MA2223: SOLUTIONS TO ASSIGNMENT 3
1. Verify that each of the following mappings are isometries on R2 .
(a) Reflection.
T : R2 → R2 , (x1 , x2 ) 7→ (x2 , x1 )
Let x = (x1 , x2 ) and y = (y1 , y2 ) be points in R2 . Recall that the
Euclidean metric on R2 is given by
d(x, y) =
p
(x1 − y1 )2 + (x2 − y2 )2
We have
d(T (x), T (y)) = d((x2 , x1 ), (y2 , y1 ))
p
=
(x2 − y2 )2 + (x1 − y1 )2
p
=
(x1 − y1 )2 + (x2 − y2 )2
= d(x, y)
(b) Translation.
T : R2 → R2 , (x1 , x2 ) 7→ (x1 + a, x2 + b)
where (a, b) ∈ R2 .
Let x = (x1 , x2 ) and y = (y1 , y2 ) be points in R2 . We have
d(T (x), T (y)) = d((x1 + a, x2 + b), (y1 + a, y2 + b))
p
=
((x1 + a) − (y1 + a))2 + ((x2 + b) − (y2 + b))2
p
=
(x1 − y1 )2 + (x2 − y2 )2
= d(x, y)
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MA2223: SOLUTIONS TO ASSIGNMENT 3
(c) Rotation.

T : R2 → R 2 , 
x1
x2


 7→ 
cos θ − sin θ

x1

sin θ

x2

cos θ
where θ ∈ R.
Let x = (x1 , x2 ) and y = (y1 , y2 ) be points in R2 . We have
d(T (x), T (y)) = d((x1 cos θ − x2 sin θ, x1 sin θ + x2 cos θ),
(y1 cos θ − y2 sin θ, y1 sin θ + y2 cos θ))
p
= ((x1 cos θ − x2 sin θ) − (y1 cos θ − y2 sin θ))2
+((x1 sin θ + x2 cos θ) − (y1 sin θ + y2 cos θ))2
p
= ((x1 − y1 ) cos θ − (x2 − y2 ) sin θ)2
+((x1 − y1 ) sin θ + (x2 − y2 ) cos θ)2
p
= (x1 − y1 )2 + (x2 − y2 )2
= d(x, y)
2. Prove that the following mappings are continuous.
(a) Contractions.
Let (X, d) and (Y, d0 ) be metric spaces and let T : X → Y be a
contraction. Then there exists α < 1 such that
d0 (T (x), T (y)) ≤ α d(x, y)
∀ x, y ∈ X
Fix x ∈ X. Given any positive real number we have
d(x, y) < =⇒ d0 (T (x), T (y)) ≤ α d(x, y) < d(x, y) < This shows that T is continuous at x.
(b) Isometries.
MA2223: SOLUTIONS TO ASSIGNMENT 3
3
Let (X, d) and (Y, d0 ) be metric spaces and let T : X → Y be an
isometry. Then
d0 (T (x), T (y)) = d(x, y)
∀ x, y ∈ X
Given any positive real number we have
d(x, y) < =⇒ d0 (T (x), T (y)) = d(x, y) < This shows that T is continuous at x.
(c) The norm on a normed vector space.
Let (X, k.k) be a normed vector space. The norm is a mapping
k.k : X → R. Note that for any pair of points x, y ∈ X we have
kxk = kx − y + yk ≤ kx − yk + kyk
=⇒ kxk − kyk ≤ kx − yk
and
kyk = kx − y − xk ≤ kx − yk + kxk
=⇒ − (kxk − kyk) ≤ kx − yk
Combining these two inequalities we have
| kxk − kyk | ≤ kx − yk
∀ x, y ∈ X
Fix x ∈ X. Given any positive real number we have
kx − yk < =⇒ | kxk − kyk | ≤ kx − yk < This shows that the norm k.k is continuous at x.
3. Let (X, k.k) be a normed vector space. Prove that a subset A of X is
bounded if and only if there exists a real number M > 0 such that
sup kxk < M
x∈A
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MA2223: SOLUTIONS TO ASSIGNMENT 3
Solution: Suppose A is bounded. Then there exists an open ball B(x0 , r)
which contains A. For each x ∈ A we have
kxk = kx − x0 + x0 k
≤ kx − x0 k + kx0 k
< r + kx0 k
Thus r + kx0 k is an upper bound for the set {kxk : x ∈ A}. Letting
M = r + kx0 k + 1 we have
sup kxk < M
x∈A
Conversely, suppose there exists M such that
sup kxk < M
x∈A
Then A is contained in the open ball B(0, M ) and so A is bounded.