MTH6127: Metric Spaces and Topology Solutions to the Exercises

MTH6127:
Metric Spaces
and Topology
Solutions to
the Exercises

, CHAPTER ww 1
Basic Definitions and Examples
w w w




Exercise 1 w



For the L1-metric defined in (1.4), compute the distance dL1 (ƒ, g) between ƒ (x) = ex
w w w w w w w w w w w w w w w w



and g(x) = 2 for [a, b] = [0, 5].
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Solution 1. We have w w w



∫ b ∫ 5
dL1 (ƒ, g) :=
w w w |ƒ (x) − g(x)|dx =
w w w w
|ex − 2|dx.
w w

a 0

Since ex − 2 changes its sign at x = ln 2 (where ex = 2), we obtain
w w w w w w w w w w w w w w w w w




∫ ln 2 ∫ 5 w


dL1 (ƒ, g) = (2 − e )dx +
x (ex — 2)dx w w w w w w w


0 ln 2 w


ln 2 5
= (2x − e x) + (ex − 2x)
w w w
w
w w w w w ww w
0 ln 2w



= (2 ln 2 − 2) − (0 − 1) + (e − 10) − (2 − 2 ln 2)
w w w w w w w w w w w
5
w w w w w w w w




= 4 ln 2 − 13 + e .
w w w w w w w
5




Solution 2. Proof. We need to check the properties (M1)–
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(M3). (M1): Obviously d∞(x, y) ≥ 0 for all x, y ∈ Rn. Moreover,
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d∞(x, y) = 0 ⇔ max |xk − yk| = 0 ⇔ |xk − yk| = 0 ∀k = 1, . . . , n
w w w w w w w w w w w w w w w w w w w w w w w w

1≤k≤n
⇔ xk = yk ∀k = 1, . . . , n ⇔ x = y.
w w w w w w w w w w w w w w w




(M2): d∞(x, y) = max1≤k≤n|xk − yk| = max1≤k≤n|yk − xk| = d∞(y, x), where we used the fa
w w w w w w w w w w w w w w w w w w


ct that |−x| = |x| (which is, in fact, (N2) for the absolute value on R).
w w w w w w w w w w w w w w w




1

,Metric Spaces and Topology
w w w Dr. Reto Buzano
w w




(M3): Using the triangle inequality for the absolute value on R (i.e.
w w w w w w w w w w w



(N3)) and the w w w



subadditivity of the maximum, we obtain for x, y, z ∈ Rnw w w w w w w w w w w




d∞(x, y) = max |xk − yk| = max |xk − zk + zk − yk|
w w w w w w w w w w w w w w w


1≤k≤n 1≤k≤n
≤ max (|xk − zk| + |zk − yk|) ≤ max |xk − zk| + max |zk − yk|
w w w w w w w w w w w w w w w w w w


1≤k≤n 1≤k≤n 1≤k≤n
= d∞(x, z) + d∞(z, y).
w w w w w




Solution 3. Proof. We need to check (M1)–
w w w w w w w




(M3) for da and db. We start with da. (M1): d(x, y) ≥ 0 ⇒ da(x, y) ≥ 0 ∀x, y ∈ X. Moreo
w w w w w w w w w w w w w w w w w w w w w w w




ver, using (M1) for d,
w w w w




da(x, y) = 0 ⇔ min{d(x, y), 1} = 0 ⇔ d(x, y) = 0 ⇔ x = y.
w w w w w w w w w w w w w w w w w w




(M2): If da(x, y) < 1, then da(x, y) = d(x, y) = d(y, x) = da(y, x) by (M2) for d. If da
w w w w w w w w w w w w w w w w w w w w w w w



(x, y) = 1, then d(x, y) ≥ 1 and hence by (M2) for d also d(y, x) ≥ 1 implying da(y, x) = 1.
w w w w w w w w w w w w w w w w w w w w w w w w




(M3): Using (M3) for d, we find
w w w w w w




da(x, y) = min{d(x, y), 1} ≤ min{d(x, z) + d(z, y), 1}
w w w w w w w w w w w w




≤ min{d(x, z), 1} + min{d(z, y), 1} = da(x, z) + da(z, y).
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For db, we obtain the following.
w w w w w




