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How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

✓ 13 results · all verified · 4 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full by a delegated reviewing agent on the owner's instruction; the judge is an additional, independent cross-model AI review of the proofs. The 9 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

R^n as a Normed Space; Vector-Valued Functions

1 · Prerequisites

2 · Summary

A note on the notation ι. A natural number here is a von Neumann natural, that is a set, so it is not an element of R. The canonical natural ι(n)=n⋅1R is the real number that n names (The canonical natural ι(n)=n⋅1F of a field, Canonical naturals are positive and strictly increasing), so 1/ι(k+1) is what an informal text writes as 1/(k+1) and ι(n) is what it writes as n inside an inequality between reals.

Objective. The published metric-spaces material already gives Rn a metric, and the published real-analysis track already gives R its calculus. This page puts the two together by adding the missing ingredient, the linear structure: a norm, the Euclidean inner product, and the observation that in finite dimensions the choice of norm never matters. It then carries limits, continuity, the derivative, the integral and series across from R to Rm, one coordinate at a time, and ends with what can honestly be said about rearranging a series of vectors.

Norms, and the seam with the published metrics. A norm on a real vector space, the induced metric, and the dictionary with the metric axioms fixes the three axioms (N1), (N2), (N3), derives nonnegativity rather than assuming it — exactly as Metric space: d(x,y)=0 iff x=y, symmetry, and the triangle inequality; pseudometric and ultrametric and Nonnegativity of a metric is a consequence of the other axioms, not an axiom do for metrics — and proves that d(u,v)=∥u−v∥ is a metric which is in addition translation invariant and absolutely homogeneous. Not every metric on a vector space arises this way, and min⁡(d,1) and d/(1+d) are metrics uniformly equivalent to d, so every metric space carries a bounded metric with the same topology is the published witness. The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn defines ⟨x,y⟩=∑k<nxkyk and proves its algebra from the finite-sum laws; Cauchy-Schwarz ∣⟨x,y⟩∣≤∥x∥2∥y∥2 with its equality case, the triangle inequality for ∥⋅∥2, the parallelogram law and polarisation restates the published The Cauchy-Schwarz inequality for finite sums in vector notation, rather than reproving it, and adds that ∥⋅∥2 is a norm, the parallelogram law and polarisation. The p-norms ∥x∥p for rational p≥1, and ∥x∥∞ introduces ∥⋅∥p for rational p≥1 and ∥⋅∥∞ for n≥1.

The seam is closed by Each ∥⋅∥p is a norm on Rn, and the induced metrics are exactly d1, d2 and d∞ of the published metric-spaces page: for n≥1 the metrics induced by ∥⋅∥1, ∥⋅∥2 and ∥⋅∥∞ are the published d1, d2 and d∞ of Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it, not merely metrics equivalent to them. Without that item the library would hold two unrelated metric structures on one set.

Equivalence of norms. Equivalent norms, and the dictionary with equivalent metrics defines equivalence and proves the dictionary: equivalent norms give Lipschitz equivalent metrics, the strongest of the three tiers of Topologically, uniformly and Lipschitz equivalent metrics on a set and Lipschitz equivalence implies uniform equivalence implies topological equivalence, hence the same open sets, convergent sequences, Cauchy sequences and uniformly continuous maps. The finite and reverse triangle inequalities for a norm; and for n≥1 every norm N on Rn satisfies N(x)≤C∥x∥1 and is Lipschitz, hence continuous, for d2 does the half that costs no compactness: the finite and reverse triangle inequalities for any norm, the bound N(x)≤C∥x∥1 from the standard basis, the comparison chain ∥x∥∞≤∥x∥2≤∥x∥1≤ι(n)∥x∥∞, and Lipschitz continuity of N for d2. For n≥1 all norms on Rn are equivalent supplies the other half by compactness of the Euclidean unit sphere, through Heine-Borel in Rn: with the Euclidean metric a subset of Rn is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line and A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value; the hypothesis n≥1 is used twice there and both uses are marked.

Sequences. For n≥1 a sequence in Rn converges iff each coordinate sequence converges, is Cauchy iff each coordinate sequence is Cauchy, and Rn is complete in every norm proves that convergence and Cauchyness in Rn are componentwise, and obtains completeness in every norm by citing the published R and Rn for n≥1 with the Euclidean metric are complete, componentwise from the Cauchy criterion in R and transporting it along norm equivalence. For n≥1 every bounded sequence in Rn has a convergent subsequence assembles Bolzano-Weierstrass in Rn from Heine-Borel in Rn: with the Euclidean metric a subset of Rn is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line and the ZF implication In any metric space compactness implies countable compactness and limit point compactness, and each of countable compactness and limit point compactness implies sequential compactness; every implication here is proved without a choice principle; it is not proved again by bisection, that work being published at order 120, and it costs no choice principle.

Vector-valued functions. Vector-valued functions f:A→Rm, their limits and continuity, with the dictionary to the metric notions defines limits and continuity for f:A→Rm and proves that they are the metric notions of Continuity of a map between metric spaces, at a point and globally, in the ε-δ form and nothing new, in the spirit of Dictionary: for A⊆R with the metric d(x,y)=∣x−y∣, continuity and uniform continuity of f:A→R agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of R is compact in the open-cover sense of R exactly when it is a compact metric subspace. A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions proves that both are componentwise and, because the domain here is a metric space rather than a subset of R, proves the algebra of sums, scalar multiples, inner products and norms directly instead of quoting Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function. The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral gives the derivative intrinsically, as a limit of difference quotients in Rm, with the componentwise formula as a consequence, and defines the integral coordinatewise with the orientation convention of The integral with oriented limits: ∫aaf:=0 and ∫baf:=−∫abf.

Three theorems about vector-valued calculus. For a≤b and f:[a,b]→Rm integrable when a<b, ∥∫abf∥2≤∫ab∥f∥2; for a<b, ∥f∥2 is integrable proves that ∥f∥2 is integrable when a<b and that ∥∫abf∥2≤∫ab∥f∥2 for a≤b, by the inner-product argument, with the case ∫abf=0 treated separately because the usual division is illegitimate there. The mean value inequality: if f:[a,b]→Rm is continuous and differentiable on (a,b) with ∥f′∥2≤M, then ∥f(b)−f(a)∥2≤M(b−a) proves ∥f(b)−f(a)∥2≤M(b−a) from the scalar mean value theorem applied to t↦⟨f(b)−f(a), f(t)⟩, with no integrability hypothesis; the equality form is false for m≥2 and the companion page carries the witness. If f:[a,b]→Rm is differentiable with integrable f′ then ∫abf′=f(b)−f(a); and a bounded derivative makes f Lipschitz gives the componentwise fundamental theorem and the Lipschitz bound, and says why the mean value inequality is proved the other way round.

Series of vectors, and how far the rearrangement question can be taken. Series of vectors in Rn, absolute convergence, rearrangement, and the set of rearrangement sums fixes partial sums, convergence, absolute convergence, rearrangement and the set S(x) of rearrangement sums, with an explicit agreement clause against Series, partial sums, convergence and the sum, divergence, and the tail series and Rearrangement of a series along a bijection of N, and unconditional convergence at n=1. An absolutely convergent series in Rn converges, and every rearrangement converges to the same sum proves that absolute convergence gives convergence, by a Cauchy estimate together with completeness, and that S(x) is then a single point, by reduction to Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum coordinatewise. The subspace Γ of directions along which a series converges absolutely, and its orthogonal complement Γ⊥ introduces Γ={a:∑k∣⟨a,xk⟩∣ converges} and Γ⊥, proves both are linear subspaces, and proves Γ=Rn exactly when the series converges absolutely; it is phrased with the inner product and not with linear functionals, because dual spaces belong to a page earlier in the plan order that is not yet built.

Steinitz's polygonal confinement theorem: finitely many vectors of norm at most 1 summing to 0 can be ordered so that every partial sum has norm at most n proves Steinitz's polygonal confinement lemma in full: finitely many vectors of norm at most 1 summing to 0 can be ordered so that every partial sum has norm at most ι(n). The proof is constructive, and it includes the support bound #supp⁡≤k−1 together with the reason equality is impossible, which is the step most write-ups omit. The set of rearrangement sums of a convergent series in Rn is a nonempty subset of the affine subspace s+Γ⊥ proves that S(x) is nonempty and contained in the affine subspace s+Γ⊥.

What this page does not settle, stated plainly. Whether S(x) is all of s+Γ⊥ when n≥2 is not settled here, and no item on this page asserts anything about it in either direction. The obstruction is machinery: every route known to this page's author needs the orthogonal decomposition of a finite-dimensional inner product space, which belongs to a page earlier in the plan order that is not yet built, and a convex-separation argument that no planned page owns. No recorded-not-proved item has been created for it. The published The same question in Rd: what the set of rearrangement sums looks like, and why that answer is not reachable at this point in the reading order raised the question and declined to state the literature's answer; this page answers the part it can and continues to decline the rest. A reader is protected from the wrong guess in the meantime: the companion page refutes the naive Rn analogue of The Riemann series theorem: a conditionally convergent real series has, for every c∈R, a rearrangement with sum c, and rearrangements diverging to +∞, to −∞, and oscillating with any prescribed lim inf⁡≤lim sup⁡ in R‾ outright, using the containment half and nothing more.

Conventions. Conventions of this page, the standing n≥1 hypothesis, and what is taken up elsewhere in the reading order is the page ledger. It records where the standing hypothesis n≥1 comes from and which items carry it and which do not, that the exponent of a p-norm is rational because real exponents do not exist at this point in the reading order (Why real exponents are deferred on the rational-powers page), that Rn is a function space so that R1 is not literally R, what is taken up elsewhere in the reading order, and which Steinitz result the confinement theorem is — it is not the Steinitz exchange lemma of linear algebra, which is published under a different id and carries the alias lem-steinitz.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicableverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

A norm on a real vector space, the induced metric, and the dictionary with the metric axioms

Definition

Throughout this page R is the complete ordered field (Complete ordered field (least-upper-bound property)) constructed in this library, in particular a field, so that "vector space" below always means vector space over R (Vector space over a field).

Let V be a vector space over R, with zero vector 0V. A norm on V is a function N:V→R such that for all u,v∈V and all λ∈R:

  • (N1) Separation. N(v)=0 if and only if v=0V.
  • (N2) Absolute homogeneity. N(λv)=∣λ∣ N(v), the absolute value being that of Absolute value in an ordered field.
  • (N3) Triangle inequality. N(u+v)≤N(u)+N(v).

A normed space is a pair (V,N) consisting of a vector space V over R and a norm N on it. When only one norm is in play we write ∥v∥ for N(v); when several are, the norm is always named.

The values of a norm are real numbers. The codomain is R, so N(v) is an honest element of the complete ordered field and no infinite value is permitted. This is the same convention Which metric axiom list this library uses, the live naming fork between semimetric and pseudometric, and why extended metrics are not treated here records for metrics.

Nonnegativity is a theorem, not an axiom

Many texts add a fourth condition N(v)≥0. It is redundant. Applying (N2) with λ=−1 gives N(−v)=∣−1∣ N(v)=N(v) (Basic properties of the absolute value, In any vector space 0Fv=0V, λ0V=0V, (−λ)v=−(λv), (−1F)v=−v, and λv=0V forces λ=0F or v=0V for (−1)v=−v), and then (N3) with u=v and −v gives

0  =  N(0V)  =  N(v+(−v))  ≤  N(v)+N(−v)  =  N(v)+N(v),

where N(0V)=0 is (N1). So N(v)+N(v)≥0, and if N(v)<0 then N(v)+N(v)<0 by addition of inequalities, which trichotomy forbids (Complete ordered field (least-upper-bound property)). Hence N(v)≥0 for every v∈V.

Consequently the verification of a candidate norm has three things to check and not four, exactly as the verification of a candidate metric has three and not four (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). No item in this library assumes nonnegativity of a norm before the argument above.

The induced metric

Let N be a norm on V and define

dN(u,v)  :=  N(u−v)(u,v∈V),

where u−v=u+(−v) (Vector space over a field). Then dN is a metric on V (Metric space: d(x,y)=0 iff x=y, symmetry, and the triangle inequality; pseudometric and ultrametric), and the three axioms are the three conditions above, in order:

A normed space is therefore a metric space, and every notion defined for metric spaces — open set (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement), convergence, Cauchyness, continuity, compactness — is available in it with no further definition. This library never introduces a second notion of any of them for normed spaces.

Two properties an arbitrary metric need not have

The metric dN satisfies, for all u,v,w∈V and λ∈R:

  • translation invariance, dN(u+w,v+w)=N((u+w)−(v+w))=N(u−v)=dN(u,v);
  • absolute homogeneity, dN(λu,λv)=N(λ(u−v))=∣λ∣ dN(u,v), by (N2).

Not every metric on a vector space arises from a norm, and homogeneity is what fails. The published bounded remetrisation min⁡(d,1) and d/(1+d) are metrics uniformly equivalent to d, so every metric space carries a bounded metric with the same topology replaces a metric d by d′=min⁡{d,1}, a metric with the same topology whose values never exceed 1; on a vector space V containing a vector v with d(v,0V)>0 this d′ cannot be dM for any norm M, since absolute homogeneity would force d′(λv,0V)=∣λ∣ d′(v,0V), which is unbounded in λ, while d′ is bounded by 1. So the passage from norms to metrics is not reversible, and a statement about a metric on a vector space is strictly weaker than the corresponding statement about a norm.

Remarks

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-29Open item page →

The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn

Definition

Let n∈N. A natural number is a von Neumann natural, that is a set, and n={0,1,…,n−1} (The natural numbers N (von Neumann), On N the order is membership: m<n  ⟺  m∈n), so

Rn  =  { x  :  x is a function n→R }

is the function space of The vector space FX of all functions X→F with pointwise operations, and Fn as the case X=n={0,1,…,n−1} at F=R and X=n, a vector space over R under the pointwise operations (Vector space over a field). We write xk:=x(k) for k<n, and two elements of Rn are equal exactly when they agree at every k<n. This is the same set that Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it calls Rn.

The Euclidean inner product of x,y∈Rn is the real number

⟨x,y⟩  :=  ∑k<nxk yk,

the finite sum of Finite sums and finite products, by recursion applied to the list k↦xkyk (extended by 0 beyond n, as every finite list in this library is). The Euclidean norm of x is

∥x∥2  :=  ⟨x,x⟩,

which is defined because ⟨x,x⟩=∑k<nxk2≥0 (a sum of nonnegative terms, Laws of finite sums and finite products clause 4 and Squares of nonzero elements are positive, the case xk=0 giving xk2=0 by Integer powers am) and every nonnegative real has a unique nonnegative square root (Square roots exist: a unique a≥0 with (a)2=a; the positives are {x2:x≠0}).

Both are defined for every n, including n=0

At n=0 the set R0 has exactly one element, the empty function, and it is the zero vector space (The standard list e:n→Fn with ei(i)=1F and ei(j)=0F for j≠i is an ordered basis of Fn; hence dim⁡FFn=n, and F0 is the zero space with basis ∅ and dimension 0 clause 5); the sum above is the empty sum, so ⟨x,y⟩=0 and ∥x∥2=0. This is the first place on this page where the two index regimes diverge, and the divergence is deliberate. The published metrics d1, d2, d∞ of Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it are defined only for n≥1, because d∞ would otherwise be a maximum over the empty index set; the algebra above needs no such restriction. The boundary in this page runs between the algebra and the metric, not where a reader would guess, and Conventions of this page, the standing n≥1 hypothesis, and what is taken up elsewhere in the reading order lists exactly which items inherit n≥1.

The algebra of the inner product

For all x,y,z∈Rn and λ∈R:

  1. Symmetry. ⟨x,y⟩=⟨y,x⟩, since xkyk=ykxk termwise.
  2. Additivity in the first argument. ⟨x+y,z⟩=⟨x,z⟩+⟨y,z⟩: the list k↦(xk+yk)zk is the termwise sum of k↦xkzk and k↦ykzk, so Laws of finite sums and finite products clause 1 applies.
  3. Homogeneity in the first argument. ⟨λx,y⟩=λ⟨x,y⟩, by Laws of finite sums and finite products clause 2.
  4. Bilinearity. Clauses 2 and 3 together with symmetry give the same two laws in the second argument.
  5. Positive definiteness. ⟨x,x⟩≥0, and ⟨x,x⟩=0 if and only if x=0. Indeed a vanishing sum of nonnegative terms has every term 0 (Laws of finite sums and finite products clause 4), so xk2=0 for every k<n, and a nonzero real has a positive square (Squares of nonzero elements are positive), whence xk=0 for every k<n and x=0.
  6. Agreement with the published Euclidean metric. For n≥1 and x,y∈Rn, ∥x−y∥2=∑k<n(xk−yk)2=d2(x,y), the two sides being the same expression (Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it). In particular ∥x∥2=d2(x,0).

That ∥⋅∥2 is a norm in the sense of A norm on a real vector space, the induced metric, and the dictionary with the metric axioms is proved in Cauchy-Schwarz ∣⟨x,y⟩∣≤∥x∥2∥y∥2 with its equality case, the triangle inequality for ∥⋅∥2, the parallelogram law and polarisation, where the triangle inequality is obtained from the Cauchy-Schwarz inequality; it is not assumed here.

