Alphabeta Math
Session-authored (Fable 5 assisted)
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)=n1R is the real number that n names (The canonical natural ι(n)=n1F 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)=uv 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,yx2y2 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 xp for rational p1, and x introduces p for rational p1 and for n1.

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 n1 the metrics induced by 1, 2 and are the published d1, d2 and d of Rn as the set of functions nR, 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 n1 every norm N on Rn satisfies N(x)Cx1 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)Cx1 from the standard basis, the comparison chain xx2x1ι(n)x, and Lipschitz continuity of N for d2. For n1 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 n1 is used twice there and both uses are marked.

Sequences. For n1 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 n1 with the Euclidean metric are complete, componentwise from the Cauchy criterion in R and transporting it along norm equivalence. For n1 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:ARm, their limits and continuity, with the dictionary to the metric notions defines limits and continuity for f:ARm 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 AR with the metric d(x,y)=xy, continuity and uniform continuity of f:AR 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 ab and f:[a,b]Rm integrable when a<b, abf2abf2; for a<b, f2 is integrable proves that f2 is integrable when a<b and that abf2abf2 for ab, 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 f2M, then f(b)f(a)2M(ba) proves f(b)f(a)2M(ba) from the scalar mean value theorem applied to tf(b)f(a),f(t), with no integrability hypothesis; the equality form is false for m2 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:ka,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 #suppk1 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 n2 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 cR, a rearrangement with sum c, and rearrangements diverging to +, to , and oscillating with any prescribed lim inflim sup in R outright, using the containment half and nothing more.

Conventions. Conventions of this page, the standing n1 hypothesis, and what is taken up elsewhere in the reading order is the page ledger. It records where the standing hypothesis n1 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:VR such that for all u,vV and all λR:

  • (N1) Separation. N(v)=0 if and only if v=0V.
  • (N2) Absolute homogeneity. N(λv)=λN(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)=1N(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 vV.

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(uv)(u,vV),

where uv=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,wV and λR:

  • translation invariance, dN(u+w,v+w)=N((u+w)(v+w))=N(uv)=dN(u,v);
  • absolute homogeneity, dN(λu,λv)=N(λ(uv))=λdN(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)=λd(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 nN. A natural number is a von Neumann natural, that is a set, and n={0,1,,n1} (The natural numbers N (von Neumann), On N the order is membership: m<n    mn), so

Rn  =  {x  :  x is a function nR}

is the function space of The vector space FX of all functions XF with pointwise operations, and Fn as the case X=n={0,1,,n1} 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 nR, and d1, d2, d are metrics on it calls Rn.

The Euclidean inner product of x,yRn is the real number

x,y  :=  k<nxkyk,

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

x2  :=  x,x,

which is defined because x,x=k<nxk20 (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 a0 with (a)2=a; the positives are {x2:x0}).

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:nFn with ei(i)=1F and ei(j)=0F for ji is an ordered basis of Fn; hence dimFFn=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 x2=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 nR, and d1, d2, d are metrics on it are defined only for n1, 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 n1 hypothesis, and what is taken up elsewhere in the reading order lists exactly which items inherit n1.

The algebra of the inner product

For all x,y,zRn and λR:

  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 kxkzk and kykzk, so Laws of finite sums and finite products clause 1 applies.
  3. Homogeneity in the first argument. λx,y=λx,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,x0, 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 n1 and x,yRn, xy2=k<n(xkyk)2=d2(x,y), the two sides being the same expression (Rn as the set of functions nR, and d1, d2, d are metrics on it). In particular x2=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,yx2y2 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,yx2y2 with its equality case, the triangle inequality for 2, the parallelogram law and polarisation

Statement

Let nN and let x,yRn, 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    x2y2, 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 nN; the metric it induces is d2 of Rn as the set of functions nR, and d1, d2, d are metrics on it whenever n1.
  3. Parallelogram law. x+y22+xy22  =  2x22+2y22.
  4. Polarisation. x,y  =  14(x+y22xy22), 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,yRn, so that x,y=k<nxkyk and x2=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<nakbkk<nak2k<nbk2.

[L2]

The inner product is symmetric, bilinear and positive definite, x,x=k<nxk20, 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 a0 with (a)2=a; the positives are {x2:x0}): every c0 has a unique s0 with s2=c, written c; hence x20 and x22=x,x (Integer powers am).

[L4]

Squaring is monotone on the nonnegatives: for a,b0, ab if and only if a2b2, 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): t0, t2=t2, and st=st.

Proof

technique · direct
1.1

Instantiating [L1] at ak:=xk and bk:=yk gives x,y2x,xy,y, with equality exactly when some (λ,μ)(0,0) has λxk=μyk for every k<n.

L1L2
1.2

Both x,y and x2y2 are nonnegative, and their squares are x,y2 and x,xy,y.

L3L5
1.3

Expanding by bilinearity and symmetry, x+y,x+y=x,x+x,y+y,x+y,y=x22+2x,y+y22.

L2L3
1.4

The same expansion at xy=x+(1)y gives xy22=x222x,y+y22.

L2L3
1.5

For a scalar λ, λx,λx=λ2x,x=λ2x22, so λx22=(λx2)2.

