Principles of Mathematics in Operations Research

Solutions 279

Problems of Chapter 11

11.1

a) (=>): Let e > 0 and XQ be given. Let b — /(xo) — e. Then, by assumption,

the set B = {x £ X : f(x) > f(xo) — e} is open. Moreover, xo £ B since

/(x 0 ) > f(x 0 ) - e. So, 36 > 0 3 Bs(x 0 ) C B; that is, x £ Bs(x 0 ) => x £ B =>

f(x) > f(x 0 ) - e.

(«=): Let b £ K be given. We will show that the set A = {x £ X : /(x) > b}

is open. If A = 0, then A is open. Assume A ^ 0, show that every point of A

is an interior point. Let XQ G A. Then, /(xo) > b. Let e = f(xo) — b. Then,

by our assumption, 35 > 0 3 x € Bs(x 0 ) =>• /(x) > /(x 0 ) -e = J^i€i

Hence, BS(XQ) C A, that is a?o S inM.

b) Similar as above.

11.2 / is continuous and X is compact =>• /(X) = B is compact in y.

q G /(X) = B — B since .B is compact, therefore closed. So, by q £ B — f{X),

we have 3p € X B q = /(p). Next, we will show that pn -* p. f : X <-+ B

is continuous, one-to-one and onto. Since X is compact, f"^1 : B H-> X is

continuous. Moreover, f(pn),q £ J3 and /(p„) —> g. Then,

f-Hf(Pn)\ , /-H9)

11.3 Let the wire be the circle Cr = {(x, y) : x^2 + 2/^2 = r-2}. For a = (x, y) G
Cr, let T(a) be the temperature at a and let / : Cr t-> K be such that
/(a) = T(a) — T(—a). Note that a and —a are diametrically opposite points.
Then, T, and hence, / are continuous.
Claim: 3a e Cr 3 f(a) = 0.
Proof: Assume not, Ma G Cr, T(a) ^ T(-a). Define A = {a £ Cr : f(a) > 0},
B — {a £ Cr : f(a) < 0}. Then, A and B are both open in Cr. Why? (since
they are the inverse images of the open sets (0, +oo) and (-co, 0) under the
continuous function /.) A n B ^ 0, because of the heated wire; AL)B = Cr,
since we assumed Va G Cr, T(a) ^ T(—a); moreover, A / 0, there is at
least one point (the point where heat is applied). Suppose not, then Cr = B,
\/a G Cr, /(a) < 0 <£> T(a) < T(-a). But, then T(-a) < T(-(-a)) = T(a),
Contradiction. Hence A ^ 0. Similarly, with the same argument, 5^0, think
of the opposite point to where heat is applied. So, A is nonempty, proper
(Ac = B =fi 0) subset of Cr which is both open and closed (Ac is open). Thus,
Cr is disconnected. Contradiction.
Another way of proving the statement is the following: Let x £ A and
y £ B, and we know that / is continuous as well as /(x) > 0 > f(y). Apply
the intermediate value theorem (Corollary 11.4.2) to conclude that 3a £ Cr 3
/(a) = 0.