Remarks

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

Cauchy-Schwarz ∣⟨x,y⟩∣≤∥x∥2∥y∥2 with its equality case, the triangle inequality for ∥⋅∥2, the parallelogram law and polarisation

Statement

Let n∈N and let x,y∈Rn, with the Euclidean inner product and the Euclidean norm as in The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn. Then:

  1. Cauchy-Schwarz. ∣⟨x,y⟩∣  ≤  ∥x∥2 ∥y∥2, with equality if and only if there is a pair (λ,μ)≠(0,0) of reals with λxk=μyk for every k<n.
  2. ∥⋅∥2 is a norm on Rn (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms), for every n∈N; the metric it induces is d2 of Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it whenever n≥1.
  3. Parallelogram law. ∥x+y∥22+∥x−y∥22  =  2∥x∥22+2∥y∥22.
  4. Polarisation. ⟨x,y⟩  =  14(∥x+y∥22−∥x−y∥22), so the inner product is recovered from the norm it induces.

Clause 1 is a citation, not a new proof. The inequality and its equality case are the published The Cauchy-Schwarz inequality for finite sums, stated there for two lists of reals; all that happens below is that it is read in the vector notation of The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn. Re-proving it here would put two proofs of one statement in the library.

Facts & Assumptions

Given: A natural number n and vectors x,y∈Rn, so that ⟨x,y⟩=∑k<nxkyk and ∥x∥2=⟨x,x⟩ (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn, Finite sums and finite products, by recursion).

[L1]

Cauchy-Schwarz for finite sums (The Cauchy-Schwarz inequality for finite sums): (∑k<nakbk)2≤(∑k<nak2)(∑k<nbk2), with equality if and only if there is (λ,μ)≠(0,0) with λak=μbk for every k<n; and the root form ∣∑k<nakbk∣≤∑k<nak2∑k<nbk2.

[L2]

The inner product is symmetric, bilinear and positive definite, ⟨x,x⟩=∑k<nxk2≥0, and ⟨x,x⟩=0 exactly when x=0 (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn, Laws of finite sums and finite products).

[L3]

Square roots (Square roots exist: a unique a≥0 with (a)2=a; the positives are {x2:x≠0}): every c≥0 has a unique s≥0 with s2=c, written c; hence ∥x∥2≥0 and ∥x∥22=⟨x,x⟩ (Integer powers am).

[L4]

Squaring is monotone on the nonnegatives: for a,b≥0, a≤b if and only if a2≤b2, and a=b if and only if a2=b2 (Squaring is monotone on the nonnegatives).

[L5]

Absolute value (Basic properties of the absolute value, Absolute value in an ordered field): ∣t∣≥0, ∣t∣2=t2, and ∣st∣=∣s∣ ∣t∣.

Proof

technique · direct
1.1

Instantiating [L1] at ak:=xk and bk:=yk gives ⟨x,y⟩2≤⟨x,x⟩ ⟨y,y⟩, with equality exactly when some (λ,μ)≠(0,0) has λxk=μyk for every k<n.

L1L2
1.2

Both ∣⟨x,y⟩∣ and ∥x∥2∥y∥2 are nonnegative, and their squares are ⟨x,y⟩2 and ⟨x,x⟩⟨y,y⟩.

L3L5
1.3

Expanding by bilinearity and symmetry, ⟨x+y,x+y⟩=⟨x,x⟩+⟨x,y⟩+⟨y,x⟩+⟨y,y⟩=∥x∥22+2⟨x,y⟩+∥y∥22.

L2L3
1.4

The same expansion at x−y=x+(−1)y gives ∥x−y∥22=∥x∥22−2⟨x,y⟩+∥y∥22.

L2L3
1.5

For a scalar λ, ⟨λx,λx⟩=λ2⟨x,x⟩=∣λ∣2∥x∥22, so ∥λx∥22=(∣λ∣∥x∥2)2.

L2L3L5
1.6

Axiom (N1) holds: ∥x∥2=0 if and only if ∥x∥22=⟨x,x⟩=0, which by positive definiteness says x=0.

L2L3L4
2.1

Comparing the squares of step 1.2 through step 1.1 and using monotonicity of squaring on the nonnegatives yields ∣⟨x,y⟩∣≤∥x∥2∥y∥2, with equality exactly in the proportional case of step 1.1; this is clause 1.

step 1.1step 1.2L4
2.2

Adding the identities of step 1.3 and step 1.4 gives ∥x+y∥22+∥x−y∥22=2∥x∥22+2∥y∥22, which is clause 3.

step 1.3step 1.4algebra
2.3

Subtracting the identity of step 1.4 from that of step 1.3 gives ∥x+y∥22−∥x−y∥22=4⟨x,y⟩, which is clause 4 after dividing by 4.

step 1.3step 1.4algebra
2.4

Both ∥λx∥2 and ∣λ∣∥x∥2 are nonnegative and by step 1.5 have equal squares, so ∥λx∥2=∣λ∣∥x∥2, which is axiom (N2).

step 1.5L3L4L5
3.1

By step 2.1 the middle term of step 1.3 satisfies 2⟨x,y⟩≤2∥x∥2∥y∥2, so ∥x+y∥22≤∥x∥22+2∥x∥2∥y∥2+∥y∥22=(∥x∥2+∥y∥2)2.

step 1.3step 2.1L5algebra
4.1

Both ∥x+y∥2 and ∥x∥2+∥y∥2 are nonnegative, so step 3.1 and monotonicity of squaring give ∥x+y∥2≤∥x∥2+∥y∥2, which is axiom (N3).

step 3.1L3L4
5.1

Steps 2.4, 1.6 and 4.1 are exactly (N1), (N2) and (N3), so ∥⋅∥2 is a norm on Rn for every n∈N, and for n≥1 the metric it induces is d2; this is clause 2, and with steps 2.1, 2.2 and 2.3 all four clauses are proved.

step 2.1step 2.2step 2.3step 2.4step 1.6step 4.1L6∎

Remarks

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-29Open item page →

The p-norms ∥x∥p for rational p≥1, and ∥x∥∞

Definition

Let n∈N and let Rn be the function space of The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn, with xk:=x(k) for k<n.

The p-norm, for a rational exponent p≥1

Let p∈Q with p≥1. For x∈Rn put

∥x∥p  :=  (∑k<n∣xk∣p)1/p,

where ∣⋅∣ is the absolute value (Absolute value in an ordered field), the sum is the finite sum of Finite sums and finite products, by recursion, and both powers are the rational powers of Rational powers ar of a positive base.

Every power written here is defined. Each base ∣xk∣ is a nonnegative real and p>0, so ∣xk∣p is given by Rational powers ar of a positive base for ∣xk∣>0 and by its supplementary clause 0p=0 for ∣xk∣=0; the sum of these nonnegative terms is nonnegative (Laws of finite sums and finite products clause 4), and 1/p is a positive rational, so the outer power is defined for the same two reasons. The value does not depend on which representative of p or of 1/p is used (Rational powers do not depend on the representative).

The exponent is a rational, and that is not a matter of taste. Rational powers ar of a positive base supplies ar for a nonnegative base and a rational exponent only; real exponents do not exist at this point in the reading order, and Why real exponents are deferred on the rational-powers page records exactly why. This is also why the published Minkowski inequality Minkowski's inequality for finite sums (rational exponent), which is what makes the triangle inequality work below, is itself stated for rational p≥1. No statement on this page is written for p ranging over a real interval.

The maximum norm

For n≥1 and x∈Rn put

∥x∥∞  :=  max⁡{ ∣xk∣  :  k<n },

the maximum of a nonempty finite set of reals, which exists and is one of its elements (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

The hypothesis n≥1 is required and propagates. At n=0 the set {∣xk∣:k<n} is empty and has no maximum (Maximum and minimum of a set). This is the same restriction the published Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it carries, for the same reason, and every statement on this page that mentions ∥⋅∥∞ inherits it. The p-norms for rational p≥1 carry no such restriction: at n=0 each is the empty sum raised to a positive rational power, hence 0.

The three cases the rest of the page uses

That each of these is a norm in the sense of A norm on a real vector space, the induced metric, and the dictionary with the metric axioms, and that the metrics they induce are exactly the published d1, d2 and d∞ of Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it, is Each ∥⋅∥p is a norm on Rn, and the induced metrics are exactly d1, d2 and d∞ of the published metric-spaces page; it is proved there and is not assumed here.

Remarks

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

Each ∥⋅∥p is a norm on Rn, and the induced metrics are exactly d1, d2 and d∞ of the published metric-spaces page

Statement

Let n∈N and let p∈Q with p≥1, with the norms of The p-norms ∥x∥p for rational p≥1, and ∥x∥∞. Then:

  1. ∥⋅∥p is a norm on Rn (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).
  2. For n≥1, ∥⋅∥∞ is a norm on Rn.
  3. The dictionary. For n≥1 and all x,y∈Rn, ∥x−y∥1=d1(x,y),∥x−y∥2=d2(x,y),∥x−y∥∞=d∞(x,y), where d1, d2, d∞ are the metrics of the published Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it. So the metric induced by each of these three norms (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms) is the correspondingly named published metric, not merely one equivalent to it.

Consequence, used repeatedly below and stated once here. By clause 3 at p=2, the metric space (Rn,d2) of the published metric-spaces page and the metric space underlying the normed space (Rn,∥⋅∥2) of this page are the same object. Hence completeness (R and Rn for n≥1 with the Euclidean metric are complete, componentwise from the Cauchy criterion in R clause 2), Heine-Borel (Heine-Borel in Rn: with the Euclidean metric a subset of Rn is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line clause 2) and the compactness equivalences (For a metric space, compact, countably compact, limit point compact, sequentially compact, and complete together with totally bounded are all equivalent, given countable choice and dependent choice) are statements about this page's normed space, with their hypothesis n≥1 inherited unchanged and not weakened. Nothing below cites any of those three theorems for n=0.

Why this lemma exists. Without it the library would hold a norm-induced metric on Rn and a separately published metric on the same set with no recorded relation, and every later citation would have to guess which was meant. The proof of clause 3 is a comparison of two written expressions; the value is that the comparison is made and recorded.

Facts & Assumptions

Given: A natural number n, a rational p≥1, vectors x,y∈Rn and a real λ; write S(x):=∑k<n∣xk∣p, so that ∥x∥p=S(x)1/p (The p-norms ∥x∥p for rational p≥1, and ∥x∥∞, Finite sums and finite products, by recursion).

[L1]

Rational powers (Rational powers ar of a positive base, Laws of rational exponents): for a,b≥0 and rationals r,s>0 one has ar≥0, (ab)r=arbr, 0r=0, and ar>0 when a>0; and for a>0, (ar)s=ars and a1=a.

[L2]

Monotonicity in the base (Monotonicity of r↦ar and of a↦ar clause 2): for a rational r>0 and reals 0≤a<b one has ar<br; hence a≤b implies ar≤br, the case a=b being trivial, and ar=0 only for a=0.

[L3]

Laws of finite sums (Laws of finite sums and finite products, Finite sums and finite products, by recursion): additivity, scaling, monotonicity; a sum of nonnegative terms is nonnegative, each single term is at most such a sum, and a sum of nonnegative terms that vanishes has every term 0.

[L4]

Minkowski's inequality for finite sums at rational p≥1 (Minkowski's inequality for finite sums (rational exponent)): (∑k<n∣ak+bk∣p)1/p≤(∑k<n∣ak∣p)1/p+(∑k<n∣bk∣p)1/p.

[L5]

Absolute value (Basic properties of the absolute value, Absolute value in an ordered field, The triangle inequality): ∣t∣≥0; ∣t∣=0 exactly when t=0; ∣st∣=∣s∣ ∣t∣; ∣s+t∣≤∣s∣+∣t∣; and ∣t∣2=t2.

[L6]

Maxima (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set): a nonempty finite set of reals has a maximum, the maximum belongs to the set and bounds it above, and a set with an upper bound belonging to it has that element as its maximum.

[L7]

Order arithmetic: multiplying an inequality by a nonnegative real preserves it (Sign rules for products and monotonicity of multiplication in its strict form, together with the case of equality settled by totality), and ≤ is transitive (Ordered field).

[L9]

The published metrics on Rn for n≥1 are d1(x,y)=∑k<n∣xk−yk∣, d2(x,y)=∑k<n(xk−yk)2 and d∞(x,y)=max⁡{∣xk−yk∣:k<n}, and each is a metric (Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it, Metric space: d(x,y)=0 iff x=y, symmetry, and the triangle inequality; pseudometric and ultrametric, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

Proof

technique · direct
1.1

Every term ∣xk∣p is nonnegative, so S(x)≥0 and ∥x∥p=S(x)1/p is defined and nonnegative.

L1L3
1.2

S(x)=0 holds exactly when ∣xk∣p=0 for every k<n, a vanishing sum of nonnegative terms having every term 0; and ∣xk∣p=0 exactly when ∣xk∣=0, that is exactly when xk=0.

L1L2L3L5
1.3

For every k<n, ∣(λx)k∣p=(∣λ∣ ∣xk∣)p=∣λ∣p∣xk∣p, so S(λx)=∣λ∣pS(x) by scaling of finite sums.

L1L3L5
1.4

Instantiating [L4] at ak:=xk and bk:=yk, and using (x+y)k=xk+yk, gives ∥x+y∥p≤∥x∥p+∥y∥p, which is axiom (N3) for ∥⋅∥p.

L4L8
1.5

Under [A1] the set {∣xk∣:k<n} is nonempty and finite, so ∥x∥∞ exists, is one of the ∣xk∣, and satisfies ∣xk∣≤∥x∥∞ for every k<n; in particular ∥x∥∞≥0.

A1L5L6
1.6

Under [A1], ∥x−y∥1=∑k<n∣xk−yk∣ by the case p=1 of the definition, and that is the written expression for d1(x,y).

L1L9
1.7

Under [A1], ∥x−y∥2=(∑k<n∣xk−yk∣2)1/2=∑k<n(xk−yk)2, using ∣t∣2=t2 and the identification of the exponent 1/2 with the nonnegative square root, and that is the written expression for d2(x,y).

L5L8L9
1.8

Under [A1], ∥x−y∥∞=max⁡{∣xk−yk∣:k<n} by definition, and that is the written expression for d∞(x,y).

L9
2.1

∥x∥p=0 holds exactly when S(x)=0, since S(x)>0 would give S(x)1/p>0 and 01/p=0.

step 1.1L1L2
2.2

Under [A1]: ∥x∥∞=0 forces ∣xk∣≤0 and ∣xk∣≥0 for every k<n, hence x=0; and ∥0∥∞=0. This is (N1) for ∥⋅∥∞.

step 1.5L5L8
2.3

Under [A1]: for every k<n, ∣(λx)k∣=∣λ∣ ∣xk∣≤∣λ∣ ∥x∥∞, and choosing j<n with ∣xj∣=∥x∥∞ gives ∣(λx)j∣=∣λ∣ ∥x∥∞; so ∣λ∣∥x∥∞ belongs to the set and bounds it above, whence ∥λx∥∞=∣λ∣∥x∥∞. This is (N2) for ∥⋅∥∞.

step 1.5L5L6L7
2.4

Under [A1]: for every k<n, ∣(x+y)k∣=∣xk+yk∣≤∣xk∣+∣yk∣≤∥x∥∞+∥y∥∞; choosing j<n with ∣(x+y)j∣=∥x+y∥∞ gives ∥x+y∥∞≤∥x∥∞+∥y∥∞, which is (N3) for ∥⋅∥∞.

step 1.5L5L6L7
3.1

By steps 2.1 and 1.2, ∥x∥p=0 exactly when xk=0 for every k<n, that is exactly when x=0; this is axiom (N1) for ∥⋅∥p.

step 2.1step 1.2L8
3.2

Steps 2.2, 2.3 and 2.4 are (N1), (N2) and (N3) for ∥⋅∥∞ under [A1], so clause 2 holds.

step 2.2step 2.3step 2.4A1L8
4.1

If λ=0 then λx=0 and both sides of (N2) are 0 by step 3.1; if λ≠0 then ∣λ∣>0, and step 1.3 with the power laws gives ∥λx∥p=(∣λ∣pS(x))1/p=(∣λ∣p)1/pS(x)1/p=∣λ∣p⋅(1/p)∥x∥p=∣λ∣ ∥x∥p; this is axiom (N2).

step 1.3step 3.1L1L5L8
5.1

Steps 3.1, 4.1 and 1.4 are (N1), (N2) and (N3) for ∥⋅∥p, so clause 1 holds.

step 1.4step 3.1step 4.1L8
6.1

Steps 1.6, 1.7 and 1.8 give clause 3, and with steps 5.1 and 3.2 all three clauses are proved; in particular the metric induced by ∥⋅∥2 on Rn for n≥1 is the published d2, which is the consequence recorded in the Statement.

step 5.1step 3.2step 1.6step 1.7step 1.8L9∎

Remarks

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-29Open item page →

Equivalent norms, and the dictionary with equivalent metrics

Definition

Let V be a vector space over R (Vector space over a field) and let M and N be norms on V (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms). M and N are equivalent when there are reals c>0 and C>0 with

c M(v)  ≤  N(v)  ≤  C M(v)for every v∈V.

The constants are not part of the data and are not unique: any smaller c and any larger C serve as well.

This is an equivalence relation on the norms on V

  • Reflexive: take c=C=1.
  • Symmetric: from cM≤N≤CM and c,C>0 one gets C−1N≤M≤c−1N, dividing by the positive constants (Inverses of positives are positive, and reciprocation reverses order).
  • Transitive: if cM≤N≤CM and c′N≤P≤C′N then c′c M≤P≤C′C M, and c′c>0, C′C>0, a product of positives being positive.

