闭区间套与极限为何存在

向一个点收拢

闭区间套定理说：若 I_1 ⊇ I_2 ⊇ I_3 ⊇ ... 是一列收缩的闭有界区间 I_n = [a_n, b_n]，每个包含下一个，则它们的交非空——至少有一个点落在每个区间里。若再有长度 b_n - a_n 收缩到 0，则该点唯一。这是完备性披上几何外衣的样子。

Proof that the intersection is nonempty.
  Nesting gives, for all m, n:   a_m <= b_n.
  (any left endpoint sits left of any right endpoint)
  So the set A = { a_n : n in N } of left endpoints
  is bounded above (every b_n is an upper bound).
  By COMPLETENESS, c = sup A exists in R.
  - c is an upper bound of A, so a_n <= c for every n.
  - each b_n is an upper bound of A, and c is the LEAST one, so c <= b_n.
  Hence a_n <= c <= b_n, i.e. c is in [a_n, b_n] for EVERY n.
  The intersection contains c, so it is nonempty.   QED

Uniqueness when lengths -> 0:
  if c and c' both lie in every I_n then |c - c'| <= b_n - a_n -> 0,
  forcing c = c'.   (closedness matters: open (0,1/n) intersect to empty)

把闭区间套归结为一个上确界。

完备性正是极限存在的根源

现在是整条轨道的回报。柯西序列是其各项最终聚拢的序列：对每个 epsilon > 0 都存在 N，使得只要 m, n ≥ N 就有 |a_m - a_n| < epsilon。这是一个只用序列本身即可检验的条件，手上不必握有候选极限。深刻的定理是：在 R 中，每个柯西序列都收敛——其极限存在。这条性质恰与完备性等价。

这正是实数存在的全部意义。在 Q 上极限可能不存在：小数截断 1, 1.4, 1.41, 1.414, ... 是一列有理柯西序列，越聚越紧，却没有有理数可供它收敛——它瞄准的是根号 2，正是完备性所填的那个洞。转到 R，洞便消失；极限早已在那里等候。完备性就是那唯一事实，它保证分析学赖以建立的极限确实存在。

Sketch: in R, every Cauchy sequence (a_n) converges.
  1. A Cauchy sequence is BOUNDED
     (beyond some N all terms lie within 1 of a_N; finitely many before).
  2. By Bolzano-Weierstrass (itself a child of completeness),
     a bounded sequence has a convergent SUBSEQUENCE a_{n_k} -> L.
  3. Cauchy + a subsequence reaching L forces the WHOLE sequence to L:
     given epsilon, pick N from the Cauchy condition (using epsilon/2),
     pick a subsequence index n_k >= N with |a_{n_k} - L| < epsilon/2.
     Then for n >= N:
       |a_n - L| <= |a_n - a_{n_k}| + |a_{n_k} - L| < epsilon/2 + epsilon/2 = epsilon.
  So a_n -> L.   QED
Every step above leans on completeness.