648 Appendix C • Answers & Hints for Selected Exercises
- (a) 0, -1
(f) 0, 0(b) 1, -1
(g) 0, 1, +ooEXERCISE SET 2.9(c) +oo, -oo
(h) 1, 0(d) 1, -1
- Redo the proof of Thm. 2.9 .7 , making appropriate changes.
(e) 3, -1
- (a) and (b) False. Take { Xn} = {l, 0, 1, 0, · · · } and {yn} = {O, 1, 0, 1, · · · }.
(c) and (d) False. Take {xn} = {-1, 0, -1, 0, · · · }, {yn} = {O, -1, 0, -1, · · · }. - Redo #6, changing sup to inf, upper bars to lower bars, etc.
- Suppose {xn}, {Yn} ~ 0, Xn-> x -j. 0, and lim Yn = y. Then y is a cluster
n-+oo
point of {Yn}, so {yn} has a subsequence Ynk -> y. Since Xn , x, Xnk , x. So
Xnk Ynk -> x y. :. x y is a cluster point of { XnYn}.
Suppose z is a cluster point of {XnYn}· Then 3 subsequence XmkYmk __, z.
Since x n -> x -j. (^0) ' ...L Xn -> .!. x > so y mk = xmk Xmk Ynmk -> x E.. · · · E. x is a cluster point
of {Yn}· By Thm. 2.9.10, y is the largest cluster point of {yn}, so ; s:; y. i.e.,
z:::; xy. :. xy is the largest cluster point of {xnYn}; i.e., lim XnYn = xy.
Chapter 3
EXERCISE SET 3.1
- Only (b), (d), and (h) are open.
n-+oo- A^0 is open, since the union of open sets is open. If U is any open subset of
A, then U is a subset of the union of all open subsets of A; i.e., U ~ A^0 • - (==>) Suppose A is open. Then \:/x EA, ::lex> 0 :::i Nc;x(x) ~A. Then,
A= LJ Nc;x(x).
xEA
( '*==) The union of any family of open sets is open, by the "open set theorem." - (b) A^0 = (-oo, 0) U (0, 1) A ext= (1, +oo) Ab= {O, l}
(d) A^0 =(-oo,1) A ext= (1, +oo) Ab= {l}
(f) A^0 =(-oo, O)U(0,1) Aext=(l, +oo) Ab={O,l}
(h) A^0 =A Aext = 0 Ab= {1, 2,3}
(j) A^0 = 0 A ext = (-oo,O) U ngl (n~l' ~) U (1,+oo) Ab= A(1) A^0 = 0 Nxt=(-oo,O)U(l, +oo) Ab=Qn[0,1]
- \:/x E JR, every nbd. of x contains irrational numbers, so x rt Q^0 • :. Q^0 = 0.
\:/x E JR, every nbd. of x contains rational numbers, so x rt Q c^0 • • •• Qext = 0.
\:/x E JR, every nbd. of x contains both rational and irrational numbers, so
x E Qb. :. Qb =JR.