The dictionary with equivalent metrics

Let dM(u,v)=M(u−v) and dN(u,v)=N(u−v) be the induced metrics (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms). Substituting v:=u−w in the displayed condition gives

c dM(u,w)  ≤  dN(u,w)  ≤  C dM(u,w)for all u,w∈V,

which is verbatim the Lipschitz equivalence of dM and dN in the sense of Topologically, uniformly and Lipschitz equivalent metrics on a set, with α=c and β=C. That is the strongest of the three tiers that item distinguishes: by Lipschitz equivalence implies uniform equivalence implies topological equivalence, Lipschitz equivalence implies uniform equivalence, which implies topological equivalence. So equivalent norms give

The last line deserves its two-line verification, since it is used constantly below and is not literally a clause of Lipschitz equivalence implies uniform equivalence implies topological equivalence. If dM(vk,v)→0 then 0≤dN(vk,v)≤C dM(vk,v), so given a rational ε>0 an index beyond which dM(vk,v)<ε/C serves for dN; the converse uses dM≤c−1dN in the same way. The Cauchy statement is the same estimate applied to dN(vk,vl). In particular (V,dM) is complete if and only if (V,dN) is.

Naming. Many texts say strongly equivalent for what Topologically, uniformly and Lipschitz equivalent metrics on a set calls Lipschitz equivalent, and simply equivalent for what it calls topologically equivalent. As there, this library always writes the qualifier for metrics. For norms there is no fork to guard against: the condition displayed above is the only one anyone calls equivalence of norms, and it is always the Lipschitz-strength one.

Remarks

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

The finite and reverse triangle inequalities for a norm; and for n≥1 every norm N on Rn satisfies N(x)≤C∥x∥1 and is Lipschitz, hence continuous, for d2

Statement

Clause 1 is about an arbitrary norm; clauses 2 to 4 are about Rn with n≥1.

  1. Finite and reverse triangle inequalities. Let V be a vector space over R and N a norm on it (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms). For every p∈N and every list u:p→V (Linear combination of a finite list, and the span span⁡(S) as the smallest linear subspace containing S), N(∑j<puj)  ≤  ∑j<pN(uj), and for all u,w∈V, ∣N(u)−N(w)∣  ≤  N(u−w).

Now let n∈N with n≥1, let Rn carry the norms of The p-norms ∥x∥p for rational p≥1, and ∥x∥∞ and write ι for the canonical natural (The canonical natural ι(n)=n⋅1F of a field).

  1. Every norm is dominated by the 1-norm. Let N be a norm on Rn and put C:=max⁡{ N(ek):k<n }, a maximum over a nonempty finite set of reals (The standard list e:n→Fn with ei(i)=1F and ei(j)=0F for j≠i is an ordered basis of Fn; hence dim⁡FFn=n, and F0 is the zero space with basis ∅ and dimension 0, Every nonempty finite set of reals has a maximum and a minimum). Then C≥0 and N(x)  ≤  C ∥x∥1for every x∈Rn.
  2. The comparison chain. For every x∈Rn, ∥x∥∞  ≤  ∥x∥2  ≤  ∥x∥1  ≤  ι(n) ∥x∥∞,∥x∥1  ≤  ι(n)  ∥x∥2. In particular ∥⋅∥1, ∥⋅∥2 and ∥⋅∥∞ are pairwise equivalent norms on Rn, with the constants displayed (Equivalent norms, and the dictionary with equivalent metrics).
  3. Every norm is Lipschitz for the Euclidean metric. With N and C as in clause 2, N:(Rn,d2)→(R,dR) is Lipschitz with constant Cι(n) (Lipschitz map, α-Hölder map for rational 0<α≤1, and contraction, Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it, The absolute value makes R a metric space: d(x,y)=∣x−y∣ is a metric, its open balls are the intervals (x−r,x+r), and it is unbounded), hence uniformly continuous and 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, Continuity of a map between metric spaces, at a point and globally, in the ε-δ form).

Where n≥1 enters. Clauses 2 and 4 need the maximum defining C to exist, and clause 3 mentions ∥⋅∥∞; at n=0 each is a maximum over the empty index set and does not exist, exactly as in Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it and The p-norms ∥x∥p for rational p≥1, and ∥x∥∞. Clause 1 carries no hypothesis on the dimension and no hypothesis on the space.

Facts & Assumptions

Given: A vector space V over R with a norm N (Vector space over a field, A norm on a real vector space, the induced metric, and the dictionary with the metric axioms); and, for clauses 2 to 4, a natural n≥1, the space Rn, a norm N on it, and vectors x,y∈Rn.

[L1]

The norm axioms: N(v)=0 exactly when v=0V; N(λv)=∣λ∣N(v); N(u+w)≤N(u)+N(w); and N(v)≥0 (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

[L3]

The induction principle (The principle of mathematical induction).

[L4]

Laws of finite sums of reals (Laws of finite sums and finite products, Finite sums and finite products, by recursion): additivity, scaling, monotonicity, ∑k<nλ=ι(n)λ, a sum of nonnegative terms is nonnegative, and every single term is at most such a sum.

[L6]

Maxima (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set): a nonempty finite set of reals has a maximum, which belongs to the set and bounds it above.

[L7]

The three norms (The p-norms ∥x∥p for rational p≥1, and ∥x∥∞, Each ∥⋅∥p is a norm on Rn, and the induced metrics are exactly d1, d2 and d∞ of the published metric-spaces page): ∥x∥1=∑k<n∣xk∣, ∥x∥2=∑k<nxk2, ∥x∥∞=max⁡{∣xk∣:k<n}, and each induces the correspondingly named published metric.

[L8]

Cauchy-Schwarz in root form (The Cauchy-Schwarz inequality for finite sums): ∣∑k<nakbk∣≤∑k<nak2∑k<nbk2.

[L9]

Square roots and squaring (Square roots exist: a unique a≥0 with (a)2=a; the positives are {x2:x≠0}, Squaring is monotone on the nonnegatives): every c≥0 has a unique c≥0 with (c)2=c; for a,b≥0, a≤b exactly when a2≤b2.

[L10]

Absolute value (Basic properties of the absolute value): ∣t∣≥0, ∣t∣2=t2, ∣st∣=∣s∣∣t∣, ∣−t∣=∣t∣, and ∣t∣ equals t or −t.

Proof

technique · direct
1.1

The finite triangle inequality holds by induction on p: at p=0 both sides are 0, since ∑j<0uj=0V and N(0V)=0 and the empty real sum is 0; and if N(∑j<puj)≤∑j<pN(uj), then N(∑j<p+1uj)=N(∑j<puj+up)≤N(∑j<puj)+N(up)≤∑j<pN(uj)+N(up)=∑j<p+1N(uj).

L1L2L3L4
1.2

For u,w∈V: N(u)=N((u−w)+w)≤N(u−w)+N(w), so N(u)−N(w)≤N(u−w); and N(w−u)=N((−1)(u−w))=∣−1∣N(u−w)=N(u−w), so the same argument with u and w exchanged gives N(w)−N(u)≤N(u−w). Since ∣N(u)−N(w)∣ is one of N(u)−N(w) and N(w)−N(u), the reverse triangle inequality follows, completing clause 1.

L1L2L10
1.3

For every j<n: xj2≤∑k<nxk2, since every single term of a sum of nonnegative terms is at most the sum; taking nonnegative square roots and using ∣xj∣2=xj2 gives ∣xj∣≤∥x∥2.

L4L7L9L10
1.4

For every j<n: ∣xj∣≤∑k<n∣xk∣=∥x∥1, again because a single term is at most the sum.

L4L7L10
1.5

∑k<n∣xk∣≤∑k<n∥x∥∞=ι(n)∥x∥∞, since ∣xk∣≤∥x∥∞ for every k<n and a constant list sums to ι(n) times its value; so ∥x∥1≤ι(n)∥x∥∞.

L4L6L7L11
1.6

Instantiating [L8] at ak:=∣xk∣ and bk:=1 gives ∥x∥1=∣∑k<n∣xk∣⋅1∣≤∑k<n∣xk∣2 ∑k<n1=∥x∥2ι(n).

L4L7L8L10
1.7

The set {N(ek):k<n} is a nonempty finite set of reals because n≥1, so C=max⁡{N(ek):k<n} exists, belongs to the set, satisfies N(ek)≤C for every k<n, and is ≥0 since every value of N is.

L1L5L6
1.8

x=∑i<nxiei, the coordinate list of x with respect to the ordered basis e being i↦x(i)=xi.

L5
2.1

∥x∥∞ is one of the numbers ∣xj∣ with j<n, so step 1.3 gives ∥x∥∞≤∥x∥2.

step 1.3L6L7
2.2

∑k<nxk2=∑k<n∣xk∣ ∣xk∣≤∑k<n∣xk∣ ∥x∥1=∥x∥1∑k<n∣xk∣=∥x∥12, using step 1.4 termwise, monotonicity and scaling; taking nonnegative square roots gives ∥x∥2≤∥x∥1.

step 1.4L4L7L9L10
2.3

Applying step 1.1 to the list i↦xiei and then (N2): N(x)=N(∑i<nxiei)≤∑i<nN(xiei)=∑i<n∣xi∣ N(ei)≤∑i<n∣xi∣ C=C∥x∥1, the last inequality by monotonicity from step 1.7. This is clause 2.

step 1.1step 1.7step 1.8L1L4L7
3.1

Steps 2.1, 2.2, 1.5 and 1.6 are the four inequalities of clause 3; since ι(n)>0 and ι(n)>0, they exhibit positive constants in both directions for each of the three pairs, so the three norms are pairwise equivalent.

step 1.5step 1.6step 2.1step 2.2L11L9
3.2

By step 1.2 applied on Rn, then step 2.3, then step 1.6: ∣N(x)−N(y)∣≤N(x−y)≤C∥x−y∥1≤Cι(n)  ∥x−y∥2.

step 1.2step 1.6step 2.3L4
4.1

Since ∥x−y∥2=d2(x,y) and ∣N(x)−N(y)∣=dR(N(x),N(y)), step 3.2 says exactly that N is Lipschitz with the nonnegative constant Cι(n), hence uniformly continuous and continuous; this is clause 4, and with steps 1.2, 2.3 and 3.1 all four clauses are proved.

step 1.2step 2.3step 3.1step 3.2L7L12∎

Remarks

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

For n≥1 all norms on Rn are equivalent

Statement

Let n∈N with n≥1. Then any two norms on Rn are equivalent (Equivalent norms, and the dictionary with equivalent metrics, A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

More precisely, for every norm N on Rn there are reals c>0 and C′>0 with

c ∥x∥2  ≤  N(x)  ≤  C′ ∥x∥2for every x∈Rn,

and the general statement follows because equivalence of norms is an equivalence relation.

Consequently all the metric notions on Rn are norm independent for n≥1: any two norms give the same open sets, the same convergent sequences with the same limits, the same Cauchy sequences and the same uniformly continuous maps (Equivalent norms, and the dictionary with equivalent metrics).

The hypothesis n≥1 is used twice in the proof and both uses are marked: once so that the constant C of The finite and reverse triangle inequalities for a norm; and for n≥1 every norm N on Rn satisfies N(x)≤C∥x∥1 and is Lipschitz, hence continuous, for d2 exists, and once so that the Euclidean unit sphere is nonempty, which is what the extreme value theorem needs. At n=0 the conclusion is true but vacuous, the zero space carrying exactly one norm (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms), and it is not obtained from the argument below.

Facts & Assumptions

[L1]

For n≥1: C:=max⁡{N(ek):k<n} exists with C≥0, N(x)≤C∥x∥1, ∥x∥1≤ι(n)∥x∥2, and N is continuous as a map (Rn,d2)→(R,dR) (The finite and reverse triangle inequalities for a norm; and for n≥1 every norm N on Rn satisfies N(x)≤C∥x∥1 and is Lipschitz, hence continuous, for d2 clauses 2, 3, 4).

[L2]

Equivalence of norms is an equivalence relation, and cM≤N≤CM with c,C>0 is what it means (Equivalent norms, and the dictionary with equivalent metrics).

[L4]

Extreme value theorem: a continuous real-valued function on a nonempty compact metric space attains a least value (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

[L5]

Continuity characterisations: a map of metric spaces continuous at every point has closed preimages of closed sets (Metric continuity characterisations, with countable choice for the sequential converse, clause (c)).

[L6]
[L7]

The norm axioms (N1) and (N2), and nonnegativity of a norm (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

[L9]

Proof

technique · direct
1.1

The singleton {1}⊆R is closed: if y≠1 then r:=∣y−1∣>0 and the ball B(y,r) omits 1, so the complement of {1} is open.

L6L9
1.2

∥⋅∥2 is itself a norm on Rn, so by [L1] applied to it, ∥⋅∥2:(Rn,d2)→(R,dR) is continuous.

L1L7
1.3

S⊆B(0,2), since x∈S gives d2(x,0)=∥x∥2=1<2; so S is bounded.

L6L7
1.4

e0∈S, because ∥e0∥2=1; this is where n≥1 is used, since for n=0 there is no index 0<n and no such vector. So S≠∅.

L8
1.5

For every a∈S and every real ε>0, a δ>0 witnessing continuity of N at a as a map on Rn also witnesses it for the restriction N∣S on the metric subspace (S,dS), because dS is the restriction of d2 and the condition is quantified over fewer points; so N∣S is continuous.

L1L10
1.6

Put C′:=Cι(n)+1, a real >0. By [L1], N(x)≤C∥x∥1≤Cι(n)∥x∥2≤C′∥x∥2, the last step because ∥x∥2≥0.

L1L7
2.1

S is the preimage of {1} under the continuous ∥⋅∥2, hence closed in Rn.

step 1.1step 1.2L5
3.1

S is a compact subset of (Rn,d2), being closed and bounded.

step 1.3step 2.1L3
4.1

By the extreme value theorem applied to the nonempty compact metric space (S,dS) and the continuous N∣S, there is xmin⁡∈S with N(xmin⁡)≤N(x) for every x∈S; put c:=N(xmin⁡).

step 1.4step 1.5step 3.1L4
5.1

c>0: from xmin⁡∈S we get ∥xmin⁡∥2=1≠0, so xmin⁡≠0 by (N1) for ∥⋅∥2, so N(xmin⁡)≠0 by (N1) for N, and N(xmin⁡)≥0; trichotomy leaves c>0.

step 4.1L7L9
5.2

Let x≠0. Then ∥x∥2>0 by (N1) and nonnegativity, so t:=1/∥x∥2>0 and u:=t x satisfies ∥u∥2=∣t∣ ∥x∥2=1 by (N2); hence u∈S and c≤N(u)=∣t∣ N(x)=N(x)/∥x∥2, that is c ∥x∥2≤N(x).

step 4.1L7L9
6.1

For x=0 both c∥x∥2 and N(x) are 0 by (N1), so c∥x∥2≤N(x) holds for every x∈Rn.

step 5.2L7
7.1

Steps 5.1, 6.1 and 1.6 give c∥x∥2≤N(x)≤C′∥x∥2 with c,C′>0, so every norm N on Rn is equivalent to ∥⋅∥2.

step 5.1step 6.1step 1.6L2
8.1

Given two norms M and N on Rn, each is equivalent to ∥⋅∥2 by step 7.1, so M is equivalent to N by symmetry and transitivity of the relation.

step 7.1L2∎

Remarks

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29Open item page →

For n≥1 a sequence in Rn converges iff each coordinate sequence converges, is Cauchy iff each coordinate sequence is Cauchy, and Rn is complete in every norm

Statement

Let n∈N with n≥1, let Rn carry the Euclidean metric d2 of Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it, and let (x(j))j∈N be a sequence in Rn (Convergence of a sequence in a metric space: xk→x iff d(xk,x)→0 in R). For k<n write (xk(j))j∈N for the k-th coordinate sequence, a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences). Then:

  1. Convergence is componentwise. For x∈Rn, x(j)→x in (Rn,d2) if and only if xk(j)→xk in R for every k<n (Limits and Cauchy sequences of reals).
  2. Cauchyness is componentwise. (x(j)) is Cauchy in (Rn,d2) (Cauchy sequence in a metric space) if and only if every coordinate sequence is Cauchy in R.
  3. Completeness in every norm. For every norm N on Rn (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms) the metric space (Rn,dN) is complete (Complete metric space: every Cauchy sequence converges in the space).

Clause 3 is obtained by citation and is not reproved here. R and Rn for n≥1 with the Euclidean metric are complete, componentwise from the Cauchy criterion in R clause 2 states that (Rn,d2) is complete, for n≥1 only, and this theorem carries that hypothesis forward without weakening it; what is added is the passage from d2 to an arbitrary norm, through For n≥1 all norms on Rn are equivalent and the dictionary of Equivalent norms, and the dictionary with equivalent metrics.

Facts & Assumptions

Given: A natural n≥1; the space Rn with the norms of The p-norms ∥x∥p for rational p≥1, and ∥x∥∞ and the metric d2; a sequence (x(j)) in Rn; a point x∈Rn; a norm N on Rn; and a rational ε>0.