L2L3L5
1.6

Axiom (N1) holds: x2=0 if and only if x22=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,yx2y2, 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+y22+xy22=2x22+2y22, 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+y22xy22=4x,y, which is clause 4 after dividing by 4.

step 1.3step 1.4algebra
2.4

Both λx2 and λx2 are nonnegative and by step 1.5 have equal squares, so λx2=λx2, which is axiom (N2).

step 1.5L3L4L5
3.1

By step 2.1 the middle term of step 1.3 satisfies 2x,y2x2y2, so x+y22x22+2x2y2+y22=(x2+y2)2.

step 1.3step 2.1L5algebra
4.1

Both x+y2 and x2+y2 are nonnegative, so step 3.1 and monotonicity of squaring give x+y2x2+y2, 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 nN, and for n1 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 xp for rational p1, and x

Definition

Let nN 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 p1

Let pQ with p1. For xRn put

xp  :=  (k<nxkp)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 xkp 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 p1. No statement on this page is written for p ranging over a real interval.

The maximum norm

For n1 and xRn 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 n1 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 nR, 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 p1 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 nR, 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 nN and let pQ with p1, with the norms of The p-norms xp for rational p1, 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 n1, is a norm on Rn.
  3. The dictionary. For n1 and all x,yRn, xy1=d1(x,y),xy2=d2(x,y),xy=d(x,y), where d1, d2, d are the metrics of the published Rn as the set of functions nR, 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 n1 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 n1 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 p1, vectors x,yRn and a real λ; write S(x):=k<nxkp, so that xp=S(x)1/p (The p-norms xp for rational p1, 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,b0 and rationals r,s>0 one has ar0, (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 rar and of aar clause 2): for a rational r>0 and reals 0a<b one has ar<br; hence ab implies arbr, 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 p1 (Minkowski's inequality for finite sums (rational exponent)): (k<nak+bkp)1/p(k<nakp)1/p+(k<nbkp)1/p.

[L5]

Absolute value (Basic properties of the absolute value, Absolute value in an ordered field, The triangle inequality): t0; t=0 exactly when t=0; st=st; s+ts+t; and t2=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 n1 are d1(x,y)=k<nxkyk, d2(x,y)=k<n(xkyk)2 and d(x,y)=max{xkyk:k<n}, and each is a metric (Rn as the set of functions nR, 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 xkp is nonnegative, so S(x)0 and xp=S(x)1/p is defined and nonnegative.

L1L3
1.2

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

L1L2L3L5
1.3

For every k<n, (λx)kp=(λxk)p=λpxkp, so S(λx)=λpS(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+ypxp+yp, 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 xkx for every k<n; in particular x0.

A1L5L6
1.6

Under [A1], xy1=k<nxkyk by the case p=1 of the definition, and that is the written expression for d1(x,y).

L1L9
1.7

Under [A1], xy2=(k<nxkyk2)1/2=k<n(xkyk)2, using t2=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], xy=max{xkyk:k<n} by definition, and that is the written expression for d(x,y).

L9
2.1

xp=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 xk0 and xk0 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=λxkλx, and choosing j<n with xj=x gives (λx)j=λx; so λx belongs to the set and bounds it above, whence λx=λx. This is (N2) for .

step 1.5L5L6L7
2.4

Under [A1]: for every k<n, (x+y)k=xk+ykxk+ykx+y; choosing j<n with (x+y)j=x+y gives x+yx+y, which is (N3) for .

step 1.5L5L6L7
3.1

By steps 2.1 and 1.2, xp=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 λ0 then λ>0, and step 1.3 with the power laws gives λxp=(λpS(x))1/p=(λp)1/pS(x)1/p=λp(1/p)xp=λxp; 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 n1 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

cM(v)    N(v)    CM(v)for every vV.

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 cMNCM and c,C>0 one gets C1NMc1N, dividing by the positive constants (Inverses of positives are positive, and reciprocation reverses order).
  • Transitive: if cMNCM and cNPCN then ccMPCCM, and cc>0, CC>0, a product of positives being positive.

The dictionary with equivalent metrics

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

cdM(u,w)    dN(u,w)    CdM(u,w)for all u,wV,

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 0dN(vk,v)CdM(vk,v), so given a rational ε>0 an index beyond which dM(vk,v)<ε/C serves for dN; the converse uses dMc1dN 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 n1 every norm N on Rn satisfies N(x)Cx1 and is Lipschitz, hence continuous, for d2

Statement

Clause 1 is about an arbitrary norm; clauses 2 to 4 are about Rn with n1.

  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 pN and every list u:pV (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,wV, N(u)N(w)    N(uw).

Now let nN with n1, let Rn carry the norms of The p-norms xp for rational p1, and x and write ι for the canonical natural (The canonical natural ι(n)=n1F 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:nFn with ei(i)=1F and ei(j)=0F for ji is an ordered basis of Fn; hence dimFFn=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 C0 and N(x)    Cx1for every xRn.
  2. The comparison chain. For every xRn, x    x2    x1    ι(n)x,x1    ι(n)  x2. 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<α1, and contraction, Rn as the set of functions nR, and d1, d2, d are metrics on it, The absolute value makes R a metric space: d(x,y)=xy is a metric, its open balls are the intervals (xr,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 n1 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 nR, and d1, d2, d are metrics on it and The p-norms xp for rational p1, 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 n1, the space Rn, a norm N on it, and vectors x,yRn.

[L1]

The norm axioms: N(v)=0 exactly when v=0V; N(λv)=λN(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 xp for rational p1, 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): x1=k<nxk, x2=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<nakbkk<nak2k<nbk2.

[L9]

Square roots and squaring (Square roots exist: a unique a0 with (a)2=a; the positives are {x2:x0}, Squaring is monotone on the nonnegatives): every c0 has a unique c0 with (c)2=c; for a,b0, ab exactly when a2b2.

