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The least-upper-bound property states that any non-empty set of real numbers that has an upper bound must have a least upper bound. 下面的證明是用柯西數列 可是我看不出哪邊有用到柯西數列 請版大指點 謝謝 Proof using Cauchy sequences It is possible to prove the least-upper-bound property using the assumption that every Cauchy sequence of real numbers converges. Let S be a nonempty set of real numbers, and suppose that S has an upper bound B1. Since S is nonempty, there exists a real number A1 that is not an upper bound for S. Define sequences A1, A2, A3, ... and B1, B2, B3, ... recursively as follows: Check whether (An + Bn) / 2 is an upper bound for S. (到這一行開始看不懂) If it is, let An+1 = An and let Bn+1 = (An + Bn) / 2. Otherwise there must be an element s in S so that s>(An + Bn) / 2. Let An+1 = s and let Bn+1 = Bn. Then A1 ≦ A2 ≦ A3 ≦ ...≦ B3 ≦ B2 ≦ B1 and |An - Bn| → 0 as n → ∞. It follows that both sequences are Cauchy and have the same limit L, which must be the least upper bound for S. -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 123.193.56.234
suhorng:An跟Bn都是 Cauchy sequence 118.166.45.226 12/01 17:37
suhorng:(An+Bn)/2 那行在做二分搜 118.166.45.226 12/01 17:38
suhorng:你注意他的選法 每次都 An < Bn 118.166.45.226 12/01 17:39
suhorng:而且 An,Bn 越來越靠近 118.166.45.226 12/01 17:39
suhorng:而且 An, Bn 都把(想像中的)sup S夾在中間 118.166.45.226 12/01 17:39
gj942l41l4:畫個圖會比較清楚 140.112.217.22 12/01 18:31