(M1): d(x, y) ≥ 0 ⇒ db(x, y) ≥ 0 ∀x, y ∈ X. Moreover, using (M1) for d,
w w w w w w w w w w w w w w w w w w




d(x, y) w


da(x, y) = 0 ⇔ w w w w w = 0 ⇔ d(x, y) = 0 ⇔ x = y.
w w w w w w w w w w

1 + d(x, y)
w w w w



d(x,y) d(y,x)
(M2): db(x, y) = w w w ww w w = ww w w = db(y, x) by (M2) for d.
w w w w w w

1+d(x,y) 1+d(y,x)
(M3): We look at the function x '→ ƒ (x) :=
w w w w w w w w w w ww
1+x
x
ww w . Since

1 x 1 h x i w



ƒ r(x) = −
w w w w w w w w w w w w w w w w


2 = 1= > 0, w

(1 + x)
w w w w w w w

1+x w w1+x 1+x w w w w w w




ƒ is monotonically increasing. By this monotonicity and (M3) for d, we therefore find
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db(x, y) = ƒ (d(x, y)) ≤ ƒ (d(x, z) + d(z, y))
w w w w w w w w w w w w



d(x, z) d(z, y) w w

= +
1 + d(x, z) + d(z, y) 1 + d(x, z) + d(z, y) w w w w w w w w w w w w


d(x, z) d(z, y) w w


≤ + = db(x, z) + db(z, y).
w w w w w w w

1 + d(x, z) 1 + d(z, y) w w w w w w w w




2

, Metric Spaces and Topology
w w w Dr. Reto Buzano
w w




Solution 4. First, we note that if x = (x k ) k∞= 1 and y = (y k ) k∞= 1 are sequences formed by
w w w w w w w w w
w
w w w
w
w w w w




only 0’s and 1’s, then 0 ≤ |xk − yk| ≤ 1 for all k ∈ N and therefore
w w w
Σ∞ 0 —k ≤ 2—k|xk − yk| ≤ 2—
w w w w w w w w w w w w w w w w w w w w w w

k
w for all k ∈ N. Since we know that Σ
w w t h e geometric series k=1 2 = 1 converges, we
w w w w w w w w w w w
w
w w w w w


∞ —k|x − y | also converges (to a value of at
obtain by the comparison test that
w
k =1 2
w
k k w w w
w
w w w w w w w w w


most 1). This shows that the function d(x, y) is indeed well-
w w w w w w w w w w w



defined and moreover all the series below converge, justifying all manipulations we perfor
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m. We can now proceed to the actual proof.
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Proof. We need to check (M1)–(M3).
w w w w w




(M1): Obviously d(x, y) ≥ 0 for all x, y ∈ X and equality holds if and only if every sum
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mand vanishes, i.e. |xk − yk| = 0 for all k ∈ N and hence x = y.
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(M2): This is obvious (it follows by (N2) for the absolute value).
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(M3): The triangle inequality for the absolute value gives for x, y, z ∈ X
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0 ≤ |xk − yk| ≤ |xk − zk| + |zk − yk| ≤ 2,
w w w w w w w w w w w w w w ∀k ∈ N. w w




Σ ∞
k =1 2 · 2
Since w
—k = 2 converges, the comparison test shows that
w w w w w w w w w w
w


∞ ∞
Σ w Σ w


2—k|xk − yk| ≤ w w w w 2—k(|xk − zk| + |zk − yk|) w w w w w w


k=1 k=1

and both series converge. Therefore,
w w w w



∞ ∞
Σ w Σ w


d(x, y) = w w w 2—k|xk − yk| ≤ w w w w 2—k(|xk − zk| + |zk − yk|) w w w w w w

k=1
∞ k=1
∞
Σ w Σ w


= w 2—k|xk − zk| + w w w w 2—k|zk − yk| = d(x, z) + d(z, y),
w w w w w w w w

k=1 k=1
Σ∞ Σ∞
by a theorem f r Σ
w w o m Convergence aΣnd ContinuΣ
ity stating that if w w w w w w w w
k =1 ak w and w
k =1 bk w both
∞ ∞
w w

∞ a +
converge, then k =1(ak + bk) = k =1 k w k =1 bk. w
w w w
w
w




3