[L10]

Absolute value (Basic properties of the absolute value): t0, t2=t2, st=st, 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,wV: N(u)=N((uw)+w)N(uw)+N(w), so N(u)N(w)N(uw); and N(wu)=N((1)(uw))=1N(uw)=N(uw), so the same argument with u and w exchanged gives N(w)N(u)N(uw). 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: xj2k<nxk2, since every single term of a sum of nonnegative terms is at most the sum; taking nonnegative square roots and using xj2=xj2 gives xjx2.

L4L7L9L10
1.4

For every j<n: xjk<nxk=x1, again because a single term is at most the sum.

L4L7L10
1.5

k<nxkk<nx=ι(n)x, since xkx for every k<n and a constant list sums to ι(n) times its value; so x1ι(n)x.

L4L6L7L11
1.6

Instantiating [L8] at ak:=xk and bk:=1 gives x1=k<nxk1k<nxk2k<n1=x2ι(n).

L4L7L8L10
1.7

The set {N(ek):k<n} is a nonempty finite set of reals because n1, 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 ix(i)=xi.

L5
2.1

x is one of the numbers xj with j<n, so step 1.3 gives xx2.

step 1.3L6L7
2.2

k<nxk2=k<nxkxkk<nxkx1=x1k<nxk=x12, using step 1.4 termwise, monotonicity and scaling; taking nonnegative square roots gives x2x1.

step 1.4L4L7L9L10
2.3

Applying step 1.1 to the list ixiei and then (N2): N(x)=N(i<nxiei)i<nN(xiei)=i<nxiN(ei)i<nxiC=Cx1, 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(xy)Cxy1Cι(n)  xy2.

step 1.2step 1.6step 2.3L4
4.1

Since xy2=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 n1 all norms on Rn are equivalent

Statement

Let nN with n1. 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

cx2    N(x)    Cx2for every xRn,

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

Consequently all the metric notions on Rn are norm independent for n1: 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 n1 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 n1 every norm N on Rn satisfies N(x)Cx1 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 n1: C:=max{N(ek):k<n} exists with C0, N(x)Cx1, x1ι(n)x2, and N is continuous as a map (Rn,d2)(R,dR) (The finite and reverse triangle inequalities for a norm; and for n1 every norm N on Rn satisfies N(x)Cx1 and is Lipschitz, hence continuous, for d2 clauses 2, 3, 4).

[L2]

Equivalence of norms is an equivalence relation, and cMNCM 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).

[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 y1 then r:=y1>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

SB(0,2), since xS gives d2(x,0)=x2=1<2; so S is bounded.

L6L7
1.4

e0S, because e02=1; this is where n1 is used, since for n=0 there is no index 0<n and no such vector. So S.

L8
1.5

For every aS 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 NS on the metric subspace (S,dS), because dS is the restriction of d2 and the condition is quantified over fewer points; so NS is continuous.

L1L10
1.6

Put C:=Cι(n)+1, a real >0. By [L1], N(x)Cx1Cι(n)x2Cx2, the last step because x20.

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 NS, there is xminS with N(xmin)N(x) for every xS; put c:=N(xmin).

step 1.4step 1.5step 3.1L4
5.1

c>0: from xminS we get xmin2=10, so xmin0 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 x0. Then x2>0 by (N1) and nonnegativity, so t:=1/x2>0 and u:=tx satisfies u2=tx2=1 by (N2); hence uS and cN(u)=tN(x)=N(x)/x2, that is cx2N(x).

step 4.1L7L9
6.1

For x=0 both cx2 and N(x) are 0 by (N1), so cx2N(x) holds for every xRn.

step 5.2L7
7.1

Steps 5.1, 6.1 and 1.6 give cx2N(x)Cx2 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 n1 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 nN with n1, let Rn carry the Euclidean metric d2 of Rn as the set of functions nR, and d1, d2, d are metrics on it, and let (x(j))jN be a sequence in Rn (Convergence of a sequence in a metric space: xkx iff d(xk,x)0 in R). For k<n write (xk(j))jN for the k-th coordinate sequence, a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences). Then:

  1. Convergence is componentwise. For xRn, 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 n1 with the Euclidean metric are complete, componentwise from the Cauchy criterion in R clause 2 states that (Rn,d2) is complete, for n1 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 n1 all norms on Rn are equivalent and the dictionary of Equivalent norms, and the dictionary with equivalent metrics.

Facts & Assumptions

Given: A natural n1; the space Rn with the norms of The p-norms xp for rational p1, and x and the metric d2; a sequence (x(j)) in Rn; a point xRn; a norm N on Rn; and a rational ε>0.

[L1]

The comparison chain for n1 (The finite and reverse triangle inequalities for a norm; and for n1 every norm N on Rn satisfies N(x)Cx1 and is Lipschitz, hence continuous, for d2 clause 3): yy2y1ι(n)y for every yRn, where y=max{yk:k<n} (The p-norms xp for rational p1, 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 n1 (For n1 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 yRn and every k<n: ykyy2, the first inequality because y bounds the set it is the maximum of.

L1
1.2

For every yRn: y2ι(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 jK is a nonempty set of naturals; let Kk be its least element, a determination rather than a selection, and put K:=max{K0,,Kn1}, a maximum of a nonempty finite set of naturals.

L3L7L8
1.4

(Rn,d2) is complete, by citation and for n1 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,vRn and k<n: ukvkd2(u,v)ι(n)max{ukvk:k<n}, by steps 1.1 and 1.2 applied to y:=uv.

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 jK; then xk(j)xkd2(x(j),x)<ε for jK, so xk(j)xk.

step 2.1L3
3.2

For jK 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,lK then xk(j)xk(l)<ε for j,lK 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,lKk and taking K:=max{K0,,Kn1} gives d2(x(j),x(l))<ε for j,lK.

