Answers & Hints for Selected Exercises 661
(b) Suppose A not open. Let g : JR ---+ JR denote the characteristic function
of A. Then
(1) for xo EA and c > 0 take 6 = 1. Then Vx EA, Ix - x 01 < 6::::?
lg(x) -g(xo)I = ll - ll = 0 < c, so g: A---+ JR is continuous. (Stmnt. #2.)
(2) since A not open, ::3xo EA 3 Vb> 0, N6(x 0 ) c/:. A. Thus, Vn EN,
::3an E Ni;n(xo) 3 an 1. A. Then, Vn E N, Ian - xol < ~ but an 1. A. So,
an ---+ xo but g(an) = 0 while g(xo) = l. :. g: JR---+ JR is not continuous at x 0.
(Stmnt. #1 not true.)
Therefore, when A is not open, Statement #2 ":/?Statement #1.
- Let A be a nonempty compact set; i.e., closed and bounded. Then ::3u =inf A,
v = supA. By Ex. 3.2.15, u,v EA. But A closed, so u,v EA.:. u = minA,
v =max A.
- By Thm. 5.3.6, f (A) is compact, hence closed and bounded. Thus, ::3u =
inf j(A) and v = sup j(A). By Ex. 3.2.15, u , v E f(A) = f(A) since f(A) is
closed.:. u = minf(A) and v = maxf(A).
- Let B = {x EA: f(x) = c} and {xn} be any convergent sequence in B, say
Xn---+ L. Since A is closed, LE A. By continuity off on A, f(xn)---+ j(L). But
{f(xn)} is the constant sequence {c}, so f(xn)---+ c. :. f(L) = c. :. LE B. By
the sequential criterion for closed sets, B is closed.
- By Thm. 5.3.6, f (A) is compact, so ::3u =inf f (A). By Cor. 5.3.7, u E f (A),
sou> 0. :. Vx EA, j(x) 2:: u > 0.
- Define {xn} by x 1 = arbitrary element of [a, b] and Xn+l E [a, b] 3
lf(Xn+i)I ~ ~lf(xn)I. Then lf(xn)I ~ 2 ~ lf(x1)I---+ 0. .'. f(xn)---+ 0.
Since { Xn} is bounded, it has a convergent subsequence; say Xnk ---+ xo.
Since {xn} is in the closed set [a,b], x 0 E [a,b]. Then f is continuous at xo, so
lim f(xnk) = f(xo). :. f(xo) = 0.
k-+oo
- x = 1.705 14. X1 = 2.092, X2 = -1.572
- Let p(x) be a polynomial of odd degree. By Thm. 4.4.24, lim p(x) = ±oo
X-+00
and lim p(x) = - lim p(x). Thus, ::3 a, b 3 p(a) < 0 and p(b) > 0. By the
x--+-oo x---+oo
intermediate value theorem, ::3c between a and b 3 p(c) = 0.
- Let h(x) = cosx - x. Then his continuous on [O, ~]and h(O) = 1 > 0,
while h( ~) = -~ < 0. Apply the intermediate value theorem.
- Let h(x) = f(x) -g(x). Then his continuous on [a, b] and h(a) = f(a) -
g(a) < 0 while h(b) = j(b) - g(b) > 0. Thus 0 is between h(a) and h(b), so
::3c E (a, b) 3 h(c) = O; i.e., f(c) = g(c).
