TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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A contraction of a nonempty complete metric space into itself has exactly one fixed point, the limit of the iterates from any starting point

Statement

Let $(X,d)$ be a nonempty complete metric space (Complete metric space: every Cauchy sequence converges in the space) and let $f : X \to X$ be a contraction with constant $q$ , so $d(f(u),f(v)) \le q\,d(u,v)$ for all $u,v \in X$ and $0 \le q < 1$ (Lipschitz map, $\alpha$ -Hölder map for rational $0 < \alpha \le 1$ , and contraction). Then:

There is exactly one $x^{*} \in X$ with $f(x^{*}) = x^{*}$ .
For every $x_0 \in X$ the sequence of iterates, defined by recursion (The recursion theorem) by $x_{n+1} := f(x_n)$ , converges to $x^{*}$ in $(X,d)$ (Convergence of a sequence in a metric space: $x_k \to x$ iff $d(x_k, x) \to 0$ in $\mathbb{R}$ ). In particular the limit does not depend on $x_0$ .
For all $m, n \in \mathbb{N}$ with $m \ge n$ , $d(x_m, x_n) \;\le\; \frac{q^{\,n}\, d(x_1,x_0)}{1-q},$ with the integer power of Integer powers $a^m$ .

Claim 3 is the quantitative form of the argument that proves claims 1 and 2, and it is recorded in the statement because the error estimates (The a priori bound $d(x^{*}, x_n) \le q^n d(x_1,x_0)/(1-q)$ and the a posteriori bound $d(x^{*}, x_{n+1}) \le q\,d(x_{n+1},x_n)/(1-q)$ ) are read off from it.

Facts & Assumptions

Given: A nonempty complete metric space $(X,d)$ , a contraction $f : X \to X$ with constant $q$ satisfying $0 \le q < 1$ , a starting point $x_0 \in X$ , and a real $\varepsilon > 0$ . Write $C := d(x_1,x_0)/(1-q)$ .

[A1]

Contraction: $d(f(u),f(v)) \le q\,d(u,v)$ for all $u,v \in X$ , with $0 \le q < 1$ (Lipschitz map, $\alpha$ -Hölder map for rational $0 < \alpha \le 1$ , and contraction).

[A2]

$(X,d)$ is complete and nonempty (Complete metric space: every Cauchy sequence converges in the space).

[L1]

Recursion: for $x_0 \in X$ and $f : X \to X$ there is a unique $n \mapsto x_n$ with $x_0$ as given and $x_{n+1} = f(x_n)$ (The recursion theorem).

[L2]

Induction on $\mathbb{N}$ (The principle of mathematical induction).

[L3]

Integer powers and their laws: $q^{0} = 1$ , $q^{n+1} = q^{n}q$ , $q^{m+n} = q^{m}q^{n}$ , and $q^{n} \ge 0$ for $q \ge 0$ (Integer powers $a^m$ , Laws of integer exponents).

[L4]

Factorisation: $1 - q^{N} = (1-q)\sum_{k<N} q^{k}$ for $N \ge 1$ , so $\sum_{k<N} q^{k} = (1-q^{N})/(1-q) \le 1/(1-q)$ when $0 \le q < 1$ (Factorisation of $b^n - a^n$ , and the resulting Lipschitz estimate, Finite sums and finite products, by recursion, Inverses of positives are positive, and reciprocation reverses order).

[L5]

Laws of finite sums: splitting, index shift, scaling and monotonicity (Laws of finite sums and finite products, Finite sums and finite products, by recursion).

[L6]

$q^{k} \to 0$ for $|q| < 1$ , and a constant multiple of a null sequence is null (For $|r| < 1$ the sequence $r^k$ is null, and for $|r| > 1$ the sequence $|r|^k$ diverges to $+\infty$ , Algebra of limits: sums, scalar multiples, products and quotients, Limits and Cauchy sequences of reals).

[L7]

Triangle inequality (M3), symmetry (M2) and separation (M1) of a metric, and nonnegativity (Metric space: $d(x,y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric, Nonnegativity of a metric is a consequence of the other axioms, not an axiom).

[L8]

A contraction is continuous, hence sequentially continuous (Contraction implies Lipschitz implies uniformly continuous implies continuous; every Hölder map is uniformly continuous, and a Lipschitz map on a bounded space is Hölder for every exponent, For a map of metric spaces the following agree: $\varepsilon$ - $\delta$ continuity everywhere, preimages of open sets are open, preimages of closed sets are closed, sequential continuity, and $f(\overline{A}) \subseteq \overline{f(A)}$ ).

[L9]

Limits in a metric space are unique, and convergence may be tested with real $\varepsilon > 0$ (A sequence in a metric space has at most one limit, Convergence of a sequence in a metric space: $x_k \to x$ iff $d(x_k, x) \to 0$ in $\mathbb{R}$ , Cauchy sequence in a metric space, The rationals embed densely in the reals).

[L10]

$1 - q > 0$ , so $1/(1-q) > 0$ and $C \ge 0$ (Inverses of positives are positive, and reciprocation reverses order, Nonnegativity of a metric is a consequence of the other axioms, not an axiom).

Proof

technique · constructive

1.1

$X$ is nonempty, so a starting point $x_0 \in X$ exists, and [L1] produces the sequence of iterates $(x_n)$ with $x_{n+1} = f(x_n)$ for every $n$ .

