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How is the sequence of upper bounds (bn)(b_n) shown to be a Cauchy sequence in the nested interval proof?

The distance between any two terms bnb_n and bmb_m (with m>nm > n) is bounded by the length of the interval [an,bn][a_n, b_n]. Because the bisection method halves the interval length at each step, this distance is bounded by (12)n−1∣b1−a1∣\left(\frac{1}{2}\right)^{n-1}|b_1 - a_1|. As nn increases, this bound approaches zero, satisfying the Cauchy condition.

Conditions

  • m>n∈Nm > n \in \mathbb{N}.
  • The sequences (an)(a_n) and (bn)(b_n) are constructed via the bisection method.
  • bnb_n is an upper bound and ana_n is a lower bound for the set MM.

Reasoning, step by step

  1. Start with the distance ∣bn−bm∣|b_n - b_m| for m>nm > n.
  2. Substitute the lower bound ana_n for bmb_m to get ∣bn−an∣|b_n - a_n|, since an≤bm≤bna_n \le b_m \le b_n.
  3. Use the nested interval property to bound ∣bn−an∣|b_n - a_n| by (12)n−1∣b1−a1∣\left(\frac{1}{2}\right)^{n-1}|b_1 - a_1|.
  4. Observe that (12)n−1\left(\frac{1}{2}\right)^{n-1} approaches 0 as n→∞n \to \infty.
  5. Conclude that ∣bn−bm∣|b_n - b_m| can be made arbitrarily small, proving (bn)(b_n) is a Cauchy sequence.

Example

The video derives the estimate ∣bn−bm∣≤(12)n−1∣b1−a1∣|b_n - b_m| \le \left(\frac{1}{2}\right)^{n-1}|b_1 - a_1|, explaining that because the interval length halves at each step, the distance between terms gets arbitrarily small, formally concluding that (bn)n∈N(b_n)_{n \in \mathbb{N}} is a Cauchy sequence.

Common misconceptions

  • Believing the absolute value is strictly necessary for the estimate; the narrator notes it is redundant because bn≥bmb_n \ge b_m for m>nm > n.
  • Assuming the sequence (bn)(b_n) is monotonically increasing; it is monotonically decreasing as the upper bounds tighten.

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Answers are generated from source material and independently checked. Consult the original video or creator if something is unclear.