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 n1 every bounded sequence in Rn has a convergent subsequence

Statement

Let nN with n1 and let (x(j))jN be a sequence in Rn whose range {x(j):jN} 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 nR, and d1, d2, d are metrics on it). Then there are a strictly increasing i:NN (Sequences of reals: bounded, eventually, frequently, tails, subsequences, A strictly increasing index map satisfies nkk) and a point pRn with

x(ij)pin (Rn,d2)

(Convergence of a sequence in a metric space: xkx iff d(xk,x)0 in R). By For n1 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 n1; a sequence (x(j)) in Rn whose range is bounded in (Rn,d2).

[L1]

Boundedness: a nonempty AX is bounded when AB(q,r) for some qX 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 akbk (k<n) the set Q={yRn:akykbk 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 qRn and a real r>0 with d2(x(j),q)<r for every jN.

L1
2.1

Put M:=r+q2, a real with M>0. By the triangle inequality for the norm, x(j)2x(j)q2+q2=d2(x(j),q)+q2<M for every j.

step 1.1L2
3.1

For every j and every k<n: xk(j)x(j)2<M, hence Mxk(j)M.

step 2.1L2
4.1

Let Q:={yRn:MykM for every k<n}. Since MM, 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:NN and pQ 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:ARm, their limits and continuity, with the dictionary to the metric notions

Definition

Throughout, mN with m1, and Rm carries the Euclidean norm 2 of The Euclidean inner product x,y=k<nxkyk on Rn and The p-norms xp for rational p1, 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 nR, 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 AX carry the restricted metric dA (Isometry, isometric embedding, and the subspace metric on a subset), let f:ARm and let aA. Then f is continuous at a when

(ε>0) (δ>0) (xA) [ 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 AR with the metric d(x,y)=xy, continuity and uniform continuity of f:AR 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)=st (The absolute value makes R a metric space: d(x,y)=xy is a metric, its open balls are the intervals (xr,x+r), and it is unbounded) and X=Rn with d2, for n1.

Limits, for a real domain

Let AR, let f:ARm, let c be a limit point of A (Limit point, isolated point, adherent point, derived set, and dense subset of R) and let LRm. We say f(x) tends to L as x tends to c, and write limxcf(x)=L, when

(ε>0) (δ>0) (xA) [ 0<xc<δ  f(x)L2<ε ].

This is the condition of The ε-δ limit limxcf(x)=L of f:AR at a limit point c of A with the absolute value in the codomain replaced by 2; as there, the puncture 0<xc 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 LL. Then ε:=LL2/2>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 xA with 0<xc<η (Limit point, isolated point, adherent point, derived set, and dense subset of R), and then

LL2    Lf(x)2+f(x)L2  <  ε+ε  =  LL2

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

Components

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

Each πi is 1-Lipschitz (Lipschitz map, α-Hölder map for rational 0<α1, and contraction): for y,zRm,

πi(y)πi(z)  =  yizi    yz2  =  d2(y,z),

the middle inequality being wiw2 at w:=yz (The finite and reverse triangle inequalities for a norm; and for n1 every norm N on Rn satisfies N(x)Cx1 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:nFn with ei(i)=1F and ei(j)=0F for ji is an ordered basis of Fn; hence dimFFn=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 mN with m1, with vector-valued functions, their components fi=πif, their limits and their continuity as in Vector-valued functions f:ARm, their limits and continuity, with the dictionary to the metric notions.

  1. Continuity is componentwise. Let (X,dX) be a metric space, AX, f:ARm and aA. Then f is continuous at a if and only if every component fi:AR (i<m) is continuous at a.
  2. Limits are componentwise. Let AR, let c be a limit point of A (Limit point, isolated point, adherent point, derived set, and dense subset of R), let f:ARm and let LRm. Then limxcf(x)=L if and only if limxcfi(x)=Li for every i<m (The ε-δ limit limxcf(x)=L of f:AR at a limit point c of A).
  3. Algebra. Let (X,dX), A, a be as in clause 1, let f,g:ARm be continuous at a and let λR. Then f+g and λf (defined pointwise) are continuous at a; the real-valued function xf(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 xN(f(x)) is continuous at a (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

Where m1 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 m1. 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 m1; a metric space (X,dX), a subset AX, a point aA and functions f,g:ARm; a real λ; and a real ε>0.

[L2]

The comparison y2y1=i<myi, and N(y)Cy1Cι(m)y2 with C:=max{N(ei):i<m}0, together with N(y)N(z)N(yz), all for m1 (The finite and reverse triangle inequalities for a norm; and for n1 every norm N on Rn satisfies N(x)Cx1 and is Lipschitz, hence continuous, for d2 clauses 1, 2, 3, The p-norms xp for rational p1, and x, The standard list e:nFn with ei(i)=1F and ei(j)=0F for ji is an ordered basis of Fn; hence dimFFn=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): t0, st=st, and s+ts+t.

Proof

technique · direct
1.1

For every yRm: yiy2 for each i<m, and y2i<myi.

L1L2
1.2

If ai<bi for every i<m and m1, then i<mai<i<mbi: the list ibiai 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,,δm1 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))2f(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=λf(x)f(a)2<ε.

L1L3L6L8
1.6

For Nf: 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 Nf.

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)2g(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 xA 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 xA and dX(x,a)<δi, and put δ:=min{δ0,,δm1}>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)Bf(x)f(a)2+f(a)2g(x)g(a)2 for every xA with dX(x,a)<δ0.

step 1.7L4L8
3.1

For xA 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)2i<mfi(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<xc<δ: step 1.1 gives fi(x)Lif(x)L2 for the forward direction, and steps 1.1, 1.2 give f(x)L2i<mfi(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 δδ0 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, mN with m1, and vector-valued functions, their components and their limits are as in Vector-valued functions f:ARm, their limits and continuity, with the dictionary to the metric notions.

