Principles of Mathematics in Operations Research

138 10 Basic Topology

Example 10.1.4 Is / : R H-» R, f(s) = s^2 bounded? (Exercise!).

/ : R i-> R, f(s) = arctan(s) — tan_1(s) is bounded. See Figure 9.4-

Definition 10.1.5 Let X = B{S) = all bounded functions f : S i-> R.

For f,g £ B(S), we define the distance asd(f,g) — sup{|/(s) — g(s)\ : s £ S}.

Proposition 10.1.6 d(f,g) > 0 is a metric, Vf,g £X = B(S).

Proof, by proving axioms of a metric:

(i) (=»)
if d(f,g) = 0 =» \f(s)-g(s)\ = 0,Vse5^ f(s) = g(s), Vs £ S => f = g.
iff = g^d(f,g)=0.
(ii) trivial,
(iii) Proposition 10.1.7 Let A ^ 0, B ^ 0 be subsets ofR. Define

A + B = {a + b: a £ A, b £ B}.

If A and B are bounded above then A + B is bounded above and

sup(A + B) < sup A + sup B.

Proof. Let x = sup A, y = sup B.

Given c e A + B, then 3a £ A, b £ B 3c= a + b. Then, c = a + b < x + y.

Moreover, sup(A + B) < x + y. •

Proposition 10.1.8 Let C, D be nonempty subsets o/R, let D be bounded

above. Suppose Vc £ C, 3d £ D 3 c < d. Then, C is also bounded above

and sup C < sup D.

Proof. Given c £ C, 3d £ D 3 c < d. So, Vc £ C, c < y = sup£>. Hence,

y is an upper bound for C. Therefore, sup C < sup D. D

Triangular Inequality: Let f,g,h £ B(S).

C = {|/(s) - g(s)\ :s£S}, then d(f,g) = supC.

A = {\f{s) - h(s)\ :s£S}, then d(f, h) = sup A

B = {\h\s) - g(s)\ :s£S}, then d(h,g) = supB.

Given x £ C, then 3s £ S 3 x — |/(s) - g(s)\

x=\f(s)-g(s)\ = \f(s)-h(s) + h(s)-g(s)\ < \f(s)-h(S)\ + \h(s)-g(s)\

=> sup C < sup(A + B) < sup A + sup B. •

Example 10.1.9 Let X = Rk, P = (Xl,...,xk)T and Q = {yi,...,yk)T €

Rfe.

di(p,q) = \xi-yi\-\ h \xk -yk\ : h metric.

<kip, q) = [(xi - 2/i)^2 H + (xfe - yk)^2 }1/2 • h metric.

doo(p,q) =max{\x 1 -yx\,...,\xk - yk\) : l^ metric.

See Figure 10.1.