[L1]

The comparison chain for n≥1 (The finite and reverse triangle inequalities for a norm; and for n≥1 every norm N on Rn satisfies N(x)≤C∥x∥1 and is Lipschitz, hence continuous, for d2 clause 3): ∥y∥∞≤∥y∥2≤∥y∥1≤ι(n)∥y∥∞ for every y∈Rn, where ∥y∥∞=max⁡{∣yk∣:k<n} (The p-norms ∥x∥p for rational p≥1, and ∥x∥∞, Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L5]

All norms on Rn are equivalent for n≥1 (For n≥1 all norms on Rn are equivalent), and equivalent norms have the same convergent sequences with the same limits and the same Cauchy sequences (Equivalent norms, and the dictionary with equivalent metrics).

[L6]

Limits in a metric space are unique, and every convergent sequence is Cauchy (A sequence in a metric space has at most one limit, Every convergent sequence in a metric space is Cauchy).

[L8]

A nonempty finite set of naturals has a greatest element, and every nonempty set of naturals has a least element (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set, The well-ordering principle).

Proof

technique · direct
1.1

For every y∈Rn and every k<n: ∣yk∣≤∥y∥∞≤∥y∥2, the first inequality because ∥y∥∞ bounds the set it is the maximum of.

L1
1.2

For every y∈Rn: ∥y∥2≤ι(n)∥y∥∞, and ∥y∥∞=∣yk0∣ for some k0<n.

L1
1.3

Conversely suppose xk(j)→xk for every k<n. Given a rational ε>0, the real ε/ι(n) is positive, so for each k<n the set of indices K such that ∣xk(j)−xk∣<ε/ι(n) for all j≥K is a nonempty set of naturals; let Kk be its least element, a determination rather than a selection, and put K:=max⁡{K0,…,Kn−1}, a maximum of a nonempty finite set of naturals.

L3L7L8
1.4

(Rn,d2) is complete, by citation and for n≥1 only.

L4
1.5

Let N be any norm on Rn. By [L5], N and ∥⋅∥2 are equivalent, so dN and d2 have the same Cauchy sequences and the same convergent sequences with the same limits.

L5
2.1

For all u,v∈Rn and k<n: ∣uk−vk∣≤d2(u,v)≤ι(n)max⁡{∣uk−vk∣:k<n}, by steps 1.1 and 1.2 applied to y:=u−v.

step 1.1step 1.2L2
2.2

Hence a Cauchy sequence in (Rn,dN) is Cauchy in (Rn,d2), converges there by step 1.4, and therefore converges in (Rn,dN) to the same point; so (Rn,dN) is complete, which is clause 3.

step 1.4step 1.5L5L6
3.1

Suppose x(j)→x in (Rn,d2) and fix k<n. Given a rational ε>0, take K with d2(x(j),x)<ε for j≥K; then ∣xk(j)−xk∣≤d2(x(j),x)<ε for j≥K, so xk(j)→xk.

step 2.1L3
3.2

For j≥K and every k<n we have ∣xk(j)−xk∣<ε/ι(n); the maximum of these n numbers is one of them, so max⁡{∣xk(j)−xk∣:k<n}<ε/ι(n) and hence d2(x(j),x)<ι(n)⋅ε/ι(n)=ε by step 2.1. Therefore x(j)→x.

step 2.1step 1.3L1L7
3.3

The same two estimates prove clause 2 with x replaced by x(l) throughout: if d2(x(j),x(l))<ε for j,l≥K then ∣xk(j)−xk(l)∣<ε for j,l≥K and every k<n; and conversely, choosing for each k<n the least Kk beyond which ∣xk(j)−xk(l)∣<ε/ι(n) for j,l≥Kk and taking K:=max⁡{K0,…,Kn−1} gives d2(x(j),x(l))<ε for j,l≥K.

step 2.1L3L7L8
4.1

Steps 3.1 and 3.2 are the two directions of clause 1.

step 3.1step 3.2
5.1

Clauses 1, 2 and 3 are steps 4.1, 3.3 and 2.2.

step 4.1step 3.3step 2.2∎

Remarks

CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

For n≥1 every bounded sequence in Rn has a convergent subsequence

Statement

Let n∈N with n≥1 and let (x(j))j∈N be a sequence in Rn whose range { x(j):j∈N } is a bounded subset of (Rn,d2) (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it). Then there are a strictly increasing i:N→N (Sequences of reals: bounded, eventually, frequently, tails, subsequences, A strictly increasing index map satisfies nk≥k) and a point p∈Rn with

x(ij)⟶pin (Rn,d2)

(Convergence of a sequence in a metric space: xk→x iff d(xk,x)→0 in R). By For n≥1 all norms on Rn are equivalent the same statement holds with d2 replaced by the metric of any norm on Rn, boundedness and convergence both being unchanged by that replacement (Equivalent norms, and the dictionary with equivalent metrics).

This is assembled from published theorems and is not proved again by bisection. The bisection is in Heine-Borel in Rn: with the Euclidean metric a subset of Rn is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, published at order 120; what is added here is the passage from compactness to sequential compactness and the reading of the conclusion in Rn.

Choice cost: none. Heine-Borel in Rn: with the Euclidean metric a subset of Rn is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line is proved by bisection and uses no choice principle, and "compact implies sequentially compact" is a theorem of ZF (In any metric space compactness implies countable compactness and limit point compactness, and each of countable compactness and limit point compactness implies sequential compactness; every implication here is proved without a choice principle). The five-way equivalence For a metric space, compact, countably compact, limit point compact, sequentially compact, and complete together with totally bounded are all equivalent, given countable choice and dependent choice is not used, precisely because it is stated under countable choice and dependent choice (The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain) and would overcharge this corollary; the arrow-by-arrow account is What each implication between the compactness properties of a metric space costs: which are theorems of ZF, which use countable choice, and which use dependent choice.

Facts & Assumptions

Given: A natural n≥1; a sequence (x(j)) in Rn whose range is bounded in (Rn,d2).

[L1]

Boundedness: a nonempty A⊆X is bounded when A⊆B(q,r) for some q∈X and real r>0 (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Open ball, closed ball and sphere in a metric space).

[L3]

Closed boxes are compact: for reals ak≤bk (k<n) the set Q={ y∈Rn:ak≤yk≤bk for every k<n } is a compact subset of (Rn,d2) (Heine-Borel in Rn: with the Euclidean metric a subset of Rn is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line clause 1, Open cover, subcover, compact metric space, and compact subset of a metric space).

[L5]

A compact subset A of X is one for which the metric subspace (A,dA) is a compact metric space, dA being the restriction of d (Open cover, subcover, compact metric space, and compact subset of a metric space, Isometry, isometric embedding, and the subspace metric on a subset).

Proof

technique · direct
1.1

The range of the sequence is nonempty and bounded, so there are q∈Rn and a real r>0 with d2(x(j),q)<r for every j∈N.

L1
2.1

Put M:=r+∥q∥2, a real with M>0. By the triangle inequality for the norm, ∥x(j)∥2≤∥x(j)−q∥2+∥q∥2=d2(x(j),q)+∥q∥2<M for every j.

step 1.1L2
3.1

For every j and every k<n: ∣xk(j)∣≤∥x(j)∥2<M, hence −M≤xk(j)≤M.

step 2.1L2
4.1

Let Q:={ y∈Rn:−M≤yk≤M for every k<n }. Since −M≤M, Q is a compact subset of (Rn,d2), and by step 3.1 every term x(j) lies in Q.

step 3.1L3
5.1

By [L5] the metric subspace (Q,dQ) is a compact metric space, and by [L4] it is sequentially compact.

step 4.1L4L5
6.1

(x(j)) is a sequence in Q, so there are a strictly increasing i:N→N and p∈Q with x(ij)→p in (Q,dQ).

step 4.1step 5.1L4L6
7.1

Since dQ is the restriction of d2 to Q×Q, the reals dQ(x(ij),p) and d2(x(ij),p) are equal for every j, so x(ij)→p in (Rn,d2) as well.

step 6.1L5L6
8.1

So the bounded sequence (x(j)) has a subsequence converging in (Rn,d2), which is the claim.

step 6.1step 7.1∎

Remarks

DefinitionDefinition: AI-adaptedProof: Not applicableverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

Vector-valued functions f:A→Rm, their limits and continuity, with the dictionary to the metric notions

Definition

Throughout, m∈N with m≥1, and Rm carries the Euclidean norm ∥⋅∥2 of The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn and The p-norms ∥x∥p for rational p≥1, and ∥x∥∞, whose induced metric is the published d2 (Each ∥⋅∥p is a norm on Rn, and the induced metrics are exactly d1, d2 and d∞ of the published metric-spaces page, Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it). A function into Rm is called vector-valued.

Continuity

Let (X,dX) be a metric space (Metric space: d(x,y)=0 iff x=y, symmetry, and the triangle inequality; pseudometric and ultrametric), let A⊆X carry the restricted metric dA (Isometry, isometric embedding, and the subspace metric on a subset), let f:A→Rm and let a∈A. Then f is continuous at a when

(∀ε>0) (∃δ>0) (∀x∈A) [ dX(x,a)<δ ⟹ ∥f(x)−f(a)∥2<ε ],

with ε,δ ranging over the positive reals, and continuous on A when it is continuous at every point of A.

This is not a new notion, and that is the point of writing it down. Since ∥f(x)−f(a)∥2=d2(f(x),f(a)) and dA is the restriction of dX, the displayed condition is verbatim the condition of Continuity of a map between metric spaces, at a point and globally, in the ε-δ form for the map of metric spaces f:(A,dA)→(Rm,d2). So every theorem about continuous maps of metric spaces applies to vector-valued functions with no translation, and this library has exactly one notion of continuity here. The same move was made once before, between the R-native and the metric notions, in Dictionary: for A⊆R with the metric d(x,y)=∣x−y∣, continuity and uniform continuity of f:A→R agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of R is compact in the open-cover sense of R exactly when it is a compact metric subspace; this item is that move one dimension up in the codomain.

The two cases used below are X=R with dR(s,t)=∣s−t∣ (The absolute value makes R a metric space: d(x,y)=∣x−y∣ is a metric, its open balls are the intervals (x−r,x+r), and it is unbounded) and X=Rn with d2, for n≥1.

Limits, for a real domain

Let A⊆R, let f:A→Rm, let c be a limit point of A (Limit point, isolated point, adherent point, derived set, and dense subset of R) and let L∈Rm. We say f(x) tends to L as x tends to c, and write lim⁡x→cf(x)=L, when

(∀ε>0) (∃δ>0) (∀x∈A) [ 0<∣x−c∣<δ ⟹ ∥f(x)−L∥2<ε ].

This is the condition of The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A with the absolute value in the codomain replaced by ∥⋅∥2; as there, the puncture 0<∣x−c∣ is what makes c a point the function need not be defined at, and the hypothesis that c is a limit point of A is what stops the condition from being satisfied vacuously.

The notation denotes: at most one L satisfies the condition. Suppose L and L′ both do and L≠L′. Then ε:=∥L−L′∥2/2>0 by (N1) for ∥⋅∥2 (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms). Take δ and δ′ for this ε and put η:=min⁡{δ,δ′}>0. Since c is a limit point of A there is x∈A with 0<∣x−c∣<η (Limit point, isolated point, adherent point, derived set, and dense subset of R), and then

∥L−L′∥2  ≤  ∥L−f(x)∥2+∥f(x)−L′∥2  <  ε+ε  =  ∥L−L′∥2

by (N3) and (N2), which trichotomy forbids. So L=L′.

Components

For i<m define the i-th coordinate projection πi:Rm→R by πi(y):=yi=y(i), and for f:A→Rm the i-th component fi:=πi∘f, a real-valued function on A.

Each πi is 1-Lipschitz (Lipschitz map, α-Hölder map for rational 0<α≤1, and contraction): for y,z∈Rm,

∣πi(y)−πi(z)∣  =  ∣yi−zi∣  ≤  ∥y−z∥2  =  d2(y,z),

the middle inequality being ∣wi∣≤∥w∥2 at w:=y−z (The finite and reverse triangle inequalities for a norm; and for n≥1 every norm N on Rn satisfies N(x)≤C∥x∥1 and is Lipschitz, hence continuous, for d2 clause 3, or directly because wi2 is one term of the sum ∑k<mwk2). Written in coordinates, f(x) is the vector whose i-th coordinate is fi(x), and f(x)=∑i<mfi(x) ei in the standard basis (The standard list e:n→Fn with ei(i)=1F and ei(j)=0F for j≠i is an ordered basis of Fn; hence dim⁡FFn=n, and F0 is the zero space with basis ∅ and dimension 0).

Remarks

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions

Statement

Let m∈N with m≥1, with vector-valued functions, their components fi=πi∘f, their limits and their continuity as in Vector-valued functions f:A→Rm, their limits and continuity, with the dictionary to the metric notions.

  1. Continuity is componentwise. Let (X,dX) be a metric space, A⊆X, f:A→Rm and a∈A. Then f is continuous at a if and only if every component fi:A→R (i<m) is continuous at a.
  2. Limits are componentwise. Let A⊆R, let c be a limit point of A (Limit point, isolated point, adherent point, derived set, and dense subset of R), let f:A→Rm and let L∈Rm. Then lim⁡x→cf(x)=L if and only if lim⁡x→cfi(x)=Li for every i<m (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A).
  3. Algebra. Let (X,dX), A, a be as in clause 1, let f,g:A→Rm be continuous at a and let λ∈R. Then f+g and λf (defined pointwise) are continuous at a; the real-valued function x↦⟨f(x),g(x)⟩ is continuous at a (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn); and for every norm N on Rm the real-valued function x↦N(f(x)) is continuous at a (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

Where m≥1 is spent. The "if" direction of clauses 1 and 2 divides ε by ι(m), which requires ι(m)≠0; and clause 3's last part quotes a bound available only for m≥1. The "only if" directions hold for every m but say nothing at m=0, there being no index i<0.

Facts & Assumptions

Given: A natural m≥1; a metric space (X,dX), a subset A⊆X, a point a∈A and functions f,g:A→Rm; a real λ; and a real ε>0.

[L2]

The comparison ∥y∥2≤∥y∥1=∑i<m∣yi∣, and N(y)≤C∥y∥1≤Cι(m)∥y∥2 with C:=max⁡{N(ei):i<m}≥0, together with ∣N(y)−N(z)∣≤N(y−z), all for m≥1 (The finite and reverse triangle inequalities for a norm; and for n≥1 every norm N on Rn satisfies N(x)≤C∥x∥1 and is Lipschitz, hence continuous, for d2 clauses 1, 2, 3, The p-norms ∥x∥p for rational p≥1, and ∥x∥∞, The standard list e:n→Fn with ei(i)=1F and ei(j)=0F for j≠i is an ordered basis of Fn; hence dim⁡FFn=n, and F0 is the zero space with basis ∅ and dimension 0).

[L5]

Laws of finite sums (Laws of finite sums and finite products, Finite sums and finite products, by recursion): additivity, scaling, monotonicity, ∑i<mμ=ι(m)μ, a sum of nonnegative terms is nonnegative, and each single term is at most such a sum.

[L7]

A nonempty finite set of reals has a minimum (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set); and a family of nonempty sets indexed by a natural number has a choice function, this being a theorem of ZF (Every natural-number-indexed list of nonempty sets has a choice function on its family of values), which is what licenses picking one δi for each i<m.

[L8]

Absolute value (Basic properties of the absolute value): ∣t∣≥0, ∣st∣=∣s∣∣t∣, and ∣s+t∣≤∣s∣+∣t∣.

Proof

technique · direct
1.1

For every y∈Rm: ∣yi∣≤∥y∥2 for each i<m, and ∥y∥2≤∑i<m∣yi∣.

L1L2
1.2

If ai<bi for every i<m and m≥1, then ∑i<mai<∑i<mbi: the list i↦bi−ai has positive terms, so its sum is at least its term at index 0, hence positive, and additivity gives the strict inequality.

L5L6
1.3

For each i<m the set of positive reals δ witnessing continuity of fi at a for a given tolerance is nonempty whenever fi is continuous at a, so a choice function on the family indexed by m produces δ0,…,δm−1 simultaneously, with no choice principle used.

L7
1.4

For f+g: given ε>0, pick δ1,δ2>0 for the tolerance ε/ι(2) at f and at g and put δ:=min⁡{δ1,δ2}; then for dX(x,a)<δ, ∥(f+g)(x)−(f+g)(a)∥2=∥(f(x)−f(a))+(g(x)−g(a))∥2≤∥f(x)−f(a)∥2+∥g(x)−g(a)∥2<ε.

L1L3L6L7
1.5

For λf: if λ=0 then λf is constant and every δ serves; otherwise ∣λ∣>0, and a δ for the tolerance ε/∣λ∣ at f gives ∥λf(x)−λf(a)∥2=∣λ∣ ∥f(x)−f(a)∥2<ε.

L1L3L6L8
1.6

For N∘f: by [L2], ∣N(f(x))−N(f(a))∣≤N(f(x)−f(a))≤Cι(m) ∥f(x)−f(a)∥2; so a δ for the tolerance ε/(Cι(m)+1) at f serves for N∘f.

L1L2L6
1.7

For ⟨f,g⟩: first take δ0>0 with ∥g(x)−g(a)∥2<1 for dX(x,a)<δ0, so that ∥g(x)∥2≤∥g(x)−g(a)∥2+∥g(a)∥2<B:=∥g(a)∥2+1 there.