The derivative

Let AR, let f:ARm and let cA 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)  :=  1xc(f(x)f(c)),

the scalar multiple being that of the vector space Rm (The vector space FX of all functions XF with pointwise operations, and Fn as the case X=n={0,1,,n1}); the division is legitimate because xc gives xc0. As in The derivative f(c)=limxcf(x)f(c)xc of f:AR at a point cA 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 limxcqf,c(x) exists in Rm, and then the derivative is

f(c)  :=  limxcqf,c(x)    Rm.

The notation denotes a single vector. At most one LRm satisfies the limit condition, as proved in Vector-valued functions f:ARm, 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))/(xc), which is the real difference quotient of fi at c (The derivative f(c)=limxcf(x)f(c)xc of f:AR at a point cA 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:ARm are differentiable at c and λR, 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,bR 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 supPL(f,P) and infPU(f,P), Darboux integrability as their equality, and the notation abf, and then

abf  :=  the function mR sending iabfi.

That really is an element of Rm. In this library Rm is the set of functions mR (The vector space FX of all functions XF with pointwise operations, and Fn as the case X=n={0,1,,n1}), 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 supPL(f,P) and infPU(f,P), Darboux integrability as their equality, and the notation abf. In the standard basis (The standard list e:nFn with ei(i)=1F and ei(j)=0F for ji is an ordered basis of Fn; hence dimFFn=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  :=  0Rm,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,μR then λf+μg is integrable with

ab(λf+μg)  =  λabf+μabg,

since each side has i-th coordinate ab(λfi+μgi) and λabfi+μabgi respectively, and those agree by Integrable functions on [a,b] form a set closed under sums and scalar multiples, and ab(λf+μg)=λabf+μabg.

Restriction and splitting. If f is integrable on [a,b] then it is integrable on every nondegenerate closed subinterval [c,d] with ac<db, 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 ab and f:[a,b]Rm integrable when a<b, abf2abf2; for a<b, f2 is integrable

Statement

Let mN with m1, let a,bR with ab 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 tf(t)2 is integrable on [a,b] (The lower and upper Darboux integrals of a bounded f on [a,b] as supPL(f,P) and infPU(f,P), Darboux integrability as their equality, and the notation abf, The p-norms xp for rational p1, and x);
  2. abf2    abf2.

The hypothesis ab 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 abf2abf2; 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 abfabf 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 φf is integrable.

Facts & Assumptions

Given: A natural m1, reals ab, a function f:[a,b]Rm that is integrable when a<b, with components f0,,fm1, and the vector v:=abfRm; 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 xp for rational p1, and x).

[L2]

Linearity of the integral: integrable functions on [a,b] are closed under sums and scalar multiples, and ab(λu+μw)=λabu+μabw (Integrable functions on [a,b] form a set closed under sums and scalar multiples, and ab(λf+μg)=λabf+μabg).

[L3]

Monotonicity of the integral: for a<b and integrable uw on [a,b], abuabw; and an integrable u0 has abu0 (If fg on [a,b] and both are integrable then abfabg; and m(ba)abfM(ba)).

[L6]

Square roots (Square roots exist: a unique a0 with (a)2=a; the positives are {x2:x0}, Squaring is monotone on the nonnegatives): every c0 has a unique c0 with (c)2=c, and ss2 is strictly increasing on the nonnegatives, hence injective there.

[L8]

Cauchy-Schwarz and the inner product: u,w=i<muiwi is bilinear and symmetric, u2=u,u, u20, and u,wu2w2 (The Euclidean inner product x,y=k<nxkyk on Rn, Cauchy-Schwarz x,yx2y2 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 u1>0, a product of nonnegatives is nonnegative, and tt (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 abf2=0 by the oriented convention, so clause 2 reads 00 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)v2f(t)2 by Cauchy-Schwarz.

L8L10
2.1

By induction on pm, 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 tv,f(t)=i<mvifi(t) is integrable, being a finite sum of scalar multiples of the integrable fi, and by linearity applied m times abv,f=i<mviabfi=i<mvivi=v,v=v22.

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 ss2 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 f2=φg 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 v22=abv,fabv2f2=v2abf2.

step 5.1step 2.2step 1.3L2L3
6.2

If v=0 then v2=0, while abf20 because f20 pointwise and a<b; so clause 2 holds in this case.

step 5.1L3L8
7.1

If v0 then v2>0, so multiplying the inequality of step 6.1 by the positive 1/v2 gives v2abf2, 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 v22v2abf2, and dividing by v2 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 φf 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 f2M, then f(b)f(a)2M(ba)

Statement

Let mN with m1, let a,bR 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:ARm, 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 MR with M0 satisfy

f(t)2    Mfor every t(a,b).

Then

f(b)f(a)2    M(ba).

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 ab and f:[a,b]Rm integrable when a<b, abf2abf2; for a<b, f2 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 m2 it is false. There need be no ξ(a,b) with f(b)f(a)=f(ξ)(ba); 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 tu,f(t), not of f.

Facts & Assumptions

Given: A natural m1, reals a<b, a function f:[a,b]Rm continuous on [a,b] and differentiable on (a,b), a real M0 bounding f2 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=w22, and u,w=i<muiwi (The Euclidean inner product x,y=k<nxkyk on Rn, The p-norms xp for rational p1, 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)(ba) (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)(ba)).

