精華區beta Math 關於我們 聯絡資訊
※ 引述《wuxr (wuxr)》之銘言: : 請問各位先進一個問題 : If (X, T) is a topological space, : A is closed iff every conv. seq. in A converges to a point in A. : 由左到右我有寫出來, 但是由右到左我被卡住了 : 還是說這個命題只有單邊的(左到右) : 求教了!謝謝^^ 有反例: Put X=|R, T={ (-n,n) | n=0,1,2,3...} u {|R}. {0} is not closed since |R-{0} = (-∞,0)u(0,∞). However, the only sequence in {0} is {0,0,...}, which indeed converges to 0. -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 59.104.172.114
wuxr :喔~~我了解了...謝謝P大...能不能再請教一個問題 12/05 12:51
ppia :什麼問題? 12/05 16:28
wuxr :{0} 在這裡面是compact 嗎? 12/05 18:06
是 compact => closed 在非Hausdroff空間中不一定對 (Hausdroff: For any points x and y in X, there exist two disjoint neighborhoods U and V of x and y, respectively.) 不過上面的例子即使在Hausdroff空間中還是可以找到 Put X=|R, T={ S | S is dense* in (infS, supS).} (* dense in the usual sense.) (X,T) is a Hausdroff topological space. Set A=|R-|N, which is not closed since |N is not dense* in |R. Let {x_n} be an arbitray sequence in |R. Seeing that |R-{x_n} is dense* in |R, which means |R-{x_n} being open, the sequence cannot converge to a point out of {x_n}. Therefore, any convergent sequence in A must converge to a point in A. ※ 編輯: ppia 來自: 59.104.172.114 (12/05 19:29)