L1L3
2.1

Suppose f is continuous at a and fix i<m. Given ε>0, take δ from the definition; for x∈A with dX(x,a)<δ, step 1.1 gives ∣fi(x)−fi(a)∣≤∥f(x)−f(a)∥2<ε. So fi is continuous at a.

step 1.1L1
2.2

Conversely suppose every fi is continuous at a. Given ε>0, the real ε/ι(m) is positive; by step 1.3 choose δi>0 for each i<m with ∣fi(x)−fi(a)∣<ε/ι(m) whenever x∈A and dX(x,a)<δi, and put δ:=min⁡{δ0,…,δm−1}>0.

step 1.3L1L6L7
2.3

By bilinearity, ⟨f(x),g(x)⟩−⟨f(a),g(a)⟩=⟨f(x)−f(a), g(x)⟩+⟨f(a), g(x)−g(a)⟩, so Cauchy-Schwarz and step 1.7 give ∣⟨f(x),g(x)⟩−⟨f(a),g(a)⟩∣≤B ∥f(x)−f(a)∥2+∥f(a)∥2 ∥g(x)−g(a)∥2 for every x∈A with dX(x,a)<δ0.

step 1.7L4L8
3.1

For x∈A with dX(x,a)<δ: each ∣fi(x)−fi(a)∣<ε/ι(m), so by steps 1.1 and 1.2, ∥f(x)−f(a)∥2≤∑i<m∣fi(x)−fi(a)∣<∑i<mε/ι(m)=ε. Hence f is continuous at a, and clause 1 is proved.

step 1.1step 1.2step 2.2L5L6
3.2

Clause 2 is the same two estimates with f(a) replaced by L, fi(a) by Li, and the condition dX(x,a)<δ by 0<∣x−c∣<δ: step 1.1 gives ∣fi(x)−Li∣≤∥f(x)−L∥2 for the forward direction, and steps 1.1, 1.2 give ∥f(x)−L∥2≤∑i<m∣fi(x)−Li∣<ε for the converse, with δ the minimum of m radii obtained as in step 2.2.

step 1.1step 1.2step 1.3L1L5L6L7
3.3

Put P:=B+∥f(a)∥2+1>0 and take δ≤δ0 positive with both ∥f(x)−f(a)∥2<ε/P and ∥g(x)−g(a)∥2<ε/P for dX(x,a)<δ; then step 2.3 bounds the difference by (B+∥f(a)∥2)ε/P<ε, so ⟨f,g⟩ is continuous at a.

step 1.7step 2.3L1L6L7
4.1

Steps 1.4, 1.5, 1.6 and 3.3 are clause 3, and with steps 3.1 and 3.2 all three clauses are proved.

step 3.1step 3.2step 1.4step 1.5step 1.6step 3.3∎

Remarks

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)verified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral

Definition

Throughout, m∈N with m≥1, and vector-valued functions, their components and their limits are as in Vector-valued functions f:A→Rm, their limits and continuity, with the dictionary to the metric notions.

The derivative

Let A⊆R, let f:A→Rm and let c∈A be a limit point of A (Limit point, isolated point, adherent point, derived set, and dense subset of R). The difference quotient of f at c is the vector-valued function

qf,c:A∖{c}→Rm,qf,c(x)  :=  1x−c (f(x)−f(c)),

the scalar multiple being that of the vector space Rm (The vector space FX of all functions X→F with pointwise operations, and Fn as the case X=n={0,1,…,n−1}); the division is legitimate because x≠c gives x−c≠0. As in The derivative f′(c)=lim⁡x→cf(x)−f(c)x−c of f:A→R at a point c∈A that is a limit point of A, and differentiability on a set, c is a limit point of A∖{c} as well, since a punctured neighbourhood of c omits c.

f is differentiable at c when lim⁡x→cqf,c(x) exists in Rm, and then the derivative is

f′(c)  :=  lim⁡x→cqf,c(x)  ∈  Rm.

The notation denotes a single vector. At most one L∈Rm satisfies the limit condition, as proved in Vector-valued functions f:A→Rm, their limits and continuity, with the dictionary to the metric notions; this is the vector-valued form of the obligation At a limit point of the domain a function has at most one limit discharges for real-valued functions and A sequence in a metric space has at most one limit for sequences.

The intrinsic form is the definition; the componentwise form is a theorem. For i<m the i-th component of qf,c(x) is (fi(x)−fi(c))/(x−c), which is the real difference quotient of fi at c (The derivative f′(c)=lim⁡x→cf(x)−f(c)x−c of f:A→R at a point c∈A that is a limit point of A, and differentiability on a set). So by A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions clause 2:

f is differentiable at c if and only if every fi is differentiable at c, and then f′(c)i=fi′(c) for every i<m.

Nothing below reverses this order of presentation: the intrinsic limit is what is defined, and the coordinates are read off it.

Algebra of derivatives. If f,g:A→Rm are differentiable at c and λ∈R, then f+g and λf are differentiable at c with (f+g)′(c)=f′(c)+g′(c) and (λf)′(c)=λf′(c): read componentwise through the displayed equivalence, these are clauses 1 and 2 of the published Sums, scalar multiples, products and quotients: (f+g)′(c)=f′(c)+g′(c), (αf)′(c)=αf′(c), (fg)′(c)=f′(c)g(c)+f(c)g′(c), and (f/g)′(c)=(f′(c)g(c)−f(c)g′(c))/g(c)2 when g(c)≠0.

The integral

Let a,b∈R with a<b and let f:[a,b]→Rm (Intervals of R: the nine order-convex forms, nondegeneracy, and length). f is integrable on [a,b] when every component fi:[a,b]→R is bounded (Lower bound, bounded below, bounded set) and Darboux integrable in the sense of The lower and upper Darboux integrals of a bounded f on [a,b] as sup⁡PL(f,P) and inf⁡PU(f,P), Darboux integrability as their equality, and the notation ∫abf, and then

∫abf  :=  the function m→R sending i↦∫abfi.

That really is an element of Rm. In this library Rm is the set of functions m→R (The vector space FX of all functions X→F with pointwise operations, and Fn as the case X=n={0,1,…,n−1}), not a set of tuples, so the displayed assignment is literally an element of it; each value ∫abfi is a single real by The lower and upper Darboux integrals of a bounded f on [a,b] as sup⁡PL(f,P) and inf⁡PU(f,P), Darboux integrability as their equality, and the notation ∫abf. In the standard basis (The standard list e:n→Fn with ei(i)=1F and ei(j)=0F for j≠i is an ordered basis of Fn; hence dim⁡FFn=n, and F0 is the zero space with basis ∅ and dimension 0) the same object is ∫abf=∑i<m(∫abfi)ei.

Oriented limits. Following The integral with oriented limits: ∫aaf:=0 and ∫baf:=−∫abf componentwise, set

∫aaf  :=  0∈Rm,∫baf  :=  −∫abf(a<b),

so that ∫uvf=−∫vuf for all u,v in an interval on which f is integrable. The clauses do not overlap with the case a<b, so nothing has to be checked for consistency, exactly as in The integral with oriented limits: ∫aaf:=0 and ∫baf:=−∫abf.

Linearity. If f,g:[a,b]→Rm are integrable and λ,μ∈R then λf+μg is integrable with

∫ab(λf+μg)  =  λ∫abf+μ∫abg,

since each side has i-th coordinate ∫ab(λfi+μgi) and λ∫abfi+μ∫abgi respectively, and those agree by Integrable functions on [a,b] form a set closed under sums and scalar multiples, and ∫ab(λf+μg)=λ∫abf+μ∫abg.

Restriction and splitting. If f is integrable on [a,b] then it is integrable on every nondegenerate closed subinterval [c,d] with a≤c<d≤b, and for a<c<b, ∫abf=∫acf+∫cbf; both are the componentwise readings of A function integrable on [a,b] is integrable on every closed subinterval and For a<c<b: f is integrable on [a,b] if and only if it is integrable on [a,c] and on [c,b], and then ∫abf=∫acf+∫cbf; with the oriented form for arbitrary a,b,c, applied to each fi and reassembled coordinate by coordinate.

Remarks

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

For a≤b and f:[a,b]→Rm integrable when a<b, ∥∫abf∥2≤∫ab∥f∥2; for a<b, ∥f∥2 is integrable

Statement

Let m∈N with m≥1, let a,b∈R with a≤b and let f:[a,b]→Rm. If a<b, assume that f is integrable (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral). Then:

  1. if a<b, the real-valued function t↦∥f(t)∥2 is integrable on [a,b] (The lower and upper Darboux integrals of a bounded f on [a,b] as sup⁡PL(f,P) and inf⁡PU(f,P), Darboux integrability as their equality, and the notation ∫abf, The p-norms ∥x∥p for rational p≥1, and ∥x∥∞);
  2. ∥∫abf∥2  ≤  ∫ab∥f∥2.

The hypothesis a≤b is not decoration. With the orientation convention of The integral with oriented limits: ∫aaf:=0 and ∫baf:=−∫abf and The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, interchanging the limits changes the sign of the right-hand side but not of the left, so for b<a the correct statement is ∥∫abf∥2≤∣∫ab∥f∥2∣; the displayed inequality as written is false in that case. This is the same trap the scalar inequality of If f,g are integrable on [a,b] then so are ∣f∣, f2, fg, max⁡(f,g) and min⁡(f,g), and ∣∫abf∣≤∫ab∣f∣ carries.

Clause 1 is a genuine obligation and is discharged before the estimate. That each fi is integrable does not by itself say that ∑i<mfi2 is; the square root has to be brought in through If f is integrable on [a,b] with values in [m,M] and φ is continuous on [m,M], then φ∘f is integrable.

Facts & Assumptions

Given: A natural m≥1, reals a≤b, a function f:[a,b]→Rm that is integrable when a<b, with components f0,…,fm−1, and the vector v:=∫abf∈Rm; write g(t):=∑i<mfi(t)2, so that ∥f(t)∥2=g(t) (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn, The p-norms ∥x∥p for rational p≥1, and ∥x∥∞).

[L2]

Linearity of the integral: integrable functions on [a,b] are closed under sums and scalar multiples, and ∫ab(λu+μw)=λ∫abu+μ∫abw (Integrable functions on [a,b] form a set closed under sums and scalar multiples, and ∫ab(λf+μg)=λ∫abf+μ∫abg).

[L3]

Monotonicity of the integral: for a<b and integrable u≤w on [a,b], ∫abu≤∫abw; and an integrable u≥0 has ∫abu≥0 (If f≤g on [a,b] and both are integrable then ∫abf≤∫abg; and m(b−a)≤∫abf≤M(b−a)).

[L6]

Square roots (Square roots exist: a unique a≥0 with (a)2=a; the positives are {x2:x≠0}, Squaring is monotone on the nonnegatives): every c≥0 has a unique c≥0 with (c)2=c, and s↦s2 is strictly increasing on the nonnegatives, hence injective there.

[L8]

Cauchy-Schwarz and the inner product: ⟨u,w⟩=∑i<muiwi is bilinear and symmetric, ∥u∥2=⟨u,u⟩, ∥u∥2≥0, and ∣⟨u,w⟩∣≤∥u∥2∥w∥2 (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn, Cauchy-Schwarz ∣⟨x,y⟩∣≤∥x∥2∥y∥2 with its equality case, the triangle inequality for ∥⋅∥2, the parallelogram law and polarisation, A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

[L10]

Order arithmetic: u>0 gives u−1>0, a product of nonnegatives is nonnegative, and t≤∣t∣ (Inverses of positives are positive, and reciprocation reverses order, Basic properties of the absolute value).

Proof

technique · direct
1.1

If a=b then ∫abf=0 and ∫ab∥f∥2=0 by the oriented convention, so clause 2 reads 0≤0 and holds, while clause 1 says nothing in that case; assume a<b from here on.

L1
1.2

Each component fi is bounded and integrable on [a,b], so each fi2 is integrable.

L1L4
1.3

Pointwise, ⟨v,f(t)⟩≤∣⟨v,f(t)⟩∣≤∥v∥2 ∥f(t)∥2 by Cauchy-Schwarz.

L8L10
2.1

By induction on p≤m, every finite sum ∑i<pfi2 is integrable, the empty sum being the constant 0 and each successor step adding one integrable function. Hence g=∑i<mfi2 is integrable.

step 1.2L2L9
2.2

The real-valued function t↦⟨v,f(t)⟩=∑i<mvifi(t) is integrable, being a finite sum of scalar multiples of the integrable fi, and by linearity applied m times ∫ab⟨v,f⟩=∑i<mvi∫abfi=∑i<mvi vi=⟨v,v⟩=∥v∥22.

step 1.2L1L2L8L9
3.1

g(t)≥0 for every t, being a finite sum of squares, and g is bounded above: each ∣fi∣ is bounded by some Bi, so g(t)≤∑i<mBi2=:K. Thus g takes its values in [0,K].

step 2.1L1L9L10
4.1

The map s↦s2 is continuous and injective on the order-convex set [0,K], with image [0,K]; by the continuous inverse theorem its inverse φ:[0,K]→[0,K], φ(u)=u, is continuous on [0,K].

step 3.1L6L7
5.1

∥f(t)∥2=g(t)=φ(g(t)) for every t∈[a,b], so ∥f∥2=φ∘g is integrable on [a,b]; this is clause 1.

step 2.1step 3.1step 4.1L5L8
6.1

Both sides of step 1.3 are integrable on [a,b], so monotonicity and linearity give ∥v∥22=∫ab⟨v,f⟩≤∫ab∥v∥2∥f∥2=∥v∥2∫ab∥f∥2.

step 5.1step 2.2step 1.3L2L3
6.2

If v=0 then ∥v∥2=0, while ∫ab∥f∥2≥0 because ∥f∥2≥0 pointwise and a<b; so clause 2 holds in this case.

step 5.1L3L8
7.1

If v≠0 then ∥v∥2>0, so multiplying the inequality of step 6.1 by the positive 1/∥v∥2 gives ∥v∥2≤∫ab∥f∥2, which is clause 2 in this case.

step 6.1L8L10
8.1

The two cases of steps 6.2 and 7.1 exhaust the possibilities for v, so clause 2 holds; with step 5.1 both clauses are proved.

step 5.1step 6.2step 7.1∎

Remarks

  • The case split at v=0 is mandatory. Step 6.1 delivers only ∥v∥22≤∥v∥2∫ab∥f∥2, and dividing by ∥v∥2 is illegitimate when that number is 0. Many textbook presentations divide without comment; the missing case is genuinely separate, and it is the one where the right-hand side has to be shown nonnegative on its own.

  • Why the inner-product route rather than a componentwise estimate. Bounding each coordinate of ∫abf separately and reassembling gives a constant depending on m; the argument above gives the sharp inequality with no constant, and it uses only bilinearity, Cauchy-Schwarz and monotonicity of the integral. The companion page checks the inequality numerically on an explicit curve and shows it is strict there.

  • Clause 1 is where the hypotheses of If f is integrable on [a,b] with values in [m,M] and φ is continuous on [m,M], then φ∘f is integrable are checked, one by one: g is integrable, its values lie in a closed bounded interval, and the outer function is continuous on that interval. The order of that theorem's hypotheses matters — continuous after integrable — and it is respected here.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

The mean value inequality: if f:[a,b]→Rm is continuous and differentiable on (a,b) with ∥f′∥2≤M, then ∥f(b)−f(a)∥2≤M(b−a)

Statement

Let m∈N with m≥1, let a,b∈R with a<b, and let f:[a,b]→Rm be continuous on [a,b] and differentiable at every point of (a,b) as a function on [a,b] (Vector-valued functions f:A→Rm, their limits and continuity, with the dictionary to the metric notions, The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, Intervals of R: the nine order-convex forms, nondegeneracy, and length). Let M∈R with M≥0 satisfy

∥f′(t)∥2  ≤  Mfor every t∈(a,b).

Then

∥f(b)−f(a)∥2  ≤  M (b−a).

No integrability of f′ is assumed, so the theorem applies to every differentiable f; that is why it is proved from the scalar mean value theorem rather than from For a≤b and f:[a,b]→Rm integrable when a<b, ∥∫abf∥2≤∫ab∥f∥2; for a<b, ∥f∥2 is integrable. If f:[a,b]→Rm is differentiable with integrable f′ then ∫abf′=f(b)−f(a); and a bounded derivative makes f Lipschitz records the comparison between the two routes.

The equality form is not asserted, and for m≥2 it is false. There need be no ξ∈(a,b) with f(b)−f(a)=f′(ξ)(b−a); the companion page carries a differentiable witness on [0,1]. The ξ produced in the proof below depends on the fixed vector u=f(b)−f(a) and is a mean value point of the real function t↦⟨u,f(t)⟩, not of f.

Facts & Assumptions

Given: A natural m≥1, reals a<b, a function f:[a,b]→Rm continuous on [a,b] and differentiable on (a,b), a real M≥0 bounding ∥f′∥2 on (a,b), the vector u:=f(b)−f(a)∈Rm, and the real-valued function φ:[a,b]→R, φ(t):=⟨u,f(t)⟩.

[L1]

The inner product is bilinear and symmetric, ⟨w,w⟩=∥w∥22, and ⟨u,w⟩=∑i<muiwi (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn, The p-norms ∥x∥p for rational p≥1, and ∥x∥∞).

[L6]

Algebra of derivatives: sums and scalar multiples of functions differentiable at a point are differentiable there, with (w+z)′(c)=w′(c)+z′(c) and (αw)′(c)=αw′(c) (Sums, scalar multiples, products and quotients: (f+g)′(c)=f′(c)+g′(c), (αf)′(c)=αf′(c), (fg)′(c)=f′(c)g(c)+f(c)g′(c), and (f/g)′(c)=(f′(c)g(c)−f(c)g′(c))/g(c)2 when g(c)≠0 clauses 1 and 2); and a differentiable function is continuous (A function differentiable at c is continuous at c).