[L9]

Order arithmetic: ba>0; a product of nonnegatives is nonnegative; and u>0 gives u1>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:AR at a point of A and on A: the ε-δ condition, its agreement with limxcf(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=u22, by bilinearity.

L1
1.4

By Cauchy-Schwarz and the bound on f2, u,f(ξ)u,f(ξ)u2f(ξ)2u2M.

L2
1.5

If u=0 then u2=0 while M(ba)0, so the conclusion holds.

L2L9
2.1

By induction on pm, each partial sum ti<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)(ba).

step 3.1L7
5.1

Combining steps 1.3 and 4.1, u22=u,f(ξ)(ba).

step 3.1step 1.3step 4.1
6.1

Since ba>0, multiplying the inequality of step 1.4 by ba and using step 5.1 gives u22u2M(ba).

step 5.1step 1.4L9
7.1

If u0 then u2>0, and multiplying step 6.1 by the positive real 1/u2 gives u2M(ba).

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)2M(ba).

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 mN with m1 and let a,bR 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 M0 satisfy f(t)2M for every t(a,b). Then f(t)f(s)2    Mtsfor 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<α1, and contraction, The absolute value makes R a metric space: d(x,y)=xy is a metric, its open balls are the intervals (xr,x+r), and it is unbounded, Rn as the set of functions nR, and d1, d2, d are metrics on it).

Facts & Assumptions

Given: A natural m1, 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 f2M, then f(b)f(a)2M(ba)): for s<t, f continuous on [s,t] and differentiable on (s,t) with f2M there, f(t)f(s)2M(ts).

[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 limxcf(x)=L of f:AR at a limit point c of A, Vector-valued functions f:ARm, 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 L0 when dY(f(x),f(x))LdX(x,x) for all x,x (Lipschitz map, α-Hölder map for rational 0<α1, 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)2M(ts)=Mts.

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=Mts; and if t<s then step 2.2 applied with the roles exchanged gives f(s)f(t)2Mst, and f(t)f(s)2=(f(s)f(t))2=f(s)f(t)2 while st=ts.

step 2.2L5
4.1

Steps 2.2 and 3.2 cover all pairs s,t[a,b], so f(t)f(s)2Mts always; since d2(f(t),f(s))=f(t)f(s)2 and dR(t,s)=ts, this is exactly the Lipschitz condition with constant M0, 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 nN with n1, so that Rn carries the Euclidean metric d2 (Rn as the set of functions nR, 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:NRn, 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: xkx iff d(xk,x)0 in R).

Partial sums and convergence

The partial sums of (xk) are

sN  :=  k<Nxk    Rn(NN),

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:nFn with ei(i)=1F and ei(j)=0F for ji is an ordered basis of Fn; hence dimFFn=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 sRn when sNs in (Rn,d2) (Convergence of a sequence in a metric space: xkx iff d(xk,x)0 in R), and then s is the sum, written k=0xk. 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 xk2 converges (Series, partial sums, convergence and the sum, divergence, and the tail series); since xk20 (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 cxk2N(xk)Cxk2 for fixed c,C>0 (For n1 all norms on Rn are equivalent, Equivalent norms, and the dictionary with equivalent metrics), so N(xk) converges exactly when xk2 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:NN be a bijection (Injection, surjection, bijection). The rearrangement of xk along σ is the series xσ(k) of the sequence kxσ(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)  :=  {sRn  :  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 1R and is not literally R. The map θ:RR1 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:nFn with ei(i)=1F and ei(j)=0F for ji is an ordered basis of Fn; hence dimFFn=n, and F0 is the zero space with basis and dimension 0), and d2(θ(s),θ(t))=st, 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 nN with n1 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=0xk.
  2. For every bijection σ:NN (Injection, surjection, bijection) the rearranged series xσ(k) converges absolutely, with k=0xσ(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 n1; a sequence (xk) in Rn with xk2 convergent; the vector partial sums sN=k<Nxk and the real partial sums TN=k<Nxk2; 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 LN: sNsL=k=LN1xk and TNTL=k=LN1xk2, 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 TNTL<ε for all N,LK.

L4L8
1.3

For every j<n and every k: 0(xk)jxk2.

L2
1.4

Likewise kxσ(k)2 is the rearrangement along σ of kxk2, a convergent series of nonnegative terms and therefore absolutely convergent, so kxσ(k)2 converges; that is, xσ(k) converges absolutely.

L7L8L1
2.1

Hence sNsL2k=LN1xk2=TNTL 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 NLK we get d2(sN,sL)=sNsL2TNTL<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 kxσ(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 sS(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 nN with n1 and let (xk) be a sequence in Rn (Series of vectors in Rn, absolute convergence, rearrangement, and the set of rearrangement sums). Define

Γ  :=  {aRn  :  ka,xk converges},Γ  :=  {yRn  :  a,y=0 for every aΓ},

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 λR and a,bΓ, bilinearity and the absolute value laws give

λa+b,xk  =  λa,xk+b,xk    λa,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 0akbk 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 WV is a linear subspace if and only if λu+vW for all λF and u,vW), Γ is a linear subspace.

Γ is a linear subspace of Rn. It contains 0, and for λR, y,zΓ and aΓ, bilinearity gives a,λy+z=λa,y+a,z=0; again One-step subspace test: a nonempty WV is a linear subspace if and only if λu+vW for all λF and u,vW applies. Equivalently Γ is the intersection of the linear subspaces {y:a,y=0} over aΓ, a nonempty family since 0Γ, 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,xka2xk2 (Cauchy-Schwarz x,yx2y2 with its equality case, the triangle inequality for 2, the parallelogram law and polarisation), and ka2xk2 converges by Convergent series add and scale termwise clause 2; the comparison test gives aΓ.

