Let f: (0, 1) → (0, 1) satisfy |f(x) − f(y)| ≤ (1/2)|x − y| for all x, y. Starting from any x₀ ∈ (0,1), the iterates x₀, f(x₀), f(f(x₀)), ... form a Cauchy sequence. Does the Contraction Mapping Theorem guarantee a fixed point in (0, 1)?
AYes — f is a contraction with k = 1/2 < 1, so the theorem applies
BNo — (0, 1) is not complete; the Cauchy sequence may converge to a limit outside the space, so no fixed point in (0, 1) is guaranteed
CYes — any bounded metric space supports the theorem if the Lipschitz constant is less than 1
DNo — the theorem requires k < 1/2, so k = 1/2 is on the boundary and excluded
This is the canonical illustration of why completeness is essential. The open interval (0, 1) is a metric space and f is indeed a contraction, but (0, 1) is not complete — Cauchy sequences can converge to 0 or 1, which are not in the space. For example, if the unique fixed point of f as a map on [0, 1] is x* = 0, the iterates approach 0 but 0 ∉ (0, 1). Completeness is not a technicality; it guarantees the limit actually lives in the space where f is defined.
Question 2 Multiple Choice
Why is the Lipschitz constant k strictly less than 1 (k < 1) required for the theorem, rather than k ≤ 1?
Ak = 1 would make convergence too slow to be practically useful
BWith k = 1, uniqueness fails: two distinct fixed points p ≠ q could satisfy d(f(p), f(q)) = d(p, q) without contradiction, so the theorem's proof breaks down
Ck = 1 makes f non-differentiable, which violates a hidden smoothness assumption
Dk ≤ 1 is actually sufficient; the strict inequality is imposed only by historical convention
The uniqueness proof is the key. If f had two fixed points p and q, then d(p, q) = d(f(p), f(q)) ≤ k · d(p, q). For k < 1, this forces d(p, q) ≤ k · d(p, q) < d(p, q) unless d(p, q) = 0, so p = q. With k = 1 (an isometry), the inequality becomes d(p, q) ≤ d(p, q), which is satisfied trivially and carries no information — two distinct fixed points remain possible. An isometry on a complete space need not have any fixed point at all.
Question 3 True / False
The Contraction Mapping Theorem gives not only existence and uniqueness of a fixed point, but also a quantitative error bound: after n iterations, the distance to the fixed point is at most kⁿ/(1−k) times the distance of the first step.
TTrue
FFalse
Answer: True
This quantitative bound is one of the theorem's most powerful practical features. It tells you precisely how many iterations are needed to achieve a desired accuracy — a feature that pure existence theorems lack. For algorithms like Newton's method (analyzed as iterated contractions), this gives explicit convergence guarantees. The bound kⁿ/(1−k) · d(x₀, f(x₀)) follows from the geometric series structure of the cumulative error across all remaining iterations.
Question 4 True / False
A contraction on a closed, bounded subset of ℝ is very likely to have a fixed point, even if the subset is not complete as a metric space.
TTrue
FFalse
Answer: False
Closed and bounded (compact) subsets of ℝ are actually complete, so this specific example works — but the claim as stated is wrong in general. In an arbitrary metric space, a closed subset need not be complete: for instance, a closed subset of an incomplete metric space inherits the incompleteness. The theorem's hypothesis is completeness, not closedness or boundedness. The key is whether Cauchy sequences in the space converge *within* the space, which requires completeness.
Question 5 Short Answer
Explain why completeness is a necessary hypothesis in the Contraction Mapping Theorem, not merely a convenient simplification.
Think about your answer, then reveal below.
Model answer: A contraction creates a Cauchy sequence of iterates: the distances between successive iterates shrink geometrically (by factor k < 1), so the sequence is Cauchy by the geometric series. But a Cauchy sequence in an incomplete space may converge to a limit that lies *outside* the space. If the limit is outside X, then f has no fixed point in X — the theorem fails. Completeness is exactly the condition that guarantees every Cauchy sequence converges to a point *within* the space, so the limit of the iterates is an actual element of X where f is defined and where f(x*) = x* can be verified.
The incompleteness counterexample is instructive: on (0,1), the map f(x) = x/2 is a contraction (k = 1/2) whose unique fixed point would be 0 — but 0 ∉ (0,1). The iterates 1/2, 1/4, 1/8, ... are Cauchy and converge, but the limit is outside the space. Completeness closes this gap. This shows completeness is load-bearing, not decorative.