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Mathematics · 术语表

什么是 complete metric space?

Definition 4.7 University Mathematics — Year 2 · 第 4 章 — Topology of Metric Spaces

A sequence (xn)(x_n) is Cauchy when sup⁡p,q≥Nd(xp,xq)→0\sup_{p, q \geq N} d(x_p, x_q) \to 0 as N→∞N \to \infty. A metric space is complete when every Cauchy sequence converges. Convergent ⇒\Rightarrow Cauchy always; closed subsets of complete spaces are complete, and complete subsets of any space are closed (same proofs as on R\R: Year 1 volume).

示例

Example 4.8 (Cauchy without a limit)

In X=QX = \Q with the usual distance, the decimal truncations of 2\sqrt2,

x0=1,x1=1.4,x2=1.41,x3=1.414,…x_0 = 1,\quad x_1 = 1.4,\quad x_2 = 1.41,\quad x_3 = 1.414, \quad\dots

satisfy ∣xp−xq∣≤10−min⁡(p,q)\abs{x_p - x_q} \leq 10^{-\min(p,q)}: Cauchy in Q\Q. A limit in Q\Q would also be the limit in R\R, namely 2∉Q\sqrt2 \notin \Q: no limit exists in XX. Incompleteness is the presence of such “phantom limits”; completeness of R\R was engineered in the Year 1 volume precisely to give every Cauchy sequence a home.

Example 4.13 (An integral equation)

On X=C([0, 1])X = C(\intcc{0}{1}) (complete, Theorem 4.9), consider T(f)(x)=1+12∫0xf(t)  ⁣dtT(f)(x) = 1 + \frac12 \int_0^x f(t)\,\dd t. For f,g∈Xf, g \in X:

∣T(f)(x)−T(g)(x)∣≤12∫0x∣f−g∣≤12 d∞(f,g),\abs{T(f)(x) - T(g)(x)} \leq \frac12 \int_0^x \abs{f - g} \leq \frac12\, d_\infty(f, g),

so TT is a 12\frac12-contraction: it has a unique continuous fixed point — the solution of f′=f2f' = \frac f2, f(0)=1f(0) = 1, namely ex/2\eu^{x/2}. This scheme, industrialized, becomes the Cauchy–Lipschitz theorem of Chapter 16.

Example 4.14 (A numerical fixed point: x=cos⁡xx = \cos x)

On the complete X=[0, 1]X = \intcc{0}{1}, the map f=cos⁡f = \cos sends XX into [cos⁡1, 1]⊆X\intcc{\cos 1}{1} \subseteq X and is a contraction: by the mean value inequality,

∣cos⁡x−cos⁡y∣≤(sup⁡[0, 1]∣sin⁡∣)∣x−y∣=(sin⁡1)∣x−y∣,sin⁡1≈0.841<1.\abs{\cos x - \cos y} \leq \bigl(\sup_{\intcc01}\abs{\sin}\bigr) \abs{x - y} = (\sin 1)\abs{x - y}, \qquad \sin 1 \approx 0.841 < 1 .

Banach: a unique solution of x=cos⁡xx = \cos x in [0, 1]\intcc01 (hence in R\R: any real fixed point lies in [−1, 1]\intcc{-1}{1}, then in [cos⁡1, 1]\intcc{\cos 1}{1} after one application), and the iteration xn+1=cos⁡xnx_{n+1} = \cos x_n converges to it from any start: x∞≈0.739085x_\infty \approx 0.739085, the famous number obtained by hammering the cosine key of a calculator. The error bound predicts (sin⁡1)n/(1−sin⁡1)(\sin1)^n/(1 - \sin1) decay — about one digit per 1313 presses; the a posteriori bound of this chapter’s weekend problem (question 14) certifies each step on the fly.

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Definition 7.1 University Mathematics — Year 3 · 第 7 章 — Complete Spaces: Baire, Ascoli, Stone–Weierstrass

A metric space is complete if every Cauchy sequence converges (Year 2: Rn\R^n is complete; C([0, 1])\mathcal C(\intcc01) with ∥⋅∥∞\norm\cdot_\infty is complete). A closed subset of a complete space is complete; a complete subset of any metric space is closed.

示例

Example 7.5 (Perturbing the identity)

Let g ⁣:Rd→Rdg \colon \R^d \to \R^d be kk-Lipschitz with k<1k < 1. Then φ=id+g\varphi = \mathrm{id} + g is a homeomorphism of Rd\R^d onto Rd\R^d. Injectivity, with a quantitative modulus:

∥φ(x)−φ(y)∥≥∥x−y∥−∥g(x)−g(y)∥≥(1−k)∥x−y∥.\norm{\varphi(x) - \varphi(y)} \geq \norm{x - y} - \norm{g(x) - g(y)} \geq (1 - k)\norm{x - y} .

Surjectivity is the fixed point theorem: solving φ(x)=y\varphi(x) = y means x=y−g(x)x = y - g(x), and x↦y−g(x)x \mapsto y - g(x) is a kk-contraction of the complete Rd\R^d — a unique solution x=ψ(y)x = \psi(y) exists for every yy. The displayed inequality makes the inverse ψ\psi Lipschitz with constant 11−k\frac1{1-k}: a homeomorphism, with explicit bounds on both moduli. This innocuous-looking statement is the engine inside the inverse function theorem (Chapter 20): near a point where DfDf is invertible, ff is an invertible linear map plus a small Lipschitz perturbation, and today’s example does the rest. It also quantifies numerical robustness: a system perturbed by less than the inverse’s margin remains solvable, with the solution moving by at most 11−k\frac{1}{1-k} times the perturbation.

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