推 annboy : The set F={f(x): x belongs to E} is bounded 12/06 00:03
→ annboy : by y0, thus f(x0) <= sup F <= y0 12/06 00:04
→ annboy : 證明應該有converse吧 看起來應該只證了一邊 12/06 00:05
推 annboy : 另外我猜想xn應該是increasing 不過THM 2.11未知 12/06 00:15
→ sogood6108 : 看不太懂耶QQ 12/06 00:35
→ annboy : 我定義的F是bounded above 所以sup F存在於R 12/06 11:32
推 PPguest : 我覺得是取極限的關係,類似於x_n→x_0, x_n<y_0 for 12/06 11:33
→ annboy : 且 sup F <= y0 另外 設q是一個limit point of F 12/06 11:33
→ annboy : 一定q <= sup F, 否則能做一個open set 導致矛盾 12/06 11:34
→ PPguest : all n, lim x_n <= y_0 12/06 11:35
→ annboy : If sup F < q, let r = (q - sup F )/2 12/06 11:40
→ annboy : But there is no point in F that also in ( q-r, q 12/06 11:41
→ annboy : +r) 12/06 11:41
→ annboy : , which contradicts to the fact q is a limit poi 12/06 11:42
→ annboy : nt of F 12/06 11:42
→ annboy : 我沒想到更簡單的解釋，不過一二樓的不等式應該正確 12/06 11:44