Closing in on a point
The nested interval theorem says: if I_1 ⊇ I_2 ⊇ I_3 ⊇ ... is a shrinking chain of closed bounded intervals I_n = [a_n, b_n], each containing the next, then their intersection is nonempty — there is at least one point lying in every interval. If in addition the lengths b_n - a_n shrink to 0, that point is unique. This is completeness wearing a geometric costume.
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)Completeness is what makes limits exist
Now the payoff for the whole track. A Cauchy sequence is one whose terms eventually cluster: for every epsilon > 0 there is an N so that |a_m - a_n| < epsilon whenever m, n ≥ N. This is a condition you can check using only the sequence itself, with no candidate limit in hand. The deep theorem is that in R, every Cauchy sequence converges — its limit exists. This property is precisely equivalent to completeness.
This is the whole point of the real numbers. Over Q the limit can fail to exist: the decimal truncations 1, 1.4, 1.41, 1.414, ... form a Cauchy sequence of rationals that clusters tighter and tighter, yet there is no rational number for it to converge to — it is aiming at the square root of 2, the very hole completeness fills. Pass to R and the hole is gone; the limit is there waiting. Completeness is the one fact that guarantees the limits analysis is built on actually exist.
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.