.3.8 If xU = {x: d(x, a) > ε}, then δ = d(x, a) - ε > 0. For yB(x, δ), we have

d(y, a) ≥ d(x, a) - d(x, y) > d(x, a) - δ = ε.

This proves that B(x, δ) ⊂ U. Therefore U is open.

2.3.9 The relation d'c d implies that Bd(1, c-1ε) ⊂ Bd'(x, ε). Then

UX is d'-open
⇒ For any xU, there is ε > 0, such that Bd'(x, ε) ⊂ U
⇒ For any xU, there is ε > 0, such that Bd(1, c-1ε) ⊂ U
UX is d-open

2.3.10 We have dd1n d and dd2 ≤ √n d. In general, ddpn1/p d. Then by Exercise 2.3.9, the metrics give the same topology.

2.3.11 By d1d on C[0,1] and Exercise 2.3.9, L1-open implies L-open. Example 2.9 shows the converse is not true.

2.3.12 Suppose d'-open subsets are d-open, then the d'-ball Bd'(x, ε) is d-open. Since xBd'(x, ε), this implies that there is δ > 0, such that Bd(x, δ) ⊂ Bd'(x, ε). This means exactly the second statement.

Conversely, suppose the second statement holds. Suppose U is d'-open. Then for any xU, there is ε > 0, such that Bd'(x, ε) ⊂ U. The second statement means exactly Bd(x, δ) ⊂ Bd'(x, ε) for some δ > 0. Thus we proved that xU implies Bd(x, δ) ⊂ U for some δ > 0. In other words, U is d-open.

By dd + d' and Exercise 2.3.9, d-open implies (d + d')-open. Therefore the third statement is the same as (d + d')-open implies d-open. By the equivalent between the first and the second statements, this means that for any x and ε, there is δ, such that d(x, y) < δ implies d(x, y) + d'(x, y) < ε. In particular, we get d(x, y) < δ implies d'(x, y) < ε, so that the third statement implies the second. Conversely, given the second statement, we have d(x, y) < δ implies d'(x, y) < ε. Then d(x, y) < min{δ, ε} implies d(x, y) + d'(x, y) < 2ε. This shows the second statement implies the third.

2.3.14 For yV, let ε = d(x, y). Then for zB(y, ε), we have d(z, y) < d(y, x). This implies zx, or equivalently zX - {x}. We have thus shown B(y, ε) ∈ X - {x} and proved the openness of X - {x}.

Deleting two points is still open because X - {x, y} = (X - {x}) ∩ (X - {x}) and the intersection of two open subsets is open.