Completeness and the Contraction Mapping Theorem
Say what a complete metric space is, and prove that a contraction on one has exactly one fixed point that iteration finds.
What a learner can do afterwards
- Give a metric space that is not complete and identify the missing limit
- Prove the contraction mapping theorem and extract the error bound iteration gives
- Apply the theorem to prove existence and uniqueness for a differential equation
1 · Read
A metric space is complete when every Cauchy sequence converges to a point inside. Gaps can fool you otherwise. Reciprocals 1, 1 over 2, 1 over 3 squeeze together yet aim at 0, missing from the half open interval that excludes it. Rationals aiming at square root of 2 miss the same way. Closed intervals keep all limits and dodge the gap.
A contraction shrinks every distance by a fixed factor k below 1. The map sending x to x plus 4 over 3 uses k equal 1 over 3. Iteration then forms a Cauchy sequence, and each step multiplies the gap by at most k. Summing the tail bounds the distance left, so halving maps turn a move of 8 into at most 4 next. The map halving then adding 1 rests at 2.
Banach fixed point theorem says a contraction on a complete space owns exactly one fixed point, and iteration from anywhere finds it. Completeness catches the orbit while shrinking forbids a rival. The same machine proves differential equations have one solution: rewrite the equation as a contraction on a complete function space and let iteration converge.
Demand k strictly below 1. A factor of exactly 1 never forces the tail to summable size. When an iteration drifts, first suspect a missing limit, then suspect a loose constant.
Completeness catches Cauchy orbits, and shrinking maps own one rest point.
2 · Watch
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Where it sits
8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.