A2L1construct

1.2

If $f(y) = y$ and $f(z) = z$ then $d(y,z) = d(f(y),f(z)) \le q\,d(y,z)$ , so $(1-q)\,d(y,z) \le 0$ ; since $1-q > 0$ this forces $d(y,z) \le 0$ , and $d(y,z) \ge 0$ gives $d(y,z) = 0$ and $y = z$ .

A1L7L10

2.1

For every $n$ , $d(x_{n+1},x_n) \le q^{\,n} d(x_1,x_0)$ , by induction on $n$ : at $n = 0$ this reads $d(x_1,x_0) \le 1 \cdot d(x_1,x_0)$ ; and if it holds at $n$ then $d(x_{n+2},x_{n+1}) = d(f(x_{n+1}),f(x_n)) \le q\,d(x_{n+1},x_n) \le q \cdot q^{\,n} d(x_1,x_0) = q^{\,n+1} d(x_1,x_0)$ .

step 1.1A1L2L3

2.2

For all $n \le m$ , $d(x_m,x_n) \le \sum_{k=n}^{m-1} d(x_{k+1},x_k)$ , by induction on $m$ from the triangle inequality: at $m = n$ both sides are $0$ , since an empty sum is $0$ ; and $d(x_{m+1},x_n) \le d(x_m,x_n) + d(x_{m+1},x_m)$ .

step 1.1L2L5L7

3.1

Combining, for $n \le m$ : $d(x_m,x_n) \le \sum_{k=n}^{m-1} q^{\,k} d(x_1,x_0) = d(x_1,x_0)\, q^{\,n} \sum_{j < m-n} q^{\,j} \le d(x_1,x_0)\, q^{\,n} \frac{1}{1-q} = q^{\,n} C$ , using the index shift $q^{\,n+j} = q^{\,n}q^{\,j}$ and monotonicity of finite sums. This is claim 3.

step 2.1step 2.2L3L4L5L10

4.1

The sequence $(q^{\,n} C)$ of reals is null by [L6], so there is $N$ with $q^{\,n} C < \varepsilon$ for every $n \ge N$ , the terms being nonnegative. Hence for all $m,n \ge N$ we get $d(x_m,x_n) \le q^{\min\{m,n\}} C < \varepsilon$ by step 3.1 and symmetry of $d$ , so $(x_n)$ is Cauchy.

step 3.1L6L7L9L10

5.1

By [A2] the sequence converges: $x_n \to x^{*}$ for some $x^{*} \in X$ .

step 4.1A2

6.1

$f$ is continuous, so $f(x_n) \to f(x^{*})$ , that is $x_{n+1} \to f(x^{*})$ ; and $x_{n+1} \to x^{*}$ as well, since given a real $\eta > 0$ an index $K$ with $d(x_k,x^{*}) < \eta$ for $k \ge K$ also serves for the shifted indices $n+1 \ge K$ when $n \ge K$ .

step 5.1L8L9

7.1

By uniqueness of limits $f(x^{*}) = x^{*}$ , so $x^{*}$ is a fixed point and the iterates from $x_0$ converge to it.

step 5.1step 6.1L9

8.1

So a fixed point exists by step 7.1 and there is at most one by step 1.2, which is claim 1; the iterates from an arbitrary $x_0$ converge to a fixed point by step 7.1, hence to the unique one, which is claim 2; and claim 3 is step 3.1.

step 3.1step 7.1step 1.2discharge-construct∎

Remarks

Every hypothesis is used, and each has a witness for its necessity. Completeness: $x \mapsto x/2$ on $(0,1]$ is a contraction with constant $1/2$ and no fixed point ( $x \mapsto x/2$ maps $(0,1]$ into itself, is a $1/2$ -contraction, and has no fixed point ↗). Nonemptiness: the empty metric space is complete and the empty map is a contraction with no fixed point, which is why the statement says nonempty. A constant $q$ strictly below $1$ : the condition $d(f(u),f(v)) < d(u,v)$ for $u \ne v$ is strictly weaker and does not suffice (FALSE: $d(fx, fy) < d(x,y)$ for all $x \ne y$ on a complete metric space forces a fixed point, $x \mapsto x + 1/x$ on $[1,\infty)$ strictly decreases every distance and has no fixed point ↗).
The proof is constructive in a usable sense. It does not merely assert a fixed point: it names the sequence that converges to it, from any starting point whatever, and step 3.1 bounds the distance to the limit in terms of the first step alone. That is what makes the theorem an algorithm rather than an existence statement, and The a priori bound $d(x^{*}, x_n) \le q^n d(x_1,x_0)/(1-q)$ and the a posteriori bound $d(x^{*}, x_{n+1}) \le q\,d(x_{n+1},x_n)/(1-q)$ is the statement of the two bounds a computation actually uses.
Where the geometric series enters. Only through [L4], which is the factorisation $1 - q^{N} = (1-q)\sum_{k<N} q^{k}$ read backwards (Factorisation of $b^n - a^n$ , and the resulting Lipschitz estimate). No theory of infinite series is used or needed: every sum in the proof is finite, and the passage to the limit is made once, at step 4.1, through the null sequence $(q^n)$ .
Uniqueness is cheaper than existence and uses neither completeness nor recursion: step 1.2 is three lines from the contraction inequality alone. It is what makes claim 2 unambiguous, since it forces the limits obtained from different starting points to coincide.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 146 results over 36 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Banach fixed-point theorem (Wikipedia) (standard reference, not scraped)
Contraction mapping (Wikipedia) (standard reference, not scraped)
Fixed-point iteration (Wikipedia) (standard reference, not scraped)