[L7]

The mean value theorem: for ψ continuous on [a,b] with a<b and differentiable on (a,b) there is ξ∈(a,b) with ψ(b)−ψ(a)=ψ′(ξ)(b−a) (The mean value theorem, as the case g(x)=x of Cauchy's: for f continuous on [a,b] with a<b and differentiable on (a,b) there is c∈(a,b) with f(b)−f(a)=f′(c)(b−a)).

[L9]

Order arithmetic: b−a>0; a product of nonnegatives is nonnegative; and u>0 gives u−1>0, so an inequality may be multiplied by a positive real (Inverses of positives are positive, and reciprocation reverses order).

Proof

technique · direct
1.1

Every component fi is continuous on [a,b] in the sense of Continuity of f:A→R at a point of A and on A: the ε-δ condition, its agreement with lim⁡x→cf(x)=f(c) at a limit point, and continuity at an isolated point and differentiable at every point of (a,b), with f′(t)i=fi′(t).

L3L4
1.2

φ(t)=∑i<muifi(t) by the coordinate formula for the inner product.

L1
1.3

φ(b)−φ(a)=⟨u,f(b)⟩−⟨u,f(a)⟩=⟨u,f(b)−f(a)⟩=⟨u,u⟩=∥u∥22, by bilinearity.

L1
1.4

By Cauchy-Schwarz and the bound on ∥f′∥2, ⟨u,f′(ξ)⟩≤∣⟨u,f′(ξ)⟩∣≤∥u∥2∥f′(ξ)∥2≤∥u∥2M.

L2
1.5

If u=0 then ∥u∥2=0 while M(b−a)≥0, so the conclusion holds.

L2L9
2.1

By induction on p≤m, each partial sum t↦∑i<puifi(t) is continuous on [a,b] and differentiable on (a,b) with derivative ∑i<puifi′(t): the empty sum is the constant 0, and each successor step adds one scalar multiple of a function that is continuous and differentiable by step 1.1.

step 1.1L5L6L8
3.1

Hence φ is continuous on [a,b], differentiable at every point of (a,b), and φ′(t)=∑i<muifi′(t)=⟨u,f′(t)⟩ for t∈(a,b).

step 1.2step 2.1L1
4.1

By the mean value theorem applied to φ there is ξ∈(a,b) with φ(b)−φ(a)=φ′(ξ)(b−a).

step 3.1L7
5.1

Combining steps 1.3 and 4.1, ∥u∥22=⟨u,f′(ξ)⟩ (b−a).

step 3.1step 1.3step 4.1
6.1

Since b−a>0, multiplying the inequality of step 1.4 by b−a and using step 5.1 gives ∥u∥22≤∥u∥2 M (b−a).

step 5.1step 1.4L9
7.1

If u≠0 then ∥u∥2>0, and multiplying step 6.1 by the positive real 1/∥u∥2 gives ∥u∥2≤M(b−a).

step 6.1L2L9
8.1

The two cases of steps 1.5 and 7.1 exhaust the possibilities for u=f(b)−f(a), so ∥f(b)−f(a)∥2≤M(b−a).

step 1.5step 7.1∎

Remarks

CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

If f:[a,b]→Rm is differentiable with integrable f′ then ∫abf′=f(b)−f(a); and a bounded derivative makes f Lipschitz

Statement

Let m∈N with m≥1 and let a,b∈R with a<b.

  1. Fundamental theorem, second part, in Rm. Let f:[a,b]→Rm be differentiable at every point of [a,b] as a function on [a,b] (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral), and suppose f′:[a,b]→Rm is integrable. Then ∫abf′  =  f(b)−f(a).
  2. A bounded derivative gives a Lipschitz function. Let f:[a,b]→Rm be continuous on [a,b] and differentiable at every point of (a,b), and let M≥0 satisfy ∥f′(t)∥2≤M for every t∈(a,b). Then ∥f(t)−f(s)∥2  ≤  M ∣t−s∣for all s,t∈[a,b], that is, f is Lipschitz with constant M as a map ([a,b],dR)→(Rm,d2) (Lipschitz map, α-Hölder map for rational 0<α≤1, and contraction, The absolute value makes R a metric space: d(x,y)=∣x−y∣ is a metric, its open balls are the intervals (x−r,x+r), and it is unbounded, Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it).

Facts & Assumptions

Given: A natural m≥1, reals a<b, and a function f:[a,b]→Rm with the hypotheses of the clause under discussion; points s,t∈[a,b].

[L1]

The vector-valued derivative and integral are componentwise: f′(c)i=fi′(c), and f is integrable exactly when every fi is, with (∫abf)i=∫abfi; equality of two elements of Rm is equality of all their coordinates (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions).

[L3]

The mean value inequality on a subinterval (The mean value inequality: if f:[a,b]→Rm is continuous and differentiable on (a,b) with ∥f′∥2≤M, then ∥f(b)−f(a)∥2≤M(b−a)): for s<t, f continuous on [s,t] and differentiable on (s,t) with ∥f′∥2≤M there, ∥f(t)−f(s)∥2≤M(t−s).

[L4]

Restricting the domain of a function preserves a limit and its value, the ε-δ condition then quantifying over fewer points; in particular if f is differentiable at c as a function on [a,b] and c is a limit point of [s,t]⊆[a,b], then the restriction of f to [s,t] is differentiable at c with the same derivative (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A, Vector-valued functions f:A→Rm, their limits and continuity, with the dictionary to the metric notions, Limit point, isolated point, adherent point, derived set, and dense subset of R, Intervals of R: the nine order-convex forms, nondegeneracy, and length).

[L6]

Lipschitz maps: f is Lipschitz with constant L≥0 when dY(f(x),f(x′))≤L dX(x,x′) for all x,x′ (Lipschitz map, α-Hölder map for rational 0<α≤1, and contraction).

Proof

technique · direct
1.1

Under the hypotheses of clause 1, each component fi is differentiable at every point of [a,b] with derivative fi′=(f′)i, and each (f′)i is integrable on [a,b].

L1
1.2

Under the hypotheses of clause 2, if s<t in [a,b] then f restricted to [s,t] is continuous on [s,t] and differentiable at every point of (s,t) with the same derivative, since (s,t)⊆(a,b) and every point of (s,t) is a limit point of [s,t].

L4
2.1

Applying [L2] to G:=fi and g:=(f′)i gives ∫ab(f′)i=fi(b)−fi(a) for every i<m.

step 1.1L2
2.2

Under the hypotheses of clause 2, for s<t in [a,b] the mean value inequality applies on [s,t] and gives ∥f(t)−f(s)∥2≤M(t−s)=M∣t−s∣.

step 1.2L3L5
3.1

The i-th coordinate of ∫abf′ is ∫ab(f′)i and the i-th coordinate of f(b)−f(a) is fi(b)−fi(a); by step 2.1 these agree for every i<m, so the two vectors are equal, which is clause 1.

step 2.1L1
3.2

If s=t then ∥f(t)−f(s)∥2=0=M∣t−s∣; and if t<s then step 2.2 applied with the roles exchanged gives ∥f(s)−f(t)∥2≤M∣s−t∣, and ∥f(t)−f(s)∥2=∥−(f(s)−f(t))∥2=∥f(s)−f(t)∥2 while ∣s−t∣=∣t−s∣.

step 2.2L5
4.1

Steps 2.2 and 3.2 cover all pairs s,t∈[a,b], so ∥f(t)−f(s)∥2≤M∣t−s∣ always; since d2(f(t),f(s))=∥f(t)−f(s)∥2 and dR(t,s)=∣t−s∣, this is exactly the Lipschitz condition with constant M≥0, which is clause 2.

step 2.2step 3.2L5L6∎

Remarks

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-29Open item page →

Series of vectors in Rn, absolute convergence, rearrangement, and the set of rearrangement sums

Definition

Let n∈N with n≥1, so that Rn carries the Euclidean metric d2 (Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it, Each ∥⋅∥p is a norm on Rn, and the induced metrics are exactly d1, d2 and d∞ of the published metric-spaces page). A sequence of vectors is a function x:N→Rn, written (xk) with xk:=x(k); as everywhere in this library N contains 0 and a sequence is indexed from 0 (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Convergence of a sequence in a metric space: xk→x iff d(xk,x)→0 in R).

Partial sums and convergence

The partial sums of (xk) are

sN  :=  ∑k<Nxk  ∈  Rn(N∈N),

the finite sum of the vector space Rn (Linear combination of a finite list, and the span span⁡(S) as the smallest linear subspace containing S), so s0=0 and sN+1=sN+xN. No third notion of finite sum is introduced: by The standard list e:n→Fn with ei(i)=1F and ei(j)=0F for j≠i is an ordered basis of Fn; hence dim⁡FFn=n, and F0 is the zero space with basis ∅ and dimension 0 clause 1 the vector sum is computed pointwise, (sN)(j)=∑k<Nxk(j) for j<n, the right-hand side being the real finite sum of Finite sums and finite products, by recursion.

The series ∑xk converges to s∈Rn when sN→s in (Rn,d2) (Convergence of a sequence in a metric space: xk→x iff d(xk,x)→0 in R), and then s is the sum, written ∑k=0∞xk. The symbol denotes a single vector, because a sequence in a metric space has at most one limit (A sequence in a metric space has at most one limit). The series diverges when (sN) does not converge.

Absolute convergence

∑xk converges absolutely when the real series ∑∥xk∥2 converges (Series, partial sums, convergence and the sum, divergence, and the tail series); since ∥xk∥2≥0 (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms), this is a statement about a series of nonnegative terms, exactly as in Absolutely convergent and conditionally convergent series, and the general starting index.

The choice of norm is immaterial. If N is any norm on Rn then c∥xk∥2≤N(xk)≤C∥xk∥2 for fixed c,C>0 (For n≥1 all norms on Rn are equivalent, Equivalent norms, and the dictionary with equivalent metrics), so ∑N(xk) converges exactly when ∑∥xk∥2 does, both being series of nonnegative terms. The notion defined above therefore depends on Rn and not on the norm chosen to test it.

Rearrangement and the set of rearrangement sums

Let σ:N→N be a bijection (Injection, surjection, bijection). The rearrangement of ∑xk along σ is the series ∑xσ(k) of the sequence k↦xσ(k), verbatim as in Rearrangement of a series along a bijection of N, and unconditional convergence one dimension down. The set of rearrangement sums of (xk) is

S(x)  :=  { s∈Rn  :  some rearrangement of ∑xk converges to s }.

Taking σ to be the identity shows that a convergent ∑xk has its own sum in S(x), so S(x)≠∅ for a convergent series.

Agreement with the one-dimensional theory

R1 is the set of functions 1→R and is not literally R. The map θ:R→R1 sending t to the function with value t at 0 is a bijection; it preserves addition and scalar multiplication, since both are computed pointwise (Vector space over a field, The standard list e:n→Fn with ei(i)=1F and ei(j)=0F for j≠i is an ordered basis of Fn; hence dim⁡FFn=n, and F0 is the zero space with basis ∅ and dimension 0), and d2(θ(s),θ(t))=∣s−t∣, so it is an isometric bijection (Isometry, isometric embedding, and the subspace metric on a subset). Under that identification, and for n=1:

Every comparison on this page between Rn and the published one-dimensional theory goes through this identification, and it is stated each time.

Remarks

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29Open item page →

An absolutely convergent series in Rn converges, and every rearrangement converges to the same sum

Statement

Let n∈N with n≥1 and let (xk) be a sequence in Rn whose series converges absolutely (Series of vectors in Rn, absolute convergence, rearrangement, and the set of rearrangement sums). Then:

  1. ∑xk converges; write s:=∑k=0∞xk.
  2. For every bijection σ:N→N (Injection, surjection, bijection) the rearranged series ∑xσ(k) converges absolutely, with ∑k=0∞xσ(k)=s.
  3. Consequently S(x)={s}: the set of rearrangement sums is a single point.

This is the Rn analogue of the published one-dimensional statements, not a generalisation of their proofs. If ∑∣ak∣ converges then ∑ak converges and Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum are proved on the real line; everything below reduces to them coordinatewise, or to completeness of (Rn,d2).

Facts & Assumptions

Given: A natural n≥1; a sequence (xk) in Rn with ∑∥xk∥2 convergent; the vector partial sums sN=∑k<Nxk and the real partial sums TN=∑k<N∥xk∥2; a bijection σ of N; a rational ε>0.

[L7]

Dirichlet's rearrangement theorem: if ∑ak converges absolutely then for every bijection σ of N the series ∑∣aσ(k)∣ converges with the same sum as ∑∣ak∣, and ∑aσ(k) converges with the same sum as ∑ak (Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum, Absolutely convergent and conditionally convergent series, and the general starting index).

[L8]

Absolute convergence implies convergence for real series, and a convergent series of nonnegative terms is absolutely convergent, its terms being their own absolute values (If ∑∣ak∣ converges then ∑ak converges, Absolutely convergent and conditionally convergent series, and the general starting index).

Proof

technique · direct
1.1

For L≤N: sN−sL=∑k=LN−1xk and TN−TL=∑k=LN−1∥xk∥2, both by splitting, the vector identity being the pointwise reading of the real one.

L1L3
1.2

The real sequence (TN) converges by hypothesis, hence is Cauchy in (R,dR): for every rational ε>0 there is K with ∣TN−TL∣<ε for all N,L≥K.

L4L8
1.3

For every j<n and every k: 0≤∣(xk)j∣≤∥xk∥2.

L2
1.4

Likewise k↦∥xσ(k)∥2 is the rearrangement along σ of k↦∥xk∥2, a convergent series of nonnegative terms and therefore absolutely convergent, so ∑k∥xσ(k)∥2 converges; that is, ∑xσ(k) converges absolutely.

L7L8L1
2.1

Hence ∥sN−sL∥2≤∑k=LN−1∥xk∥2=TN−TL by the finite triangle inequality.

step 1.1L2
2.2

By step 1.3 and the comparison test, the real series ∑k∣(xk)j∣ converges for every j<n; so each coordinate series ∑k(xk)j converges absolutely.

step 1.3L6L8
3.1

By steps 2.1 and 1.2, for N≥L≥K we get d2(sN,sL)=∥sN−sL∥2≤∣TN−TL∣<ε, and the same bound with N and L exchanged; so (sN) is Cauchy in (Rn,d2).

step 2.1step 1.2L4
3.2

Fix a bijection σ. For every j<n the sequence k↦(xσ(k))j is the rearrangement along σ of the sequence k↦(xk)j; by step 2.2 the latter series converges absolutely, so Dirichlet's theorem gives that ∑k(xσ(k))j converges with the same sum as ∑k(xk)j.

step 2.2L7
4.1

Since (Rn,d2) is complete, the Cauchy sequence (sN) converges; that is, ∑xk converges, which is clause 1. Write s for its sum.

step 3.1L4
5.1

By clause 1 applied to the sequence k↦xσ(k), which converges absolutely by step 1.4, the series ∑xσ(k) converges; and by step 3.2 each coordinate of its sum equals the corresponding coordinate of s, so its sum is s. This is clause 2.

step 4.1step 3.2step 1.4L5
6.1

By clause 2 every rearrangement of ∑xk converges to s, and the identity bijection shows s∈S(x); so S(x)={s}, which is clause 3.

step 4.1step 5.1L1∎

Remarks

DefinitionDefinition: AI-adaptedProof: Not applicableverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

The subspace Γ of directions along which a series converges absolutely, and its orthogonal complement Γ⊥

Definition

Let n∈N with n≥1 and let (xk) be a sequence in Rn (Series of vectors in Rn, absolute convergence, rearrangement, and the set of rearrangement sums). Define

Γ  :=  { a∈Rn  :  ∑k∣⟨a,xk⟩∣ converges },Γ⊥  :=  { y∈Rn  :  ⟨a,y⟩=0 for every a∈Γ },

the inner product being the Euclidean one (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn) and the series that of Series, partial sums, convergence and the sum, divergence, and the tail series. Elements of Γ are the summing directions of (xk): those a for which the real series of the projections ⟨a,xk⟩ converges absolutely (Absolutely convergent and conditionally convergent series, and the general starting index). Both sets depend on the sequence (xk); when several are in play the notation is Γ(x) and Γ(x)⊥.

Phrased with the inner product, deliberately. Abstract linear maps are already defined in Linear map between vector spaces over the same field, so a linear functional can be read as a linear map into R. This library does not yet define the dual space or prove that every such functional on Rn is represented by an inner product with a vector. Writing Γ with Euclidean directions avoids presupposing that agreement, and nothing on this page depends on it.

Both are linear subspaces

Γ is a linear subspace of Rn (Linear subspace of a vector space). It is nonempty: ⟨0,xk⟩=0 for every k by bilinearity, and the series with all terms 0 converges. For λ∈R and a,b∈Γ, bilinearity and the absolute value laws give

∣⟨λa+b,xk⟩∣  =  ∣λ⟨a,xk⟩+⟨b,xk⟩∣  ≤  ∣λ∣ ∣⟨a,xk⟩∣+∣⟨b,xk⟩∣

(Basic properties of the absolute value), and the series of the right-hand side converges by Convergent series add and scale termwise clauses 1 and 2, so the left-hand series converges by the comparison test (If 0≤ak≤bk eventually, convergence of ∑bk gives convergence of ∑ak, and divergence of ∑ak gives divergence of ∑bk, the terms being nonnegative). By the one-step subspace test (One-step subspace test: a nonempty W⊆V is a linear subspace if and only if λu+v∈W for all λ∈F and u,v∈W), Γ is a linear subspace.

