Principles of Mathematics in Operations Research

164 11 Continuity

then for this specific e > 0, 35 > 0 9 W with x < t < x + 5, we have

\f(t)-f(x)\<e.

CASE 1: X G Q.

FindteR\Q3x<t<x + S \f(t)-f(x)\ <e= &=^.

But 4|/(£) - f(x)\ = |1 - t - x\ = |2x - 1 + t - x| = |(2x - 1) - (x - t)\.

Since \a — b\ > | |a| — |6| |,

|/(t) - /(x)| > | |2x - 1| - |x - 1| | > \2x -l\-\x-t\> \2x -l\-6.

Then, we have

\2x-l\~8<\f(t)-f(x)\<^1 -^^.

=>• (5 > 2~ ', Contradiction since 5 > 0 con fee iafcen as small as we want.

CASE 2: X G R \ <Q>. Proceed in similar way, but choose t as rational.

11.5 Monotonic Functions

Definition 11.5.1 Let f : (a,b) t-» R. / is said to 6e monotonically increas-

ing (decreasing) on (a, 6) if and only if

a < xi < x 2 < b => f(Xl) < f{x 2 ) (/(aJi) > f(x 2 )).

Proposition 11.5.2 Let f : (a, b) i-> R 6e monotonically increasing on (a,b).

Then, Vx 6 (a, fr), f(x+) and f{x—) exist and

sup f(t) = f(x-)<f(x)<f(x+)= inf /(*).

a<t<x x<t<b

Furthermore, a < x\ < x 2 < b =$• f(xi+) < f{x 2 —).

Theorem 11.5.3 Let f : (a,b) H-> R be monotonically decreasing on (a,b),

then Vx G (a, 6), f(x+) and f(x—) exisi and

inf /(*) = /(x-) > /(x) > f(x+) = sup /(f).

a<t<x x<t<b

Furthermore, a < x\ < x 2 < b =>• /(xi+) > f(x 2 — ).

Proof. Let x G (a, 6) be arbitrary. Vf with 0 < £ < x, we have /(£) > /(x). So,
{/(i) : a < £ < x} is bounded below by /(x). Let ^4 = ini{f(t) : a < t < x}.
We will show A — f(x\—):
Let e > 0 be given. Then, A + e is no longer lower bound of {f(t) : a < t < x}.
Hence, 3t 0 G (a, x) 9 /(i 0 ) < A + e. Let <5 = x - t 0 Vt3x-5 = t 0 <t<
x => f(x) < f(t 0 ) < A + e and f(t) > A > A - e. Hence, Vr G (x - £, x) we
have A - s < f(t) < A + e =>• |/(r) - ^| < e. Thus, 4 = f(x-). Therefore,
info<t<x f(t) = A = /(x-) > f(x). Similarly, supa;<t<b /(x) = /(x+) < f(x).
Let a < xi < x 2 < b, apply first part b <- x 2 and x <r- x. /(xi+) =
SUPar^^xa /() ^ i^iKKi)! /(<) = f{2 + )- •