Re: [分析] closed set

[sato186 (銀色轟炸機)]

※ 引述《Madroach (∞)》之銘言：
: 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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◆ From: 111.248.9.101
→ josh28      :如果考慮連續函數f(x,y)=x+y 然後假設A+B是range上的
→ josh28      :一個子集這樣呢?
推 ss1132      :A=正整數 B=-n-1/n  n=2,3,4......
→ ss1132      :A+B不是closed,因為有1/2,1/3,1/4......
→ Madroach    :謝謝!
　　ss大大提供的集合沒錯, 可是原因好像有點怪, 應該是這樣:

　　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.


(題目來源: 台灣聯合大學系統 99學年度碩士班考題 高等微積分 #2)

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: 數學到底有什麼技巧呢?
靈性, 信念, 經驗

: 想不出來做不出來是真的不會嗎?
我覺得 這是緣分的問題 (茶)

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※ 編輯: sato186         來自: 114.33.209.112       (02/10 04:08)
推 Madroach    :謝謝!