Γ⊥ is a linear subspace of Rn. It contains 0, and for λ∈R, y,z∈Γ⊥ and a∈Γ, bilinearity gives ⟨a,λy+z⟩=λ⟨a,y⟩+⟨a,z⟩=0; again One-step subspace test: a nonempty W⊆V is a linear subspace if and only if λu+v∈W for all λ∈F and u,v∈W applies. Equivalently Γ⊥ is the intersection of the linear subspaces {y:⟨a,y⟩=0} over a∈Γ, a nonempty family since 0∈Γ, and The intersection of a nonempty family of linear subspaces of V is a linear subspace of V gives the same conclusion.

Γ is everything exactly when the series converges absolutely

If ∑xk converges absolutely then Γ=Rn. For any a, Cauchy-Schwarz gives ∣⟨a,xk⟩∣≤∥a∥2∥xk∥2 (Cauchy-Schwarz ∣⟨x,y⟩∣≤∥x∥2∥y∥2 with its equality case, the triangle inequality for ∥⋅∥2, the parallelogram law and polarisation), and ∑k∥a∥2∥xk∥2 converges by Convergent series add and scale termwise clause 2; the comparison test gives a∈Γ.

Conversely, if Γ=Rn then ∑xk converges absolutely. Each standard basis vector ej lies in Γ, and ⟨ej,xk⟩=(xk)j (The standard list e:n→Fn with ei(i)=1F and ei(j)=0F for j≠i is an ordered basis of Fn; hence dim⁡FFn=n, and F0 is the zero space with basis ∅ and dimension 0, The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn), so each real series ∑k∣(xk)j∣ converges. A finite sum of convergent series converges, by Convergent series add and scale termwise clause 1 and induction on the number of summands (The principle of mathematical induction, Laws of finite sums and finite products, Finite sums and finite products, by recursion), so ∑k∑j<n∣(xk)j∣=∑k∥xk∥1 converges; and ∥xk∥2≤∥xk∥1 (The finite and reverse triangle inequalities for a norm; and for n≥1 every norm N on Rn satisfies N(x)≤C∥x∥1 and is Lipschitz, hence continuous, for d2 clause 3, The p-norms ∥x∥p for rational p≥1, and ∥x∥∞), so ∑k∥xk∥2 converges by the comparison test.

That equivalence is what makes the containment theorem below contain An absolutely convergent series in Rn converges, and every rearrangement converges to the same sum as a special case: absolute convergence gives Γ=Rn, hence Γ⊥={0} (any y∈Γ⊥ satisfies ⟨y,y⟩=0 and so y=0 by positive definiteness), and the affine subspace below collapses to a point.

Affine subspaces

At this point in the reading order the general definition is not yet available, so the Euclidean instance is fixed here; the later Affine subspaces as translates x+U of linear subspaces supplies the general definition. For a linear subspace W⊆Rn and s∈Rn, the affine subspace through s with direction W is the coset

s+W  :=  { s+w  :  w∈W }.

A coset is determined by W together with any one of its points. If p∈s+W, say p=s+w0 with w0∈W, then p+W=s+W: every p+w=s+(w0+w) lies in s+W because W is closed under addition, and every s+w=p+(w−w0) lies in p+W because W is closed under addition and under multiplication by −1 (Linear subspace of a vector space, Vector space over a field). In particular s+W=s′+W if and only if s−s′∈W.

Remarks

  • 0∈Γ always, so Γ is never empty and Γ⊥ is never larger than Rn by accident. At the other extreme, if Γ={0} then Γ⊥=Rn, the condition on y being vacuous apart from a=0.

  • The definition does not presuppose convergence of ∑xk, and neither Γ nor Γ⊥ mentions the sum. Convergence is a hypothesis of the theorems that use them, not of the definition.

  • No orthogonal decomposition is claimed. Nothing here asserts that Rn is the direct sum of Γ and Γ⊥, or that (Γ⊥)⊥=Γ. Those are statements of the theory of inner product spaces and orthogonality, which is planned for a page earlier in the plan order that is not yet built, and no item on this page uses them. What is used is only that Γ⊥ is a linear subspace and that ⟨a,y⟩=0 for a∈Γ, y∈Γ⊥.

  • The name. Γ is the set of directions in which the series is absolutely summable; along a direction outside Γ the projected real series converges conditionally at best, and it is exactly there that rearrangement can move the sum.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29Open item page →

Steinitz's polygonal confinement theorem: finitely many vectors of norm at most 1 summing to 0 can be ordered so that every partial sum has norm at most n

Statement

Let n∈N with n≥1, let m∈N and let v:m→Rn be a finite list of vectors with

∥vi∥2≤1  for every i<m,∑i<mvi=0.

Then there is a bijection π:m→m (Injection, surjection, bijection) such that

∥∑j<kvπ(j)∥2  ≤  ι(n)for every k≤m,

where ι is the canonical natural of R (The canonical natural ι(n)=n⋅1F of a field) and the sums are the finite sums of the vector space Rn (Linear combination of a finite list, and the span span⁡(S) as the smallest linear subspace containing S).

The bound depends only on the dimension, not on m. That is the whole content: the triangle inequality alone gives only ι(k), which grows with the number of vectors used.

Which Steinitz result this is. This is Steinitz's polygonal confinement lemma, the rearrangement lemma of his 1913 paper on conditionally convergent series. It is not the Steinitz exchange lemma of linear algebra, which is published in this library as thm-steinitz-exchange and carries the alias lem-steinitz. The two are unrelated results by the same author, and no item on this page uses the bare alias.

Facts & Assumptions

Given: Naturals n≥1 and m; a list v:m→Rn with ∥vi∥2≤1 for i<m and ∑i<mvi=0. Every finite list below is extended by 0 beyond its range, so that the finite sums of Finite sums and finite products, by recursion apply verbatim; a list into Rn is summed in the vector space Rn (Linear combination of a finite list, and the span span⁡(S) as the smallest linear subspace containing S).

[L2]

Laws of finite sums of reals (Laws of finite sums and finite products, Finite sums and finite products, by recursion): additivity, scaling, splitting ∑i<rbi=∑i<qbi+∑i=qr−1bi for q≤r with ∑i=qr−1bi=∑l<r−qbq+l, monotonicity, ∑j<pλ=ι(p)λ, and the fact that a single term of a sum of nonnegative terms is at most the sum.

[L4]

The induction principle (The principle of mathematical induction) and the well-ordering principle: every nonempty subset of N has a least element (The well-ordering principle).

[L7]

The canonical natural (The canonical natural ι(n)=n⋅1F of a field, Canonical naturals are positive and strictly increasing): ι(0)=0 by the recursion clause, ι(p+q)=ι(p)+ι(q) for p,q≥1 by claim 3 there and trivially when p=0 or q=0, ι is strictly increasing, and ι(p)>0 for p≥1.

[L8]

Order arithmetic: u>0 gives u−1>0; an inequality may be multiplied by a nonnegative real; and trichotomy (Inverses of positives are positive, and reciprocation reverses order).

Proof

technique · constructive
1.1

Deleting one entry from a finite sum. Let b:N→R, let r≥1, let q<r, and let b∧q be the list with bi∧q:=bi for i<q and bi∧q:=bi+1 for q≤i<r−1. Then ∑i<rbi=∑i<r−1bi∧q+bq: splitting the left side at q and again at q+1 gives ∑i<qbi+bq+∑l<r−1−qbq+1+l, and splitting the right side at q gives ∑i<qbi+∑l<r−1−qbq+l+1, and the two agree.

L2
1.2

The easy case m≤n. Take π to be the identity of m, a bijection. For k≤m the finite triangle inequality and ∥vj∥2≤1 give ∥∑j<kvj∥2≤∑j<k∥vj∥2≤∑j<k1=ι(k)≤ι(n), since k≤m≤n and ι is increasing. So the theorem holds in this case, and we assume m>n from here on.

constructL1L2L7
1.3

Stage data. For n≤k≤m call a pair (b,μ) admissible at k when b:k→m is injective, μ:N→R vanishes at every j≥k, satisfies 0≤μj≤1 for j<k, and satisfies ∑j<kμjvb(j)=0 and ∑j<kμj=ι(k−n).

construct
1.4

Stage m is admissible. Take bm:= the identity of m and μjm:=ι(m−n)/ι(m) for j<m, μjm:=0 for j≥m; here ι(m)>0 because m>n≥1, and 0≤ι(m−n)≤ι(m) gives 0≤μjm≤1. Then ∑j<mμjmvj=(ι(m−n)/ι(m))∑j<mvj=0 and ∑j<mμjm=ι(m)⋅ι(m−n)/ι(m)=ι(m−n).

constructL2L7L8
1.5

The estimate for k<n, for an arbitrary ordering. For every bijection ρ:m→m and every k<n, the finite triangle inequality gives ∥∑j<kvρ(j)∥2≤∑j<k∥vρ(j)∥2≤∑j<k1=ι(k)≤ι(n).

L1L2L7
2.1

The reindexing identity. For every k∈N, every r∈N, every injective f:r→k and every c:N→R vanishing at every j<k outside the image of f, one has ∑j<kcj=∑i<rcf(i). This is proved by induction on k, with r, f and c universally quantified. At k=0 the only injective f:r→0 has r=0 and both sums are empty. At k+1, write ∑j<k+1cj=∑j<kcj+ck: if k is not in the image of f then ck=0 and f maps into k, so the inductive hypothesis applies directly; and if k=f(q) for the unique such q<r, then r≥1 and the list g:=f∧q of step 1.1 is an injective map r−1→k off whose image c vanishes on {j:j<k}, so the inductive hypothesis gives ∑j<kcj=∑i<r−1cg(i), while step 1.1 applied to bi:=cf(i) gives ∑i<rcf(i)=∑i<r−1cg(i)+cf(q); adding ck=cf(q) to the first identity yields the claim.

step 1.1L2L4
2.2

The feasible set at k−1 is nonempty. Let (b,μ) be admissible at k with n<k≤m, and let Λ be the set of all μ′:N→R vanishing at every j≥k, with 0≤μj′≤1 for j<k, ∑j<kμj′vb(j)=0 and ∑j<kμj′=ι(k−1−n). The scalar ρ:=ι(k−1−n)/ι(k−n) is defined and lies in [0,1], since ι(k−n)>0 and 0≤ι(k−1−n)≤ι(k−n); and ρμ lies in Λ.

step 1.3L2L7L8
3.1

Both identities hold verbatim for lists with values in Rn, since a vector identity is the conjunction of its n coordinate identities and the coordinates of a vector finite sum are the real finite sums of the coordinates.

step 1.1step 2.1L3
3.2

The minimal number of fractional coordinates. Call μ′∈Λ r-simple when there is an injective f:r→k with μj′∈{0,1} for every j<k outside the image of f. The set R:={ r∈N:some μ′∈Λ is r-simple } contains k, taking f to be the identity of k, so R is a nonempty set of naturals and has a least element r0; fix μ∈Λ and an injective f:r0→k witnessing it.

step 2.2L4
4.1

Two consequences used repeatedly. Taking k=r and f a bijection of k in step 2.1 gives ∑j<kcf(j)=∑j<kcj for every c; and taking c to vanish off the image of an injective f:r→k gives ∑j<kcj=∑i<rcf(i), both in R and in Rn.

step 2.1step 3.1
4.2

Every marked coordinate is strictly fractional. For every i<r0 one has 0<μf(i)<1: otherwise μf(i)∈{0,1}, and then f∧i, an injective map r0−1→k off whose image μ takes values in {0,1}, would witness that μ is (r0−1)-simple, contradicting minimality of r0.

step 1.1step 3.2
4.3

Suppose r0≥n+2, towards a contradiction. Define w:r0→Rn+1 by wi(t):=(vb(f(i)))(t) for t<n and wi(n):=1.

step 3.2
5.1

The list w is linearly dependent: there is λ:r0→R, not identically 0, with ∑i<r0λiwi=0. If w is not injective, say wi1=wi2 with i1≠i2, take λi1:=1, λi2:=−1 and λi:=0 otherwise; the list i↦λiwi then vanishes off {i1,i2} and sums to wi1−wi2=0 by step 4.1. If w is injective, its image is a subset of Rn+1 equinumerous with r0≥n+2, hence not linearly independent by [L6]; so some injective list h:p→im⁡(w) is linearly dependent, giving ν:p→R not identically 0 with ∑l<pνlh(l)=0, and setting λi:=νl when wi=h(l) and λi:=0 otherwise turns that into ∑i<r0λiwi=0 by step 4.1, the list i↦λiwi vanishing off the image of the injective map l↦ the unique i with wi=h(l).

step 4.3L6
5.2

The step length. Let i0 be the least i<r0 with λi≠0, which exists because λ is not identically 0. Define s:r0→R by si:=(1−μf(i))/λi if λi>0, by si:=μf(i)/(−λi) if λi<0, and by si:=si0 if λi=0; every si is a positive real by step 4.2. Put t∗:=min⁡{s0,…,sr0−1}, a minimum over a nonempty finite set of reals, so t∗>0 and t∗=si for some i<r0; choosing that i if λi≠0 and i0 otherwise, there is i∗<r0 with λi∗≠0 and t∗=si∗.

step 4.2L4L5L8
6.1

Reading the coordinates of step 5.1. The coordinate n gives ∑i<r0λi=0, and the coordinates t<n give ∑i<r0λivb(f(i))=0 in Rn.

step 4.3step 5.1L3
6.2

The moved point. Define μ′:N→R by μj′:=μj+t∗λi if j=f(i) for the unique i<r0 with that property, and μj′:=μj otherwise. Then 0≤μj′≤1 for every j<k: outside the image of f nothing changes; at j=f(i) with λi>0 one has μf(i)<μj′≤μf(i)+siλi=1; with λi<0 one has 0=μf(i)+siλi≤μj′<μf(i); and with λi=0 the value is unchanged.

step 4.2step 5.2L8
7.1

The moved point is feasible. The list j↦μj′−μj vanishes at every j<k off the image of f and takes the value t∗λi at f(i), so step 4.1 gives ∑j<k(μj′−μj)=∑i<r0t∗λi=t∗⋅0=0; likewise the Rn-valued list j↦(μj′−μj)vb(j) vanishes off that image and takes the value t∗λivb(f(i)) at f(i), so ∑j<k(μj′−μj)vb(j)=t∗∑i<r0λivb(f(i))=0. Hence ∑j<kμj′=ι(k−1−n) and ∑j<kμj′vb(j)=0, so μ′∈Λ.

step 4.1step 6.1step 6.2L2L3
8.1

The contradiction. By step 5.2, μf(i∗)′=μf(i∗)+t∗λi∗∈{0,1}. So f∧i∗, an injective map r0−1→k, witnesses that μ′ is (r0−1)-simple: off the image of f the value μj′=μj lies in {0,1}, and at f(i∗) it lies in {0,1} as just shown. This contradicts the minimality of r0, so the supposition of step 4.3 is untenable and r0≤n+1.

step 1.1step 3.2step 5.2step 6.2step 7.1
9.1

The support bound. There is j0<k with μj0=0. Suppose instead that μj>0 for every j<k; then off the image of f the value μj lies in {0,1} and is positive, hence equals 1. Put νj:=1−μj for j<k and νj:=0 for j≥k, so ν vanishes at every j<k off the image of f and satisfies 0<νf(i)<1 for i<r0 by step 4.2, while ∑j<kνj=ι(k)−ι(k−1−n)=ι(n+1) by [L7].

step 4.2step 8.1L2L7
10.1

By step 4.1, ∑j<kνj=∑i<r0νf(i). If r0=0 this is the empty sum 0, contradicting ι(n+1)>0. If r0≥1 then every term of ∑i<r0(1−νf(i)) is positive, so that sum is at least its term at index 0 and hence positive, whence ∑i<r0νf(i)=ι(r0)−∑i<r0(1−νf(i))<ι(r0)≤ι(n+1) using step 8.1. Either way ι(n+1)<ι(n+1) or ι(n+1)=0, both impossible; so some μj0 is 0.

step 4.1step 8.1step 9.1L2L7L8
11.1

Descending one stage. With j0 as in step 9.1, put b′:=b∧j0:k−1→m and μ′′:=μ∧j0, extended by 0 beyond k−1. Then b′ is injective with image im⁡(b)∖{b(j0)}, 0≤μj′′≤1 for j<k−1, and by step 1.1 in both its real and its vector form, ∑j<k−1μj′′=∑j<kμj−μj0=ι(k−1−n) and ∑j<k−1μj′′vb′(j)=∑j<kμjvb(j)−μj0vb(j0)=0. So (b′,μ′′) is admissible at k−1.

constructstep 1.1step 3.1step 3.2step 10.1
12.1

Iterating. Starting from the admissible pair of step 1.4 at k=m and applying step 11.1 once for each k from m down to n+1, one obtains admissible pairs (bk,μk) for every k with n≤k≤m, with im⁡(bk−1)⊆im⁡(bk) and im⁡(bk)∖im⁡(bk−1) a single element. This is a recursion of length m−n, each stage determined by the previous one together with finitely many determinations (a least natural, a minimum of a finite set of reals), so no choice principle is involved.

constructstep 1.4step 11.1L4L5
13.1

The ordering. Define π:m→m by π(j):=bn(j) for j<n and, for each k with n<k≤m, π(k−1):= the unique element of im⁡(bk)∖im⁡(bk−1). The images im⁡(bk) increase from im⁡(bn), of size n, to im⁡(bm)=m, gaining exactly one element at each stage, so π is injective with image m, that is a bijection, and for every k with n≤k≤m the set {π(j):j<k} is exactly im⁡(bk).

constructstep 12.1L4L6
14.1

Both enumerations give the same partial sum. Fix k with n≤k≤m and let c:N→Rn be ci:=vi for i∈im⁡(bk) and ci:=0 otherwise. Then c vanishes at every i<m off the image of the injective list j↦π(j) on k, and also off the image of bk, so step 4.1 applied twice gives ∑j<kvπ(j)=∑i<mci=∑j<kvbk(j).

step 4.1step 13.1
15.1

The estimate for n≤k≤m. Since ∑j<kμjkvbk(j)=0, additivity gives ∑j<kvbk(j)=∑j<k(1−μjk)vbk(j); each coefficient 1−μjk is nonnegative, so the finite triangle inequality and ∥vi∥2≤1 give ∥∑j<kvbk(j)∥2≤∑j<k(1−μjk)∥vbk(j)∥2≤∑j<k(1−μjk)=ι(k)−ι(k−n)=ι(n).

step 12.1step 14.1L1L2L7
16.1

By steps 14.1 and 15.1 the bound ∥∑j<kvπ(j)∥2≤ι(n) holds for n≤k≤m, and by step 1.5 it holds for k<n; together with the case m≤n of step 1.2, the required bijection π has been exhibited in every case.

step 1.2step 13.1step 14.1step 15.1step 1.5discharge-construct∎

Remarks

  • The support bound of steps 9.1 and 10.1 is the step most write-ups omit. From r0≤n+1 one gets only that the support of μ has at most (k−1−n)+(n+1)=k elements, which is no information at all. What rules out equality is that the quantities 1−μf(i) would then be strictly positive at each of at most n+1 marked indices while summing to ι(n+1); that is exactly the computation in steps 9.1 and 10.1, and without a coordinate μj0=0 the descending construction does not start.

