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

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.

For $n \geq 1$ all norms on $R^{n}$ are equivalent

Statement

Let $n \in N$ with $n \geq 1$ . Then any two norms on $R^{n}$ 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 $R^{n}$ there are reals $c > 0$ and $C^{'} > 0$ with

$c ∥ x ∥_{2} \leq N (x) \leq C^{'} ∥ x ∥_{2} for every x \in R^{n},$

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

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

The hypothesis $n \geq 1$ is used twice in the proof and both uses are marked: once so that the constant $C$ of The finite and reverse triangle inequalities for a norm; and for $n \geq 1$ every norm $N$ on $R^{n}$ satisfies $N (x) \leq C ∥ x ∥_{1}$ and is Lipschitz, hence continuous, for $d_{2}$ 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

Given: A natural $n \geq 1$ , the space $R^{n}$ with the norms of The $p$ -norms $∥ x ∥_{p}$ for rational $p \geq 1$ , and $∥ x ∥_{\infty}$ and the published metric $d_{2}$ ( $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it, Each $∥ \cdot ∥_{p}$ is a norm on $R^{n}$ , and the induced metrics are exactly $d_{1}$ , $d_{2}$ and $d_{\infty}$ of the published metric-spaces page), and a norm $N$ on $R^{n}$ ; write $S : = {x \in R^{n} : ∥ x ∥_{2} = 1}$ .

[L1]

For $n \geq 1$ : $C : = \max {N (e_{k}) : k < n}$ exists with $C \geq 0$ , $N (x) \leq C ∥ x ∥_{1}$ , $∥ x ∥_{1} \leq \sqrt{ι (n)} ∥ x ∥_{2}$ , and $N$ is continuous as a map $(R^{n}, d_{2}) \to (R, d_{R})$ (The finite and reverse triangle inequalities for a norm; and for $n \geq 1$ every norm $N$ on $R^{n}$ satisfies $N (x) \leq C ∥ x ∥_{1}$ and is Lipschitz, hence continuous, for $d_{2}$ clauses 2, 3, 4).

[L2]

Equivalence of norms is an equivalence relation, and $c M \leq N \leq C M$ with $c, C > 0$ is what it means (Equivalent norms, and the dictionary with equivalent metrics).

[L3]

Heine-Borel in $R^{n}$ for $n \geq 1$ : a subset of $(R^{n}, d_{2})$ is compact if and only if it is closed and bounded (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ 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, Open cover, subcover, compact metric space, and compact subset of a metric space, 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, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

[L4]

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

[L5]

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

[L6]

Balls, openness and boundedness (Open ball, closed ball and sphere in a metric space, 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, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space): $U$ is open when every point of $U$ has a ball inside $U$ ; $A$ is bounded when $A = \emptyset$ or $A \subseteq B (x_{0}, r)$ for some $x_{0}$ and real $r > 0$ .

[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).

[L8]

The standard basis vector $e_{0} \in R^{n}$ exists for $n \geq 1$ , with $e_{0} (0) = 1$ and $e_{0} (j) = 0$ for $0 \neq j < n$ (The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ); hence $∥ e_{0} ∥_{2} = \sqrt{1} = 1$ (The $p$ -norms $∥ x ∥_{p}$ for rational $p \geq 1$ , and $∥ x ∥_{\infty}$ , Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

[L9]

Inverses: $u > 0$ gives $u^{- 1} > 0$ , and trichotomy of the order of $R$ (Inverses of positives are positive, and reciprocation reverses order, Complete ordered field (least-upper-bound property)).

[L10]

Continuity at a point in the $ε$ - $δ$ form, and the metric subspace $(A, d_{A})$ with $d_{A}$ the restriction of $d$ (Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form, Isometry, isometric embedding, and the subspace metric on a subset, Open cover, subcover, compact metric space, and compact subset of a metric space).

Proof

technique · direct

1.1

The singleton ${1} \subseteq R$ is closed: if $y \neq 1$ then $r : = ∣ y - 1 ∣ > 0$ and the ball $B (y, r)$ omits $1$ , so the complement of ${1}$ is open.

L6L9

1.2

$∥ \cdot ∥_{2}$ is itself a norm on $R^{n}$ , so by [L1] applied to it, $∥ \cdot ∥_{2} : (R^{n}, d_{2}) \to (R, d_{R})$ is continuous.

L1L7

1.3

$S \subseteq B (0, 2)$ , since $x \in S$ gives $d_{2} (x, 0) = ∥ x ∥_{2} = 1 < 2$ ; so $S$ is bounded.

L6L7

1.4

$e_{0} \in S$ , because $∥ e_{0} ∥_{2} = 1$ ; this is where $n \geq 1$ is used, since for $n = 0$ there is no index $0 < n$ and no such vector. So $S \neq \emptyset$ .

