Topology. SM464. HW 21. Solutions.
Due: Friday, September/05.
Subject: Metric Spaces.
(2) Let X = C([0, 1]) be the set of continuous functions on [0, 1]. For
f, g ∈ X, put
∫ 1
d(f, g) =
|f (x) − g(x)|dx.
0
Prove that d is a metric.
Proof. Axiom 1. If f = g, then |f (x) − g(x)| = 0 for all x. Hence,
∫1
∫1
d(f, g) = 0 |f (x)−g(x)|dx = 0. Conversely, assume that d(f, g) = 0 |f (x)−
g(x)|dx = 0. We claim that h(x) = |f (x) − g(x)| ≡ 0. We will prove it by
contradiction. Assume that ϵ = h(x0 ) > 0 for some x0 ∈ [0, 1]. Since h is a
continuous function, there exists δ > 0 such that whenever x ∈ [0, 1] is such
that |x0 − x| < δ, |h(x0 ) − h(x)| < ε/2. Unfolding the last inequality, we see
that h(x) > h(x0 ) − ε/2 = ε/2. It follows that
∫
1
∫
h(x)dx ≥
∫
min{1,x0 +δ}
h(x)dx ≥
max{0,x0 −δ}
0
min{1,x0 +δ}
max{0,x0 −δ}
ε
dx ≥ δε/2 > 0.
2
This contradicts the assumption
∫
1
h(x)dx = 0.
0
Axiom 2.
∫
d(f, g) =
1
∫
1
|f (x) − g(x)|dx =
0
|g(x) − f (x)|dx = d(g, f ).
0
Axiom 3.
∫
1
d(f, g) =
|f (x) − g(x)|dx
0
∫
≤
1
|f (x) − h(x) + h(x) − g(x)|dx
0
∫
≤
1
|f (x) − h(x)| + |h(x) − g(x)|dx
0
= d(f, h) + d(h, g).
1
2
(3) Let X = Z. For a, b ∈ Z, put
( )max{n≥0 : 3n divides
1
d(a, b) =
3
a−b}
if a ̸= b; and d(a, b) = 0 if a = b. Prove that (X, d) is a metric space.
Proof. We only need to check the triangle inequality. Let a, b, c ∈ Z.
Without loss of generality we can assume that the set {a, b, c} has no repeated elements, as in this case the triangle inequality becomes a trivial
identity.
Note that there exist nab ≥ 0 and kab ∈ Z such that
a − b = 3nab kab
and kab is not divisible by 3. Since, by assumption, a ̸= b, kab ̸= 0. Similarly,
we can write down
a − c = 3nac kac
and
c − b = 3ncb kcb .
Note that, by definition of the metric d, we have that d(a, b) = 3−nab ,
d(a, c) = 3−nac , and d(c, b) = 3−ncb . Consider
3nab kab = a − b = (a − c) + (c − b) = 3nac kac + 3ncb kcb .
Consider the case when nac ≥ ncb . Then the last equation can be rewritten
as
a − b = 3ncb (3nac −ncb kac + kcb ).
Since nab is the maximum integer with the property that a − b = 3nab kab ,
we get that nab ≥ ncb . It follows that
d(a, b) = 3−nab ≤ 3−ncb = d(c, b).
In the case nac ≤ ncb , using similar arguments, we get
d(a, b) ≤ d(a, c).
In either case,
d(a, b) ≤ max(d(a, c), d(c, b)) ≤ d(a, c) + d(c, b).