Principles of Mathematics in Operations Research

162 11 Continuity

Proof. Assume that f{E) is not connected, i.e.

3nonemptyA,Bcy9Aflfl = f), AnB = 0, f{E) = A U B.

LetG = EDf~^1 (A), H = Enf~^1 (B), A ± 0 =* 3q G A C /(£?) =• q = f(p)

for some_p E E =$ p e f~l{A)_^ p £ G =$• G J± $.

Assume G n H ^ 0. Let p G G n J? =*• p G if = £7 n /_1(^) =*

/(p) GB,pGG = £n/-^1 (yl)c /-'(A) (*)

Aci4 /_1(A) : closed =» /-x(>0 C /_1(3) =^p6 /_1(3) =»

/(P) G A (**)

(*)+(**)=> /(p) einfl/l, Contradiction. Thus, G n ff = 0. Similarly,

GO# = 0.

£ c r'iHE)) = /-^ UB) = r

X

(A) u r

1

^)

£ = £n t/-

1

^) u rHB)] = [En f~\A)] u [£ n r\B)) = GUH,

meaning that E is not connected. Contradiction! •

Corollary 11.4.2 (Intermediate Value Theorem) Let f : [a, b] —» K be

continuous and assume f(a) < f(b). Let c G R be such that

f{a) <c< f(b) =>ce f([a,b]), i.e. 3x G (a,b) 3 f(x) = c.

Proof. [a,b] is connected, so /([a, 6]) is connected; thus /([a, 6]) is an interval

[a,0].f(a),f(b)e[a,l3]=*cef([a,b]),

3x G [a, b] 3 f(x) = c, /(a) < c =$> x ^ a and /(ft) > c =>• x ^ b.

Thus, x G (a, 6). •

Example 11.4.3 Let I=[0,l],f:I—>Ibe continuous. Let us show that
3x G I 3 f{x) = x. Let g{x) = f{x) —x be continuous. Show 3x £ I 3 g{x) =
O.If$ such x => Vx G I we have g(x) > 0 or g(x) < 0.

(i) g(x) > 0, Vi€/^ f(x) > x, Vx G /. Then, /(l) > 1; a Contradiction,

(ii) g(x) < 0, Vx G 7 => /(x) < x,Vx G J. TTien, /(0) < 0; a Contradiction.

Definition 11.4.4 (Discontinuities) Let f : (a,b) -» X w/iere (-X", d) is a
metric space. Let xbe3a<x<b and q € X. We say, f(x+) — q or
limj-yj,.). f(t) = q if Ve > 0 35 > 0 9 V£ wii/i x<£<x + (5we /lave
d(f(t),f(x)) < e. f(x+) = q <$ V subsequence {tn} with x < tn < 6,Vn
and lim^oo bn = x we have limn>oo f(t) - q. f(x-) = lim^z- f(t) is
defined analogously. Let x G (a, b) => limt,.x f(t) exists •& f(x+) = f(x-) =
lim*-^ f(t). Suppose f is discontinuous at some x G (a, b).