  • Where the dimension enters, and only there. The single place the number n is used is step 5.1, where n+2 vectors in Rn+1 are linearly dependent. The extra coordinate constantly 1 is what converts the constraint ∑iλi=0 into a linear condition, so that one dependence delivers both identities of step 6.1 at once.

  • No choice principle is used. The construction is a recursion of length m−n; at each stage the objects produced are a least natural number (The well-ordering principle) and a minimum of a nonempty finite set of reals (Every nonempty finite set of reals has a maximum and a minimum), both determined rather than selected, and the pair (μ,f) of step 3.2 is a single selection from a nonempty set at each of finitely many stages.

  • The reindexing identity of step 2.1 is proved here rather than cited. Laws of finite sums and finite products is stated for sums ∑k<nak over an initial segment of N and carries no invariance clause, and no lemma available to this page gives the form step 2.1 needs — an injective f:r→k with the summand vanishing at every j<k off its image. That form is therefore proved here. Step 2.1 contains permutation invariance as the special case r=k with f a bijection.

  • The constant ι(n) is not claimed to be optimal. What is proved is that some ordering keeps every partial sum inside the ball of radius ι(n); on an explicit list of six unit vectors in R2 the companion page exhibits one ordering that meets the bound — with room to spare, so the bound is not attained there — and another that violates it, so the theorem is seen to say something.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29Open item page →

The set of rearrangement sums of a convergent series in Rn is a nonempty subset of the affine subspace s+Γ⊥

Statement

Let n∈N with n≥1, let (xk) be a sequence in Rn whose series converges (Series of vectors in Rn, absolute convergence, rearrangement, and the set of rearrangement sums) and write s:=∑k=0∞xk. Let Γ and Γ⊥ be as in The subspace Γ of directions along which a series converges absolutely, and its orthogonal complement Γ⊥. Then:

  1. Nonemptiness. s∈S(x), so S(x)≠∅.
  2. Containment. S(x)  ⊆  s+Γ⊥, the affine subspace through s with direction Γ⊥ (The subspace Γ of directions along which a series converges absolutely, and its orthogonal complement Γ⊥). Equivalently, t−s∈Γ⊥ for every rearrangement sum t.
  3. The absolutely convergent case. If ∑xk converges absolutely then Γ=Rn, Γ⊥={0}, the affine subspace is the single point {s}, and S(x)={s}.
  4. The one-dimensional conditionally convergent case. Let n=1 and identify R1 with R as in Series of vectors in Rn, absolute convergence, rearrangement, and the set of rearrangement sums. If ∑xk converges conditionally (Absolutely convergent and conditionally convergent series, and the general starting index) then Γ={0}, Γ⊥=R1, and the containment of clause 2 is an equality, S(x)=s+Γ⊥=R1, by the published The Riemann series theorem: a conditionally convergent real series has, for every c∈R, a rearrangement with sum c, and rearrangements diverging to +∞, to −∞, and oscillating with any prescribed lim inf⁡≤lim sup⁡ in R‾.

What this theorem does not say, stated here and repeated in the Remarks. It proves a containment and nothing more. Whether S(x) is all of s+Γ⊥ when n≥2 is not settled anywhere on this page, and no item on this page asserts anything about it in either direction. Clause 4 is the case n=1, where the answer is supplied by a published theorem about the real line; it is not evidence for any statement in higher dimensions.

Facts & Assumptions

Given: A natural n≥1; a sequence (xk) in Rn with ∑xk convergent of sum s; a bijection σ of N; a vector a∈Γ; the partial sums sN=∑k<Nxk and sNσ=∑k<Nxσ(k).

[L2]

Γ and Γ⊥ are linear subspaces; a∈Γ means ∑k∣⟨a,xk⟩∣ converges; Γ=Rn exactly when ∑xk converges absolutely; and s+W denotes the coset of a linear subspace W (The subspace Γ of directions along which a series converges absolutely, and its orthogonal complement Γ⊥, Linear subspace of a vector space).

[L5]

Dirichlet's rearrangement theorem: an absolutely convergent real series has, for every bijection σ of N, a rearrangement converging to the same sum (Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum, Absolutely convergent and conditionally convergent series, and the general starting index).

[L8]

An absolutely convergent series in Rn converges, every rearrangement converges to the same sum, and S(x) is then a single point (An absolutely convergent series in Rn converges, and every rearrangement converges to the same sum).

[L9]

Absolute value and order arithmetic: ∣uv∣=∣u∣∣v∣, ∣u∣≥0, and u>0 gives u−1>0 (Basic properties of the absolute value, Inverses of positives are positive, and reciprocation reverses order).

Proof

technique · direct
1.1

The identity map of N is a bijection and the rearrangement along it is the original series, so s∈S(x) and clause 1 holds.

L1
1.2

For every a∈Rn and every finite list u:p→Rn, ⟨a,∑j<puj⟩=∑j<p⟨a,uj⟩: at p=0 both sides are 0, and the successor step is additivity of the inner product in its second argument.

L3L4
1.3

If uN→u in (Rn,d2) then ⟨a,uN⟩→⟨a,u⟩ in R, since ∣⟨a,uN⟩−⟨a,u⟩∣=∣⟨a,uN−u⟩∣≤∥a∥2 ∥uN−u∥2, so a tolerance ε/(∥a∥2+1) on the right serves for ε on the left.

L3L7L9
1.4

Now let n=1 and suppose ∑xk converges conditionally, so the real series ∑k(xk)0 converges and ∑k∣(xk)0∣ diverges. For a∈R1, ⟨a,xk⟩=a0(xk)0 and ∣⟨a,xk⟩∣=∣a0∣ ∣(xk)0∣; if a0≠0 then convergence of ∑k∣a0∣∣(xk)0∣ would give convergence of ∑k∣(xk)0∣ after multiplying by the positive 1/∣a0∣, which is false, so a∈Γ forces a0=0; and a=0 does lie in Γ. Hence Γ={0}.

L1L2L9
2.1

Let t∈S(x), say sNσ→t for a bijection σ, and let a∈Γ. By steps 1.2 and 1.3, ⟨a,t⟩=lim⁡N⟨a,sNσ⟩=lim⁡N∑k<N⟨a,xσ(k)⟩, so the real series ∑k⟨a,xσ(k)⟩ converges with sum ⟨a,t⟩.

step 1.2step 1.3L1L7
2.2

In the same way ⟨a,s⟩=lim⁡N⟨a,sN⟩=lim⁡N∑k<N⟨a,xk⟩, so ∑k⟨a,xk⟩ converges with sum ⟨a,s⟩.

step 1.2step 1.3L7
2.3

With Γ={0} the condition defining Γ⊥ is ⟨0,y⟩=0, which holds for every y, so Γ⊥=R1 and s+Γ⊥=R1.

step 1.4L2L3
3.1

The real sequence k↦⟨a,xσ(k)⟩ is the rearrangement along σ of the sequence k↦⟨a,xk⟩, and the latter series converges absolutely because a∈Γ; so by Dirichlet's theorem the two series have the same sum.

step 2.1step 2.2L2L5
3.2

By the Riemann series theorem applied to the conditionally convergent real series ∑k(xk)0, every real c is the sum of some rearrangement of it; transporting along the identification of R with R1, every element of R1 lies in S(x). So S(x)=R1=s+Γ⊥, which with steps 1.4 and 2.3 is clause 4.

step 1.4step 2.3L1L6L7
4.1

Combining steps 2.1, 2.2 and 3.1 gives ⟨a,t⟩=⟨a,s⟩, hence ⟨a,t−s⟩=0 by bilinearity.

step 2.1step 2.2step 3.1L3
5.1

Since a∈Γ was arbitrary, t−s∈Γ⊥, that is t∈s+Γ⊥; as t∈S(x) was arbitrary, clause 2 holds.

step 4.1L2
6.1

Suppose ∑xk converges absolutely. Then Γ=Rn, so any y∈Γ⊥ satisfies ⟨y,y⟩=0 and hence y=0; thus Γ⊥={0} and s+Γ⊥={s}. Moreover S(x)={s} by [L8], so clause 3 holds and the containment of clause 2 is an equality in this case.

step 5.1L2L3L8
7.1

Clauses 1, 2, 3 and 4 are steps 1.1, 5.1, 6.1 and 3.2.

step 1.1step 5.1step 6.1step 3.2∎

Remarks

  • This theorem proves containment only, and the reverse inclusion is not proved, assumed, or asserted anywhere on this page. For n≥2 the question whether every point of s+Γ⊥ is a rearrangement sum remains open within this page; The same question in Rd: what the set of rearrangement sums looks like, and why that answer is not reachable at this point in the reading order records that proof boundary. Finite-dimensional orthogonal decomposition and convex-separation results are now published elsewhere in this library, but this page supplies no reverse-inclusion rearrangement construction or convergence estimates. See Conventions of this page, the standing n≥1 hypothesis, and what is taken up elsewhere in the reading order.

  • The title claims exactly clause 2 and clause 1, and no more. A title asserting that S(x) is the affine subspace would assert the reverse inclusion, which is not proved here.

  • Clause 4 is the published one-dimensional dichotomy seen from this page. Over R a convergent series is either absolutely convergent, and then Γ is everything and S is a point (clause 3), or conditionally convergent, and then Γ is {0} and S is the whole line (clause 4). Both extremes are consistent with clause 2, and both are equalities; that is a fact about dimension 1, where a linear subspace of R1 is {0} or everything and there is no room in between.

  • What the containment already rules out. Even without the reverse inclusion, clause 2 forbids a rearrangement sum from leaving the affine subspace. That is enough to refute the naive Rn analogue of the Riemann series theorem, and the companion page does so with an elementary witness, using clause 2 and nothing further.

RemarkRemark: AI-generatedProof: Not applicableverified 2026-09-26 (gpt-6-sol)Open item page →

Conventions of this page, the standing n≥1 hypothesis, and what is taken up elsewhere in the reading order

1. The standing hypothesis $n \ge 1$, and exactly where it comes from

The published Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it defines Rn together with the metrics d1, d2, d∞ only for n≥1, and says why: at n=0 the value d∞(x,y) would be a maximum over the empty index set, which does not exist. Everything downstream of that item inherits the hypothesis, and this page inherits it too. In particular R and Rn for n≥1 with the Euclidean metric are complete, componentwise from the Cauchy criterion in R and Heine-Borel in Rn: with the Euclidean metric a subset of Rn is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line are stated for n≥1 and are never cited here for all n.

The boundary runs between the algebra and the metric, not where a reader would guess. The following items of this page carry no hypothesis on the dimension:

The remaining items all carry n≥1 (or m≥1 for the codomain of a vector-valued function), and each states it in its own Statement: The p-norms ∥x∥p for rational p≥1, and ∥x∥∞ for ∥⋅∥∞; clauses 2 and 3 of Each ∥⋅∥p is a norm on Rn, and the induced metrics are exactly d1, d2 and d∞ of the published metric-spaces page; clauses 2, 3, 4 of The finite and reverse triangle inequalities for a norm; and for n≥1 every norm N on Rn satisfies N(x)≤C∥x∥1 and is Lipschitz, hence continuous, for d2; For n≥1 all norms on Rn are equivalent; For n≥1 a sequence in Rn converges iff each coordinate sequence converges, is Cauchy iff each coordinate sequence is Cauchy, and Rn is complete in every norm; For n≥1 every bounded sequence in Rn has a convergent subsequence; Vector-valued functions f:A→Rm, their limits and continuity, with the dictionary to the metric notions; A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions; The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral; For a≤b and f:[a,b]→Rm integrable when a<b, ∥∫abf∥2≤∫ab∥f∥2; for a<b, ∥f∥2 is integrable; The mean value inequality: if f:[a,b]→Rm is continuous and differentiable on (a,b) with ∥f′∥2≤M, then ∥f(b)−f(a)∥2≤M(b−a); If f:[a,b]→Rm is differentiable with integrable f′ then ∫abf′=f(b)−f(a); and a bounded derivative makes f Lipschitz; Series of vectors in Rn, absolute convergence, rearrangement, and the set of rearrangement sums; An absolutely convergent series in Rn converges, and every rearrangement converges to the same sum; The subspace Γ of directions along which a series converges absolutely, and its orthogonal complement Γ⊥; Steinitz's polygonal confinement theorem: finitely many vectors of norm at most 1 summing to 0 can be ordered so that every partial sum has norm at most n; and The set of rearrangement sums of a convergent series in Rn is a nonempty subset of the affine subspace s+Γ⊥.

Where a statement about n=0 is nevertheless true, it is proved here from scratch rather than imported: see the second remark of For n≥1 a sequence in Rn converges iff each coordinate sequence converges, is Cauchy iff each coordinate sequence is Cauchy, and Rn is complete in every norm for completeness of R0.

2. The exponent of a $p$-norm is rational

Rational powers ar of a positive base supplies ar for a positive base and any rational exponent, together with 0r for rational r>0; real exponents do not exist at this point in the reading order; Why real exponents are deferred on the rational-powers page records why. Consequently The p-norms ∥x∥p for rational p≥1, and ∥x∥∞ defines ∥⋅∥p for rational p≥1 only, and the published Minkowski inequality it rests on is itself stated for rational p. No statement on this page is written with p ranging over a real interval, and the phrase "for p∈[1,∞)" appears nowhere.

3. $\mathbb{R}^{n}$ is a function space

4. What is taken up elsewhere in the reading order

Each item below is a statement about where material sits in this library's reading order, and none of them is a claim about mathematics that this library denies.

5. The open half of the rearrangement question

The set of rearrangement sums of a convergent series in Rn is a nonempty subset of the affine subspace s+Γ⊥ proves that the set S(x) of rearrangement sums of a convergent series in Rn is nonempty and contained in the affine subspace s+Γ⊥, and Steinitz's polygonal confinement theorem: finitely many vectors of norm at most 1 summing to 0 can be ordered so that every partial sum has norm at most n proves Steinitz's polygonal confinement lemma in full. The reverse inclusion is not proved on this page, and this page asserts nothing about it in either direction, for any n≥2. No recorded-not-proved item has been created for it either.

The reverse inclusion still needs an argument beyond the containment proof on this page. Orthogonal decomposition and convex separation are now available elsewhere in the library, but their existence alone does not establish the rearrangement construction. Any discharge here must supply that construction and its convergence estimates explicitly.

The published The same question in Rd: what the set of rearrangement sums looks like, and why that answer is not reachable at this point in the reading order raised this question on the series page and declined to state what the literature answers; this page answers the part it can and continues to decline the rest. What a reader is protected from meanwhile is the wrong guess: the companion page refutes outright the naive Rn analogue of the Riemann series theorem, using the containment half and nothing more.

6. A naming collision worth stating once

Steinitz's polygonal confinement theorem: finitely many vectors of norm at most 1 summing to 0 can be ordered so that every partial sum has norm at most n is Steinitz's polygonal confinement lemma from his 1913 paper on conditionally convergent series. It is not the Steinitz exchange lemma of linear algebra, which is published in this library under the id thm-steinitz-exchange and additionally carries the alias lem-steinitz. The two are different theorems by the same author; the ids do not collide, and no item on this page uses the bare alias.

5 · Examples, counterexamples and false statements

None yet.

Sources