1.5

For every $a \in S$ and every real $ε > 0$ , a $δ > 0$ witnessing continuity of $N$ at $a$ as a map on $R^{n}$ also witnesses it for the restriction $N ∣_{S}$ on the metric subspace $(S, d_{S})$ , because $d_{S}$ is the restriction of $d_{2}$ and the condition is quantified over fewer points; so $N ∣_{S}$ is continuous.

L1L10

1.6

Put $C^{'} : = C \sqrt{ι (n)} + 1$ , a real $> 0$ . By [L1], $N (x) \leq C ∥ x ∥_{1} \leq C \sqrt{ι (n)} ∥ x ∥_{2} \leq C^{'} ∥ x ∥_{2}$ , the last step because $∥ x ∥_{2} \geq 0$ .

L1L7

2.1

$S$ is the preimage of ${1}$ under the continuous $∥ \cdot ∥_{2}$ , hence closed in $R^{n}$ .

step 1.1step 1.2L5

3.1

$S$ is a compact subset of $(R^{n}, d_{2})$ , being closed and bounded.

step 1.3step 2.1L3

4.1

By the extreme value theorem applied to the nonempty compact metric space $(S, d_{S})$ and the continuous $N ∣_{S}$ , there is $x_{\min} \in S$ with $N (x_{\min}) \leq N (x)$ for every $x \in S$ ; put $c : = N (x_{\min})$ .

step 1.4step 1.5step 3.1L4

5.1

$c > 0$ : from $x_{\min} \in S$ we get $∥ x_{\min} ∥_{2} = 1 \neq 0$ , so $x_{\min} \neq 0$ by (N1) for $∥ \cdot ∥_{2}$ , so $N (x_{\min}) \neq 0$ by (N1) for $N$ , and $N (x_{\min}) \geq 0$ ; trichotomy leaves $c > 0$ .

step 4.1L7L9

5.2

Let $x \neq 0$ . Then $∥ x ∥_{2} > 0$ by (N1) and nonnegativity, so $t : = 1 / ∥ x ∥_{2} > 0$ and $u : = t x$ satisfies $∥ u ∥_{2} = ∣ t ∣ ∥ x ∥_{2} = 1$ by (N2); hence $u \in S$ and $c \leq N (u) = ∣ t ∣ N (x) = N (x) / ∥ x ∥_{2}$ , that is $c ∥ x ∥_{2} \leq N (x)$ .

step 4.1L7L9

6.1

For $x = 0$ both $c ∥ x ∥_{2}$ and $N (x)$ are $0$ by (N1), so $c ∥ x ∥_{2} \leq N (x)$ holds for every $x \in R^{n}$ .

step 5.2L7

7.1

Steps 5.1, 6.1 and 1.6 give $c ∥ x ∥_{2} \leq N (x) \leq C^{'} ∥ x ∥_{2}$ with $c, C^{'} > 0$ , so every norm $N$ on $R^{n}$ is equivalent to $∥ \cdot ∥_{2}$ .

step 5.1step 6.1step 1.6L2

8.1

Given two norms $M$ and $N$ on $R^{n}$ , each is equivalent to $∥ \cdot ∥_{2}$ by step 7.1, so $M$ is equivalent to $N$ by symmetry and transitivity of the relation.

step 7.1L2∎

Remarks

What the proof spends, and where it stops. The only nonelementary ingredients are compactness of the Euclidean unit sphere, obtained from Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ 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 extreme value theorem A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value. Heine-Borel in $R^{n}$ is proved by bisection and uses no choice principle, and the extreme value theorem is a theorem of ZF (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), so this theorem costs no choice either.
The sphere is where the argument is finite-dimensional. The step that fails outside $R^{n}$ is step 3.1: closed and bounded gives compact by Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ 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, a theorem about $R^{n}$ for a natural $n$ and about nothing else. The companion page exhibits a real vector space carrying two inequivalent norms, and the same space with a closed bounded set that is not compact. This remark is a statement about this proof and those witnesses; it makes no broader classification claim about normed spaces.
The constants are not canonical. Nothing in the statement fixes $c$ or $C^{'}$ , and the proof produces one admissible pair, not the best one. Sharp constants for the three named norms on $R^{2}$ are computed on the companion page.

Depends on

Used by

Dependency tree · two levels

105 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Norm (mathematics) (Wikipedia) (standard reference, not scraped)
Heine-Borel theorem (Wikipedia) (standard reference, not scraped)
J. Demmel, MA221 Lecture 3: Vector Norms (standard reference, not scraped)
G. Zitelli, Math 641 Functional Analysis, Part I (standard reference, not scraped)

For n≥1n≥1 all norms on RnRn are equivalent