Prove: (a) \\bar{A∪B} = \\bar{A}∪\\bar{B}; (b) \\bar{\\bar{A}} = \\bar{A}
(\\bar{A} 就是在A上方加一條橫線,這裡代表closure of A,其定義請見下方。)
因為有兩小題,所以解答贈點為20點喔。^_^
註:
metric function 和 metric space 的定義:
Suppose that there exists a real-valued function ρ defined for every ordered pair (x, y) of points of the universal set X, having the following properties:
(i) ρ(x, y) ≧ 0, and ρ(x, y) = 0 if and only if x = y.
(ii) ρ(x, y) = ρ(y, x) (symmetry).
(iii) ρ(x, z) ≦ ρ(x, y) + ρ(y, z) (the triangle inequality).
The function ρ is then called a metric function (or a distance function) on X, and the pair (X, ρ) is called a metric space.
open ball 的定義:
For any x∈X, ε>0, the set
B(y, ε) = {y; ρ(x, y) < ε}
is called the open ball with center y and radius ε, or an ε-neighborhood of y.
開集合的定義:
A set E is called an open set if for any y∈E there is a ball B(y, ε) that is contained in E.
閉集合的定義:
A set is said to be closed if its complement is an open set.
accumulation point 和 closure的定義:
A point y is called a point of accumulation of a set D if any ball B(y, ε) contains points of D other than y.
Denote by D\' the set of all the points of accumulation of D.
The set D∪D\' is called the closure of D, and it is denoted by \\bar{D}.
2006-08-25 13:26:16 · 1 個解答 · 發問者 維正 2 in 科學 ➔ 數學
由於closure的符號太難打,我把它簡化成'先給個Lemma:If A包含於B,則A'包含於B'證明:這很trivial!,Let ε>0 be given,x 屬於A',there exists y屬於A,such that ρ(x,y)<ε,so y屬於A,y屬於B,that is , there exists y屬於B,such that ρ(x,y)<ε,Hence x屬於B' ie A'包含於B'(a)since A包含於A∪B,B包含於A∪B,so A'包含於(A∪B)',B'包含於(A∪B)'=>A'∪B'包含於(A∪B)'Suppose that x屬於(A∪B)',let ε>0 be given,there exists y屬於A∪B,such that ρ(x,y)<εthat is there exists y屬於A,or y屬於B,such that ρ(x,y)<εx屬於A' or 屬於B'=>(A∪B)'包含於(A'∪B')=>(A∪B)'=A'∪B'(b)We know A包含於A',so A'包含於(A')'Now,geven any ε>0,x屬於(A')',there exists y屬於A' such that ρ(x,y)<ε/2But y屬於A',there exists z屬於A such that ρ(y,z)<ε/2so,there exists z屬於A such that ρ(x,z)≦ρ(x,y)+ρ(y,z)<ε/2+ε/2=ε=>x屬於A'=>(A')'包含於A'=>(A')'=A'
2006-08-25 18:48:23 · answer #1 · answered by ? 7 · 0⤊ 0⤋