Introduction to Topology – Solutions Manual (Tej Bahadur Singh) | Complete Worked Solutions

Tej Bahadur Singh




Solutions Manual to
Introduction to Topology
(by the author)

,
,Chapter 1
Topological Spaces




Section 1.1
1. For x, y, z ∈ X, if d(x, z) = 1, then x 6= z so that d(x, y) + d(y, z) is at
least 1.
2. (a) Positivity and symmetry are clear. For triangle inequality, let
x, y, z ∈ X. Then d(x, z) ≤ d(x, y) + d(y, z). If d(x, z) > 0, then
(1 + d(x, z))/d(x, z) ≥ [1 + d(x, y) + d(y, z)]/[d(x, y) + d(y, z)] ⇒
d(x, z)/(1 + d(x, z)) ≤ [d(x, y) + d(y, z)] /[1 + d(x, y) + d(y, z)] ≤
d(x, y)/(1 + d(x, y)) + d(y, z)/(1 + d(y, z)), since d(x, y) ≥ 0 and
d(y, z) ≥ 0. If d(x, z) = 0, the inequality still holds good, so
d′ (x, z) ≤ d′ (x, y) + d′ (y, z). Clearly, d′ (x, y) ≤ 1 for all x, y ∈ X
and, therefore, diam(X) ≤ 1.
(b) d1 (x, y) = min{1, d(x, y)} ≥ 0 and d1 (x, y) = 0 ⇔ d(x, y) =
0 ⇔ x = y. Symmetry is obvious. For triangle inequality, con-
sider points x, y, z ∈ X. If d(y, z) < 1 and d(x, y) < 1, then
d(x, z) ≤ d(x, y) + d(y, z) implies d1 (x, z) ≤ d1 (x, y) + d1 (y, z).
If d(x, y) ≥ 1 or d(y, z) ≥ 1, then d1 (x, y) + d1 (y, z) is at least 1,
while d1 (x, z) ≤ 1. Also diam(X) ≤ 1.
3. (a) If x 6= 0, then xi 6= 0 for some i. So k x k= max1≤i≤n | xi |> 0.
(b) k xa k= maxi | xi a |=| a | maxi | xi |=| a |k x k.
(c) k x + y k= maxi | xi + yi |≤ maxi (| xi | + | yi |) ≤ maxi | xi |
+maxi | yi |≤k x k + k y k.
4. (a) d∗ (x, y) = max | xi − yi | is obviously nonnegative and d∗ (x, y) =
i
0 ⇔ x = y. The symmetry is also obvious. For triangle inequality,
let x = (xi ), y = (yi ), z = (zi ) be any three elements of Rn . Then,
for each i, | xi − zi |≤| xi − yi | + | yi − zi |≤ d∗ (x, y) + d∗ (y, z). So
d∗ (x, z) ≤ d∗ (x, y) + d∗ (y, z) and d∗ is a metric.
(b) Let x = (xi ), y = (yi ) and z = (zi ) be elements of Rn . Clearly
d+ (x, y) ≥ 0 and d+ (x, y) = 0 ⇔ x = y. Also, the equality
d+ (x, y) = d+ (y, x) is obvious. For the triangle inequality, we have
Pn Pn
1 | xi − zi | ≤ P1 (| xi − yi | + P | yi − zi |)
n n
= 1 | x i − y i | + 1 | y i − zi | .

1

, 2 Introduction to Topology

5. B(x, r) in (R2 , d∗ ) is the square without its boundary centered at x
and side 2r, B(x, r) in (R2 , d+ ) is the rhombus without its boundary
and B(x, r) in (R2 , d) is the disc without its boundary centered at x
with radius r. In R3 , Bd∗ (x, r) is a cube without its faces, Bd+ (x, r) is
a tetragonal bipyramid without its faces and Bd (x, r) is a (solid) ball
without its surface.
6. The properties of a metric for the function d in Ex 1.1.4 follow from the
properties of Riemann integral. For the function d∗ in Ex 1.1.5, we see
that the set {| f (x) − g(x) |: x ∈ X} bounded for any f, g ∈ B(X), for
| f (x) − g(x) |≤| f (x) | + | g(x) | ∀ x ∈ X. So d∗ (f, g) is a real number
for all f, g ∈ B(X). The verification of the properties of a metric are
routine.

7. Choose a fixed point x0 ∈ X. Then for every x ∈ X,

d(f (x), g(x)) ≤ d(f (x), f (x0 )) + d(f (x0 ), g(x0 ))+
d(g(x0 ), g(x))
≤ diam (f (X)) + d(f (x0 ), g(x0 )) + diam (g(X)).
∗
So d (f, g) ≤ diam (f (X)) + d (f (x0 ), g(x0 )) + diam (g(X)) < ∞.
∗ ∗
Since d(y, y ′ ) ≥ 0 ∀ y, y ′ ∈ Y , d (f, g) ≥ 0 and d (f, g) = 0 ⇔
d(f (x), g(x)) = 0 ∀ x ∈ X ⇔ f = g. If f, g, h ∈ B(X, Y ), then
∗ ∗
d(f (x), g(x)) ≤ d(f (x), h(x)) + d(h(x), g(x)) ≤ d (f, h) + d (h, g). This
∗ ∗ ∗
implies d (f, g) ≤ d (f, h) + d (h, g). Symmetry is obvious.

8. (a) If f (x) = x + v, then kf (x) − f (y)k = kx − yk. So f sends B(x; ǫ)
into B(f (x); ǫ).
(b) Suppose g(x) = rx. If r = 0, then g is a constant map. If r 6= 0, then
kx − yk < ǫ/r ⇒ kg(x) − g(y)k < ǫ.

9. This is immediate from the inequality kxk − kyk ≤ kx − yk.

10. Let x, y ∈ X be arbitrary. Then, for every a ∈ A, d(x, a) ≤ d(x, y) +
d(y, a) ⇒ inf a∈A d(x, a) ≤ d(x, y) + inf a∈A d(y, a) ⇒ d(x, A) − d(y, A) ≤
d(x, y). Interchanging the roles of x and y, we obtain d(y, A) − d(x, A) ≤
d(x, y), and hence d(x, A) − d(y, A) ≤ d(x, y). This clearly implies that
f is continuous.

11. If there exists an x ∈ X and a real K such that d(x, a) ≤ K for all
a ∈ A, then d(a, b) ≤ 2K for all a, b ∈ A. So diam(A) ≤ 2K.

12. (a) Choose x0 ∈ A ∩ B. Let x, y ∈ A ∪ B. If x ∈ A, y ∈ B, then
d(x, y) ≤ d(x, x0 ) + d(x0 , y) ≤ diam(A) + diam(B). If both x, y are
in A, then d(x, y) ≤ diam(A), and if both of them are in B, then
d(x, y) ≤ diam(B). So diam(A ∪ B) ≤ diam(A) + diam(B).