Principles of Mathematics in Operations Research

166 11 Continuity

f(t) > Cm + /(«)• Fix f(s) => cm + f(s) is a bound for all f(t)'s. So,

take the infimum over t's. f(xm+) > f(s) + cm •& f(xm+) — cm >

f(s). If we take supremum over s's, we will have f(xm+) — cm >

f{xm-) => f(xm+) - f{xm-) > cm. Therefore, f(xm+) ^ f(xm-) (In

fact, f(xm+) - f(xm-) = cm).

(c) f is continuous at every x £ (a,b) \ E:

Let x £ (a,b)\E be fixed. We will show that f is continuous at x. Let e > 0

be given, since ^ c„ converges, 3N 3 XI^LJV+I c„ < oo CJV+I +SJV = s =>

rN+i — s — sjv- Let 5' = Min{\x — a^l,..., |x — XN\,X — a,b — x}. Let

u 2.

Claim (i) If x < xn < x + 5 then n > N + 1. If n < N + 1, then

\x — XN\ > 5' = 25, Contradiction.

Claim (ii) If x — 5 < xn < x, then n > N + 1. f(x) — e < f(x —

5), f(x + 5)<f(X)+S, f{x)-f(x-S) = Y/„eNxCn-Tln€N:c_sCn =

12neNx\Nx-S Cn - Sr=w+i c" < £- For the second claim, f(x + 5) -

f(x) — J2n£Nx+s cn-2^neNx C" ~ 12n&Nx+s\Nx C« - z2n=N+l C« < £-

Lett be 3 \t-x\ < 6, i.e. x-5 < t < x+6 => f(x-6) < f{t) < F(x+5).

Hence, f{x) - e < f(t) < f(x) + e, \f(t) - f(x)\ < e.

Problems

11.1. Let (X, d) be a metric space. A function / : X t-t R is called lower
semi-continuous (Isc) if V6 £ R the set {x £ X : f{x) > b} is open in X;
upper semi-continuous (use) if V6 £ R the set {x £ X : f(x)<b} is open in
X. Show that
a) / is Isc <£> Ve > 0, Vx 0 ; 35 > 0 3 x £ Bs(x 0 ) => f(x) > f(x 0 ) - e.
b) / is use <£> Ve > 0, Vx 0 ; 35 > 0 3 x £ B 5 (x 0 ) => f(x) < f(x 0 ) + s.

11.2. Let (X, dx) be a compact metric space, (Y, dy) be a metric space and
let / : X t-> Y be continuous and one-to-one. Assume for some sequence {pn}
in X and for some q £Y, linXr^oo f(pn) = q. Show that

3p £ X 3 lim pn = P and f(p) = q.

x—>oo

11.3. Give a mathematical argument to show that a heated wire in the shape
of a circle (see Figure 11.5) must always have two diametrically opposite points
with the same temperature.

Web material

http://archives.math.utk.edu/visual.calculus/1/continuous.7/