Answers to Homework 4, Math 4111
(1) Prove that the following examples from class are indeed metric
spaces. You only need to verify the triangle inequality.
(a) Let C be the set of continuous functions from [0, 1] → R
with the sup norm: ||f || = supx∈[0,1] {|f (x)|}. (You may
use any fact that you have studied in Calculus.)
Let f, g, h ∈ C. We want to prove that d(f, g) ≤ d(f, h) +
d(h, g). That is, ||f − g|| ≤ ||f − h|| + ||h − g||. Writing
f − h = u, h − g = v, we have u + v = f − g. So, we wish
to show that ||u + v|| ≤ ||u|| + ||v||. For any x ∈ [0, 1],
|(u + v)(x)| = |u(x) + v(x)| ≤ |u(x)| + |v(x)|
by triangle inequality. Since |u(x)| ≤ supx∈[0,1] {|u(x)|} =
||u|| and similarly |v(x)| ≤ ||v|| for all x ∈ [0, 1], we get
that |(u + v)(x)| ≤ ||u|| + ||v|| for all x ∈ [0, 1] and thus,
||u + v|| = supx∈[0,1] {|(u + v)(x)|} ≤ ||u|| + ||v||.
(b) Q with the p-adic norm for a prime p. Let me recall it here.
If 0 6= r ∈ Q, we can write it as pn (a/b) with a, b non-zero
integers relatively prime to p and a unique integer n which
we call vp (r), the valuation of r. Define ||r|| = 0 if r = 0
and = p−vp (r) if r 6= 0.
Exactly as in the previous problem, suffices to prove that
for any two rational numbers r, s, ||r + s|| ≤ ||r|| + ||s||. If
r, s or r + s is zero, this is immediate, so we may assume
that none of these are zero. If we write r = pn (a/b), s =
pm (c/d) as in the definition, we may assume without loss of
generality, n ≥ m. Then, we have r = pm (pn−m a/b) with
pn−m a ∈ Z. So,
n−m
n−m
p
a c
p
ad + bc
m
m
r+s=p
+
=p
.
b
d
bd
Since the denominator bd is relatively prime to p, we see
n−m
that p bdad+bc = pk (e/f ) with k ≥ 0 and e, f integers
relatively prime to p. Thus vp (r + s) = m + k ≥ m. So,
||r + s|| = p−(m+k) ≤ p−m ≤ p−n + p−m = ||r|| + ||s||.
(Notice that we have actually proved something stronger.
We have proved ||r + s|| ≤ max{||r||, ||s||}.)
(c) The space `2 of all sequences {xn } where
pP xn ∈ R with
P 2
x2n . First prove
xn < ∞ with the norm ||{xn }|| =
2
that if {xn }, {yn } ∈ ` then so is {xn + yn } and thus the
1
2
metric d({xn }, {yn }) = ||{xn − yn }|| is well defined and
then it is indeed a metric.
If we P
show that the monotonic non-decreasing sequence
An = nk=1 (xk +yk )2 is bounded above by (||{xn }|| + ||{yn }||)2 ,
then, it will prove that {An } has a limit and this limit once
defined is precisely ||{xn +yn }||2 and as before, the triangle
inequality will follow.
An =
n
X
2
(xk + yk ) =
k=1
n
X
x2k
+
k=1
n
X
yn2
+2
n
X
k=1
xk y k
k=1
≤ ||{xn }||2 + ||{yn }||2 + 2
n
X
|xk yk |
k=1
n
X
k=1
!2
|xk yk |
≤
n
X
!
x2k
k=1
n
X
!
yk2
k=1
2
≤ ||{xn }|| ||{yn }||2
by Cauchy-Scwartz inequality. Thus, we get,
An ≤ ||{xn }||2 + ||{yn }||2 + 2||{xn }|| · ||{yn }||
= (||{xn }|| + ||{yn }||)2
This is what we set out to prove.
(2) Let d, d0 be two metrics on a set M . We say they are equivalent
if a subset U ⊂ M is open with respect to the d-metric if and
only if it is open with respect to the the d0 -metric.
(a) Prove that d, d0 are equivalent if and only if for any a ∈
M, r > 0, there exists an s > 0 such that Bd0 (a, s) ⊂
Bd (a, r) (open balls of radius s, r with center a in the d0 , dmetric respectively) and given r0 > 0 there exists s0 > 0
such that Bd (a, s0 ) ⊂ Bd0 (a, r0 ).
If d, d0 are equivalent, Bd (a, r) is open with respect to d0
and a ∈ Bd (a, r). So there exists an s > 0 such that
Bd0 (a, s) ⊂ Bd (a, r). The other part is similar.
Now assume the conditions on open balls. Let U be open
with respect to the d-metric. Then if a ∈ U , there exists
an r > 0 and Bd (a, r) ⊂ U , by definition of open set.
By hypothesis, we have an s > 0 such that Bd0 (a, s) ⊂
Bd (a, r) ⊂ U . Thus, U is open with respect to the d0 metric. Identical argument will show that if U is open
with respect to d0 , then it is with respect to d.
3
(b) Prove that if d is a metric and A is a positive real number,
then d0 = Ad defined as d0 (a, b) = Ad(a, b) is also a metric
and equivalent to d.
Verification that d0 is a metric is straightforward.
Since Bd (a, r) = Bd0 (a, Ar), we see that the metrics are
equivalent.
(c) Prove that, if d is a metric, then d0 defined as,
d0 (a, b) =
(1)
d(a, b)
1 + d(a, b)
is also a metric and equivalent to d. (This is often called
the bounded metric associated to d. Note that d0 (a, b) ≤ 1
for all a, b.)