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:nFn with ei(i)=1F and ei(j)=0F for ji is an ordered basis of Fn; hence dimFFn=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 kj<n(xk)j=kxk1 converges; and xk2xk1 (The finite and reverse triangle inequalities for a norm; and for n1 every norm N on Rn satisfies N(x)Cx1 and is Lipschitz, hence continuous, for d2 clause 3, The p-norms xp for rational p1, and x), so kxk2 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 yΓ 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 WRn and sRn, the affine subspace through s with direction W is the coset

s+W  :=  {s+w  :  wW}.

A coset is determined by W together with any one of its points. If ps+W, say p=s+w0 with w0W, 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+(ww0) 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 ssW.

Remarks

  • 0Γ 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 aΓ, yΓ.

  • 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 nN with n1, let mN and let v:mRn be a finite list of vectors with

vi21  for every i<m,i<mvi=0.

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

j<kvπ(j)2    ι(n)for every km,

where ι is the canonical natural of R (The canonical natural ι(n)=n1F 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 n1 and m; a list v:mRn with vi21 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=qr1bi for qr with i=qr1bi=l<rqbq+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)=n1F 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,q1 by claim 3 there and trivially when p=0 or q=0, ι is strictly increasing, and ι(p)>0 for p1.

[L8]

Order arithmetic: u>0 gives u1>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:NR, let r1, let q<r, and let bq be the list with biq:=bi for i<q and biq:=bi+1 for qi<r1. Then i<rbi=i<r1biq+bq: splitting the left side at q and again at q+1 gives i<qbi+bq+l<r1qbq+1+l, and splitting the right side at q gives i<qbi+l<r1qbq+l+1, and the two agree.

L2
1.2

The easy case mn. Take π to be the identity of m, a bijection. For km the finite triangle inequality and vj21 give j<kvj2j<kvj2j<k1=ι(k)ι(n), since kmn 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 nkm call a pair (b,μ) admissible at k when b:km is injective, μ:NR vanishes at every jk, satisfies 0μj1 for j<k, and satisfies j<kμjvb(j)=0 and j<kμj=ι(kn).

construct
1.4

Stage m is admissible. Take bm:= the identity of m and μjm:=ι(mn)/ι(m) for j<m, μjm:=0 for jm; here ι(m)>0 because m>n1, and 0ι(mn)ι(m) gives 0μjm1. Then j<mμjmvj=(ι(mn)/ι(m))j<mvj=0 and j<mμjm=ι(m)ι(mn)/ι(m)=ι(mn).

constructL2L7L8
1.5

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

L1L2L7
2.1

The reindexing identity. For every kN, every rN, every injective f:rk and every c:NR 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:r0 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 r1 and the list g:=fq of step 1.1 is an injective map r1k off whose image c vanishes on {j:j<k}, so the inductive hypothesis gives j<kcj=i<r1cg(i), while step 1.1 applied to bi:=cf(i) gives i<rcf(i)=i<r1cg(i)+cf(q); adding ck=cf(q) to the first identity yields the claim.

step 1.1L2L4
2.2

The feasible set at k1 is nonempty. Let (b,μ) be admissible at k with n<km, and let Λ be the set of all μ:NR vanishing at every jk, with 0μj1 for j<k, j<kμjvb(j)=0 and j<kμj=ι(k1n). The scalar ρ:=ι(k1n)/ι(kn) is defined and lies in [0,1], since ι(kn)>0 and 0ι(k1n)ι(kn); 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 μΛ r-simple when there is an injective f:rk with μj{0,1} for every j<k outside the image of f. The set R:={rN:some μΛ 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 μΛ and an injective f:r0k 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:rk 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 fi, an injective map r01k off whose image μ takes values in {0,1}, would witness that μ is (r01)-simple, contradicting minimality of r0.

step 1.1step 3.2
4.3

Suppose r0n+2, towards a contradiction. Define w:r0Rn+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:r0R, not identically 0, with i<r0λiwi=0. If w is not injective, say wi1=wi2 with i1i2, take λi1:=1, λi2:=1 and λi:=0 otherwise; the list iλiwi then vanishes off {i1,i2} and sums to wi1wi2=0 by step 4.1. If w is injective, its image is a subset of Rn+1 equinumerous with r0n+2, hence not linearly independent by [L6]; so some injective list h:pim(w) is linearly dependent, giving ν:pR 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 iλiwi 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 λi0, which exists because λ is not identically 0. Define s:r0R by si:=(1μf(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,,sr01}, a minimum over a nonempty finite set of reals, so t>0 and t=si for some i<r0; choosing that i if λi0 and i0 otherwise, there is i<r0 with λi0 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:NR by μj:=μj+tλi if j=f(i) for the unique i<r0 with that property, and μj:=μj otherwise. Then 0μj1 for every j<k: outside the image of f nothing changes; at j=f(i) with λi>0 one has μf(i)<μjμf(i)+siλi=1; with λi<0 one has 0=μf(i)+siλiμj<μf(i); and with λi=0 the value is unchanged.

step 4.2step 5.2L8
7.1

The moved point is feasible. The list jμjμj vanishes at every j<k off the image of f and takes the value tλi at f(i), so step 4.1 gives j<k(μjμj)=i<r0tλi=t0=0; likewise the Rn-valued list j(μjμj)vb(j) vanishes off that image and takes the value tλivb(f(i)) at f(i), so j<k(μjμj)vb(j)=ti<r0λivb(f(i))=0. Hence j<kμj=ι(k1n) and j<kμjvb(j)=0, so μΛ.

step 4.1step 6.1step 6.2L2L3
8.1

The contradiction. By step 5.2, μf(i)=μf(i)+tλi{0,1}. So fi, an injective map r01k, witnesses that μ is (r01)-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 r0n+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:=1μj for j<k and νj:=0 for jk, 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(k1n)=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 r01 then every term of i<r0(1νf(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(1νf(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:=bj0:k1m and μ:=μj0, extended by 0 beyond k1. Then b is injective with image im(b){b(j0)}, 0μj1 for j<k1, and by step 1.1 in both its real and its vector form, j<k1μj=j<kμjμj0=ι(k1n) and j<k1μjvb(j)=j<kμjvb(j)μj0vb(j0)=0. So (b,μ) is admissible at k1.

