Completeness and the Contraction Mapping Theorem · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Shrinking maps settle once

Mathematics · Topology · ages 21-22
Name ______________________   Date ____________

  1. Map halves its input then adds 1. What is its fixed point?

    Answer: ______________

  2. Half open interval with 0 excluded and 1 included uses usual distance. Which sequence shows incompleteness?

    • Reciprocals 1, 1/2, 1/3, dots
    • Constant 1, 1, 1, dots
    • Constant 1/2, 1/2, dots
    • Terms 0.9, 0.99, 0.999, dots

  3. The half open interval excludes 0 and keeps 1. Which sequence shows incompleteness?

    • Reciprocals 1, 1 over 2, 1 over 3, onward
    • Constant 1, 1, 1, onward
    • Terms 0.9, 0.99, 0.999, onward

  4. Rationals miss irrationals, so a rational sequence can aim at square root of 2 and miss. The rationals are incomplete.

    Circle one:   True   False

  5. With contraction factor k, what control do you hold over iteration error?

    • The error after n steps obeys a geometric bound in k
    • The error grows with each new step
    • Iteration needs no starting point at all

  6. Banach fixed point theorem gives a unique fixed point plus an error estimate that shrinks geometrically with the contraction constant. Is this correct?

    Circle one:   True   False

  7. Map sends x to (x plus 4) divided by 3. What is a valid contraction constant?

    • 1
    • 3
    • 1/3
    • 4

  8. A map sends x to x plus 4 over 3. What is a valid contraction constant?

    • 1
    • 3
    • 1 over 3

  9. A contraction runs on a complete space. What does Banach promise?

    • At least three fixed points in every ball
    • Exactly one fixed point, found by iterating anywhere
    • No fixed point but a shrinking orbit

  10. Map sends x to (2x plus 9) divided by 5. What is its fixed point?

    Answer: ______________

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Answer key

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Shrinking maps settle once W1-mt_DYsLzg-IaK-s1

  1. 2 · Solve x equals x divided by 2 plus 1, giving x equals 2.
  2. Reciprocals 1, 1/2, 1/3, dots · Reciprocals 1, 1/2, 1/3, dots is Cauchy with limit 0 outside the space, while the constant sequences and the 0.9, 0.99 progression converge inside.
  3. Reciprocals 1, 1 over 2, 1 over 3, onward · Reciprocals squeeze together yet aim at 0 outside the space.
  4. True · Decimal approximations of square root of 2 are Cauchy with no rational limit.
  5. The error after n steps obeys a geometric bound in k · Each step multiplies the gap by at most k, and the tail sums geometrically.
  6. True · Iteration converges to one fixed point with geometric error bound in powers of the constant.
  7. 1/3 · 1/3 is correct since distance shrinks by factor 1/3, while 1 is not below 1 and 3 and 4 expand.
  8. 1 over 3 · Gaps shrink to a third, which sits below 1.
  9. Exactly one fixed point, found by iterating anywhere · Completeness catches the Cauchy orbit and shrinking forbids a second rest point.
  10. 3 · Solve x equals (2x plus 9) divided by 5, giving 5x equals 2x plus 9, so 3x equals 9, hence 3.
Worksheet · LightMySky