Let us check that this is indeed a metric. As usual, we will
check only the triangle inequality, since others are obvious.
For any three points a, b, c, let α = d(a, b), β = d(b, c) and
γ = d(a, c). Then these are non-negative and α ≤ β + γ
since d is a metric. We wish to show that,
α
β
γ
≤
+
.
1+α
1+β 1+γ
α(1 + β)(1 + γ) = α + α(β + γ) + αβγ
≤ β + γ + α(β + γ) + 2βγ + 2αβγ
= β(1 + α)(1 + γ) + γ(1 + α)(1 + β)
Dividing both sides by (1 + α)(1 + β)(1 + γ), we get the
inequality (1).
Next we check that the two metrics are equivalent. We will
r
check that Bd0 (a, 1+r
) ⊂ Bd (a, r) for any a ∈ M and any
r
r > 0. Similarly, if 0 < r < 1, then Bd (a, 1−r
) ⊂ Bd0 (a, r).
Again, we will prove only one of these, the other being
similar. (Why is this enough?) This is just elementary
algebra.
r
), then 1 ≤ 1 +
So, assume 0 < r < 1. If x ∈ Bd (a, 1−r
r
1
1
d(a, x) < 1 + 1−r = 1−r which implies 1+d(a,x) > 1 − r.
d(a, x)
1 + d(a, x)
1
=1−
1 + d(a, x)
< 1 − (1 − r) = r.
d0 (a, x) =
4
The other inclusion is equally easy.
(d) Define for x = (x1 , . . . , xn ), y = (y1 , . . . , yn ) ∈ Rn , d(x, y) =
|x1 − y1 | + · · · + |xn − yn |. Decide whether this is a metric
and if so, whether it is equivalent to the Euclidean metric.
This is a metric is easy, since the triangle inequality follows
easily from the triangle inequality for R1 .
This metric is equivalent to the Euclidean metric, which
we call E. We will verify the conditions
in Problem 2
√
(a) above. We show that Bd (a, r/ n) ⊂ BE (a, r) for any
a ∈ Rn , r > 0 and BE (a, r/n) ⊂ Bd (a, r). We will show
one of these, the other
√ being similar.
So, let x ∈ Bd (a, r/ n). As before, we write x = (x1 , . . . , xn )
and a = (a1 , . . . , an ). Then,
r
|x1 − a1 | + · · · + |xn − an | < √ .
n
√
This implies |xk − ak | <Pr/ n for 1 ≤ k ≤ n and so
(xk − ak )2 < r2 /n. Thus (xk − ak )2 < r2 and so
qX
dE (a, x) =
(xk − ak )2 < r.
So, x ∈ BE (a, r).
(3) Let f : (M, d) → (N, d0 ) be a function. Prove that the following
statements are equivalent.
(a) f −1 (U ) is open for any U ⊂ N open.
(b) f −1 (Z) is closed for any Z ⊂ N closed.
(c) If f (a) = b for any r > 0, there exists s > 0 such that
f (Bd (a, s)) ⊂ Bd0 (b, r).
(d) For any subset S ⊂ M and a an adherent point of S, f (a)
is an adherent point of f (S).
(a) ⇔ (b): This is easy by noticing that for any set S ⊂ N ,
f −1 (N − S) = M − f −1 (S).
(a) ⇒ (c): Let f (a) = b and for any r > 0, consider
Bd0 (b, r). Then this set is open in N and thus by (a),
f −1 (Bd0 (b, r)) is open in M . It also contains a and so by
definition of open sets, there exists an s > 0 such that
Bd (a, s) ⊂ f −1 (Bd0 (b, r)) . This means, f (Bd (a, s)) ⊂
Bd0 (b, r).
(c) ⇒ (a): Let U ⊂ N be open and let a ∈ f −1 (U ).
Then f (a) = b ∈ U and thus there exists an r > 0
such that Bd0 (b, r) ⊂ U . By (c), we have an s > 0 such
that f (Bd (a, s)) ⊂ Bd0 (b, r) ⊂ U and thus Bd (a, s) ⊂
5
f −1 (Bd0 (b, r) ⊂ f −1 (U ). Since this is true for any a ∈
f 1 (U ), this set is open in M .
(c) ⇒ (d): Let S ⊂ M and a an adherent point. If
f (a) is not an adherent point of f (S), there exists a ball
Bd0 (f (a), r) which does not intersect f (S). But, by (c),
there exists a Bd (a, s) such that f (Bd (a, s)) ⊂ Bd0 (f (a), r)
and thus f (Bd (a, s)) ∩ f (S) = ∅. But, Bd (a, s) ∩ S 6= ∅ and
that leads to a contradiction.
(d) ⇒ (a): Let U be open in N and assume that f −1 (U )
is not open. Then there exists an a ∈ f −1 (U ) such that
for any n ∈ N, Bd (a, 1/n) is not contained in f −1 (U ). So,
pick xn ∈ Bd (a, 1/n) − f −1 (U ) and let S = {xn }. If r >
0, choosing an n ∈ N such that 1/n < r, we see that
Bd (a, 1/n) ⊂ Bd (a, r) and thus Bd (a, r) ∩ S 6= ∅ for any
r > 0. So, a is an adherent point of S. By (d), f (a)
must be an adherent point of f (S). Since a ∈ f −1 (U ),
f (a) ∈ U and since U is open, we can find an r > 0 such
that Bd0 (f (a), r) ⊂ U . But, f (a) is an adherent point of
f (S) implies for some n ∈ N, f (xn ) ∈ Bd0 (f (a), r) which
implies f (xn ) ∈ U and thus xn ∈ f −1 (U ), contradicting
our choice of xn .