作者sato186 (银色轰炸机)
看板Math
标题Re: [分析] closed set
时间Thu Feb 10 02:31:00 2011
※ 引述《Madroach (∞)》之铭言：
: A and B are nonempty closed subset in R.
: Define A+B = {a+b | a in A, b in B}
: A*B = { ab | a in A, b in B}.
: Prove or disprove A+B, A*B are closed.
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1F：→ josh28 :如果考虑连续函数f(x,y)=x+y 然後假设A+B是range上的 02/09 20:28
2F：→ josh28 :一个子集这样呢? 02/09 20:28
3F：推 ss1132 :A=正整数 B=-n-1/n n=2,3,4...... 02/09 20:50
4F：→ ss1132 :A+B不是closed,因为有1/2,1/3,1/4...... 02/09 20:51
5F：→ Madroach :谢谢! 02/09 21:17
ss大大提供的集合没错, 可是原因好像有点怪, 应该是这样:
For all positive integer n ≧ 2, we see that n belong A and
1 1
-1
–n–── belong B, so n ＋ (–n–──) ＝
── are in A + B for all n ≧ 2.
n n
n
1
But ｌｉｍ ── ＝
0 does not belong A + B. ( A + B has no integers.)
n→∞ n
至於 A ×B = { ab | a in A, b in B.} 是不是 closed 呢? 答案
不是. 反例如下:
1 +
Set A ＝ Ｚ, and
B ＝ {──｜n is in Ｚ .} ∪ {0}. We have already known
n
that both A and B are closed in
|R. But A ×B ＝
q __
{──｜p, q are both in Ｚ, p ≠0.} ＝ Ｑ is not closed in
|R due to
Ｑ
p
＝
|R.
(题目来源: 台湾联合大学系统 99学年度硕士班考题 高等微积分 #2)
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: 数学到底有什麽技巧呢?
灵性, 信念, 经验
: 想不出来做不出来是真的不会吗?
我觉得
这是缘分的问题 (茶)
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※ 发信站: 批踢踢实业坊(ptt.cc)
※ 编辑: sato186 来自: 114.33.209.112 (02/10 04:08)
6F：推 Madroach :谢谢! 02/10 08:34