Answers & Hints for Selected Exercises 649
- (==>) Suppose A is open. Let x EA. Then :le> 0 3 Nc;(x) ~A, so Nc;(x)
contains no points of Ac, so x ¢:. Ab .. ·. A contains none of its boundary points.
(-¢=) Suppose A contains no boundary points. Then Vx E A, x ¢:. Ab, so
some nbd. N of x contains no points of Ac; i.e., N ~A. :. A is open.
- Apply Def. 3.1.15.
- (a) 6 (b) none (c) 1,2,3,4,5,6,7,8,9 (d) none (e) Z
(f), (g), (h) none (i), (j) { ~ : n EN} (k), (1) none
- If x is an isolated point of A , then x has a nbd N containing no points of
A other than x. Then N contains x, a point of A, and points of Ac, so x E Ab.
The converse is false since 1 is a boundary point of [O, 1] but not an isolated pt.
- (a) x E (AnB)^0 :::>:le> 0 3 Nc;(x) ~ AnB::::? :le> 0 3 Nc;(x) ~A and
Nc;(x) ~ B :::> x E A^0 n B^0 •
x E A^0 nB^0 ==> :le1,e2 > 0 3 Nc; 1 (x) ~A and Nc; 2 (x) ~ B. Let€=
min{e 1 ,e 2 }. Then€> 0 and Nc;(x) ~ AnB, so x E (AnB)^0.
(b) x E A^0 U B^0 ::::? 3e 1 > 0 3 Ne;, (x) ~A~ AU B or :le2 > 0 3 Nc; 2 (x) ~
B ~AUE:::> :le> 0 3 Nc;(x) ~AUE:::> x E (AUB)^0 •
(c) Let A= Q and B = Qc. Then A^0 U B^0 = 0 while (AU B)^0 =IR..
- A is dense in JR<:::? Va< b E JR, (a, b) n A# 0 (see Defn. 1.5.6). Note that
Vx E (a, b), :le> 0 3 Nc;(x) ~ (a, b). Thus, A is dense in JR<:::? Vx E JR, Ve> 0,
Nc;(x) n A# 0.
EXERCISE SET 3.2
- 0 c = JR, open; {aY = (-oo, a) U (a, +oo), open; (-oo, a]c = (a, +oo),
open; etc.
- (b),(d),(h) open; (c),(e),(g),(j) closed; (a),(f),(i),(k),(1) neither.
- (a) [3, 5] U {6} (b),(d),(f),(g) (-oo, 1] (c),(e) none (h),(k) JR (i),(j) {O}
(1) [0,1 ]
- No; e.g., sup{l, 2} = 2, but 2 is not a cluster point of {1, 2}.
Suppose u = sup A ¢:_ A. By the €-criterion for sup A, Ve :> 0, :la E A 3
u - e < a < u (since u ¢:. A). Thus, every Nc;(u) contains a point of A other
than u.
- (b) If x E Ab but x ¢:_ A, then every nbd. of x contains a point of A other
than x (since x ¢:_A).
(c) If x is a cluster point of A but x ¢:. A , then every nbd. of x contains a
point of A and also contains a point (x) of Ac.
- Let A be a finite set.
