652 Appendix C • Answers & Hints for Selected Exercises
EXERCISE SET 3.4- C ~ [O, 1], so it is bounded. C is the intersection of a family of closed sets,
so it is closed. - C contains no open intervals, so C^0 = 0. By Exercise 3.2.23, Cb= C-C^0 •
Since C is closed C = C. :. Cb = C - 0 = C. - In stage 1 of defining the Cantor set, C 1 is formed by removing from
[O, 1] every number whose ternary decimal requires "l" as first digit. [e.g.,
~ = 0.10000 · · · = 0.02222 · · · remains.] In stage 2, C 2 is formed by removing
from C 1 every number whose ternary decimal requires "1" as second digit. In
general, Cn+l is formed by removing from Cn every number whose ternary
decimal requires "1" as n+ pt digit. What remains in the Cantor set are those
numbers in [O, 1] whose ternary decimal can be written using only O's and 2's. - (a) C 1 = [o, ~] u [~, 1] = [0.0,0.l] u [0.2, LO] in base-3, so L 1 = {0.0,0.2}
in base-3.
C2 = [O,i] u [~,~] u [~,~] u [~,1], so L2 = {0.00, 0.02,0.20,0.22} in
base-3.
In the general induction step, the typical interval comprising Ck is of the form
[O.d1d2 · .. dk, O.d~d~ · · · d~] or [O.d1d2 .. · dk, 1.00 · · · O],
where di E {O, 2}. Thus, the two left endpoints contributed to Lk+ 1 by re-
moving the middle third from this interval are, in base-3, O.d 1 d 2 · · · dkO and
O.d1d2 · · · dk2.
(b) By Thm. 3.4.10, every x E C is a ternary decimal x == I::, 1 * where
di E {O, 2}. That is , x = n-+oo lim I:~=l · Since each I:: 1 is a terminating
base-3 decimal consisting of O's and 2's, it is a member of L. Thus, every x E C
is the limit of a sequence of members of L.
- ( =?-) A perfect =?-A' = A =?-A closed (Exercise 3 .2.21) and A has no isolated
points since {isolated points} = A - A' by Exercise 3.2.24.
( {:::) A closed with no isolated points =?-A = A and A - A' = 0 =?-A' ~ A
and A - A' = 0 =?-A = A'.
11. A nowhere dense=?-A contains no nonempty open intervals, =?-A
0
= 0 =?-
A0 = 0 =?-Acclc = 0 (Exercise 3.2.23) =?-Ace! =JR=?-Ac dense in R The
converse is not true since Q c is dense in JR, but Q is not nowhere dense.
- Q is dense in JR but μ(Q) = 0 since Q is countable. (Thm. 3.4.20)
- (a) A= (A - B) U (An B), a union of disjoint sets, so by (μ4),
μ(A) =μ(A - B) +μ(An B).
(b) AU B = (A - B) U B , a union of disjoint sets, so by (μ4),
μ(Au B) =μ(A - B) + μ(B) =μ(A) - μ(An B) + μ(B).