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 nkm, with im(bk1)im(bk) and im(bk)im(bk1) a single element. This is a recursion of length mn, 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:mm by π(j):=bn(j) for j<n and, for each k with n<km, π(k1):= the unique element of im(bk)im(bk1). 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 nkm 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 nkm and let c:NRn be ci:=vi for iim(bk) and ci:=0 otherwise. Then c vanishes at every i<m off the image of the injective list jπ(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 nkm. Since j<kμjkvbk(j)=0, additivity gives j<kvbk(j)=j<k(1μjk)vbk(j); each coefficient 1μjk is nonnegative, so the finite triangle inequality and vi21 give j<kvbk(j)2j<k(1μjk)vbk(j)2j<k(1μjk)=ι(k)ι(kn)=ι(n).

step 12.1step 14.1L1L2L7
16.1

By steps 14.1 and 15.1 the bound j<kvπ(j)2ι(n) holds for nkm, and by step 1.5 it holds for k<n; together with the case mn 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 r0n+1 one gets only that the support of μ has at most (k1n)+(n+1)=k elements, which is no information at all. What rules out equality is that the quantities 1μf(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 mn; 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:rk 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 nN with n1, 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=0xk. 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. sS(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, tsΓ 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 cR, a rearrangement with sum c, and rearrangements diverging to +, to , and oscillating with any prescribed lim inflim 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 n2 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 n1; a sequence (xk) in Rn with xk convergent of sum s; a bijection σ of N; a vector aΓ; the partial sums sN=k<Nxk and sNσ=k<Nxσ(k).

[L2]

Γ and Γ are linear subspaces; aΓ means ka,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=uv, u0, and u>0 gives u1>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 sS(x) and clause 1 holds.

L1
1.2

For every aRn and every finite list u:pRn, a,j<puj=j<pa,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 uNu in (Rn,d2) then a,uNa,u in R, since a,uNa,u=a,uNua2uNu2, so a tolerance ε/(a2+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 aR1, a,xk=a0(xk)0 and a,xk=a0(xk)0; if a00 then convergence of ka0(xk)0 would give convergence of k(xk)0 after multiplying by the positive 1/a0, which is false, so aΓ forces a0=0; and a=0 does lie in Γ. Hence Γ={0}.

L1L2L9
2.1

Let tS(x), say sNσt for a bijection σ, and let aΓ. By steps 1.2 and 1.3, a,t=limNa,sNσ=limNk<Na,xσ(k), so the real series ka,xσ(k) converges with sum a,t.

step 1.2step 1.3L1L7
2.2

In the same way a,s=limNa,sN=limNk<Na,xk, so ka,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 ka,xσ(k) is the rearrangement along σ of the sequence ka,xk, and the latter series converges absolutely because aΓ; 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,ts=0 by bilinearity.

step 2.1step 2.2step 3.1L3
5.1

Since aΓ was arbitrary, tsΓ, that is ts+Γ; as tS(x) was arbitrary, clause 2 holds.

step 4.1L2
6.1

Suppose xk converges absolutely. Then Γ=Rn, so any yΓ 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 n2 the question whether every point of s+Γ is a rearrangement sum is open as far as this library is concerned. It is not open in the mathematical literature, and this page deliberately states nothing about what the literature says, exactly as 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 declines to. What is missing here is machinery, not effort: every route known to the author of this page passes through the orthogonal decomposition of a finite-dimensional inner product space and through a separation argument for convex sets, and neither exists in this library — the first belongs to a page earlier in the plan order that is not yet built, and the second to no planned page at all. See Conventions of this page, the standing n1 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 applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)verified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

Conventions of this page, the standing n1 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 nR, and d1, d2, d are metrics on it defines Rn together with the metrics d1, d2, d only for n1, 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 n1 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 n1 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 n1 (or m1 for the codomain of a vector-valued function), and each states it in its own Statement: The p-norms xp for rational p1, 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 n1 every norm N on Rn satisfies N(x)Cx1 and is Lipschitz, hence continuous, for d2; For n1 all norms on Rn are equivalent; For n1 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 n1 every bounded sequence in Rn has a convergent subsequence; Vector-valued functions f:ARm, 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 ab and f:[a,b]Rm integrable when a<b, abf2abf2; for a<b, f2 is integrable; The mean value inequality: if f:[a,b]Rm is continuous and differentiable on (a,b) with f2M, then f(b)f(a)2M(ba); 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 n1 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 xp for rational p1, and x defines p for rational p1 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 n2. No recorded-not-proved item has been created for it either.

The obstruction is machinery and not effort. Every route to the reverse inclusion known to the author of this page reduces first to the case Γ={0} by an orthogonal projection, which needs the orthogonal decomposition named in §4, and then runs a separation argument for convex sets in Rn, which exists nowhere in this library and is owned by no planned page. When both exist, the discharge is an addition to this page, not a new page.

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