向一個點收攏
閉區間套定理說：若 I_1 ⊇ I_2 ⊇ I_3 ⊇ ... 是一列收縮的閉有界區間 I_n = [a_n, b_n]，每個包含下一個，則它們的交非空——至少有一個點落在每個區間裡。若再有長度 b_n - a_n 收縮到 0，則該點唯一。這是完備性披上幾何外衣的樣子。
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)完備性正是極限存在的根源
現在是整條軌道的回報。柯西序列是其各項最終聚攏的序列：對每個 epsilon > 0 都存在 N，使得只要 m, n ≥ N 就有 |a_m - a_n| < epsilon。這是一個只用序列本身即可檢驗的條件，手上不必握有候選極限。深刻的定理是：在 R 中，每個柯西序列都收斂——其極限存在。這條性質恰與完備性等價。
這正是實數存在的全部意義。在 Q 上極限可能不存在：小數截斷 1, 1.4, 1.41, 1.414, ... 是一列有理柯西序列，越聚越緊，卻沒有有理數可供它收斂——它瞄準的是根號 2，正是完備性所填的那個洞。轉到 R，洞便消失；極限早已在那裡等候。完備性就是那唯一事實，它保證分析學賴以建立的極限確實存在。
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.