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TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)
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A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions

Statement

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

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

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

Facts & Assumptions

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

[L2]

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

[L5]

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

[L7]

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

[L8]

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

Proof

technique · direct
1.1

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

L1L2
1.2

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

L5L6
1.3

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

L7
1.4

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

L1L3L6L7
1.5

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

L1L3L6L8
1.6

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

L1L2L6
1.7

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

L1L3
2.1

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

step 1.1L1
2.2

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

step 1.3L1L6L7
2.3

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

step 1.7L4L8
3.1

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

step 1.1step 1.2step 2.2L5L6
3.2

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

step 1.1step 1.2step 1.3L1L5L6L7
3.3

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

step 1.7step 2.3L1L6L7
4.1

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

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

Remarks

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