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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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For $n \geq 1$ a sequence in $R^{n}$ converges iff each coordinate sequence converges, is Cauchy iff each coordinate sequence is Cauchy, and $R^{n}$ is complete in every norm

Statement

Let $n \in N$ with $n \geq 1$ , let $R^{n}$ carry the Euclidean metric $d_{2}$ of $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it, and let $(x^{(j)})_{j \in N}$ be a sequence in $R^{n}$ (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ). For $k < n$ write $(x_{k}^{(j)})_{j \in N}$ for the $k$ -th coordinate sequence, a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences). Then:

Convergence is componentwise. For $x \in R^{n}$ , $x^{(j)} \to x$ in $(R^{n}, d_{2})$ if and only if $x_{k}^{(j)} \to x_{k}$ in $R$ for every $k < n$ (Limits and Cauchy sequences of reals).
Cauchyness is componentwise. $(x^{(j)})$ is Cauchy in $(R^{n}, d_{2})$ (Cauchy sequence in a metric space) if and only if every coordinate sequence is Cauchy in $R$ .
Completeness in every norm. For every norm $N$ on $R^{n}$ (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms) the metric space $(R^{n}, d_{N})$ 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 $R^{n}$ for $n \geq 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $R$ clause 2 states that $(R^{n}, d_{2})$ is complete, for $n \geq 1$ only, and this theorem carries that hypothesis forward without weakening it; what is added is the passage from $d_{2}$ to an arbitrary norm, through For $n \geq 1$ all norms on $R^{n}$ are equivalent and the dictionary of Equivalent norms, and the dictionary with equivalent metrics.

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 metric $d_{2}$ ; a sequence $(x^{(j)})$ in $R^{n}$ ; a point $x \in R^{n}$ ; a norm $N$ on $R^{n}$ ; and a rational $ε > 0$ .

[L1]

The comparison chain for $n \geq 1$ (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 $d_{2}$ clause 3): $∥ y ∥_{\infty} \leq ∥ y ∥_{2} \leq ∥ y ∥_{1} \leq ι (n) ∥ y ∥_{\infty}$ for every $y \in R^{n}$ , where $∥ y ∥_{\infty} = \max {∣ y_{k} ∣ : k < n}$ (The $p$ -norms $∥ x ∥_{p}$ for rational $p \geq 1$ , and $∥ x ∥_{\infty}$ , Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L2]

The dictionary $d_{2} (u, v) = ∥ u - v ∥_{2}$ , and $(u - v)_{k} = u_{k} - v_{k}$ (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, 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 $0$ ).

[L3]

Convergence and Cauchyness in a metric space, and their agreement on $R$ with the real notions (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ , Cauchy sequence in a metric space, Limits and Cauchy sequences of reals, The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded); rational and real $ε$ may be used interchangeably in both.

[L4]

$(R^{n}, d_{2})$ is complete for $n \geq 1$ ( $R$ and $R^{n}$ for $n \geq 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $R$ clause 2, Complete metric space: every Cauchy sequence converges in the space).

[L5]

All norms on $R^{n}$ are equivalent for $n \geq 1$ (For $n \geq 1$ all norms on $R^{n}$ 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).

[L7]

$ι (n) > 0$ for $n \geq 1$ , and $u > 0$ gives $u^{- 1} > 0$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L8]

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

Proof

technique · direct

1.1

For every $y \in R^{n}$ and every $k < n$ : $∣ y_{k} ∣ \leq ∥ y ∥_{\infty} \leq ∥ y ∥_{2}$ , the first inequality because $∥ y ∥_{\infty}$ bounds the set it is the maximum of.

1.2

For every $y \in R^{n}$ : $∥ y ∥_{2} \leq ι (n) ∥ y ∥_{\infty}$ , and $∥ y ∥_{\infty} = ∣ y_{k_{0}} ∣$ for some $k_{0} < n$ .

1.3

Conversely suppose $x_{k}^{(j)} \to x_{k}$ 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 $∣ x_{k}^{(j)} - x_{k} ∣ < ε / ι (n)$ for all $j \geq K$ is a nonempty set of naturals; let $K_{k}$ be its least element, a determination rather than a selection, and put $K : = \max {K_{0}, \dots, K_{n - 1}}$ , a maximum of a nonempty finite set of naturals.

L3L7L8

1.4

$(R^{n}, d_{2})$ is complete, by citation and for $n \geq 1$ only.

1.5

Let $N$ be any norm on $R^{n}$ . By [L5], $N$ and $∥ \cdot ∥_{2}$ are equivalent, so $d_{N}$ and $d_{2}$ have the same Cauchy sequences and the same convergent sequences with the same limits.

2.1

For all $u, v \in R^{n}$ and $k < n$ : $∣ u_{k} - v_{k} ∣ \leq d_{2} (u, v) \leq ι (n) \max {∣ u_{k} - v_{k} ∣ : k < n}$ , by steps 1.1 and 1.2 applied to $y : = u - v$ .

step 1.1step 1.2L2

2.2

Hence a Cauchy sequence in $(R^{n}, d_{N})$ is Cauchy in $(R^{n}, d_{2})$ , converges there by step 1.4, and therefore converges in $(R^{n}, d_{N})$ to the same point; so $(R^{n}, d_{N})$ is complete, which is clause 3.

step 1.4step 1.5L5L6

3.1

Suppose $x^{(j)} \to x$ in $(R^{n}, d_{2})$ and fix $k < n$ . Given a rational $ε > 0$ , take $K$ with $d_{2} (x^{(j)}, x) < ε$ for $j \geq K$ ; then $∣ x_{k}^{(j)} - x_{k} ∣ \leq d_{2} (x^{(j)}, x) < ε$ for $j \geq K$ , so $x_{k}^{(j)} \to x_{k}$ .

step 2.1L3

3.2

For $j \geq K$ and every $k < n$ we have $∣ x_{k}^{(j)} - x_{k} ∣ < ε / ι (n)$ ; the maximum of these $n$ numbers is one of them, so $\max {∣ x_{k}^{(j)} - x_{k} ∣ : k < n} < ε / ι (n)$ and hence $d_{2} (x^{(j)}, x) < ι (n) \cdot ε / ι (n) = ε$ by step 2.1. Therefore $x^{(j)} \to x$ .

step 2.1step 1.3L1L7

3.3

The same two estimates prove clause 2 with $x$ replaced by $x^{(l)}$ throughout: if $d_{2} (x^{(j)}, x^{(l)}) < ε$ for $j, l \geq K$ then $∣ x_{k}^{(j)} - x_{k}^{(l)} ∣ < ε$ for $j, l \geq K$ and every $k < n$ ; and conversely, choosing for each $k < n$ the least $K_{k}$ beyond which $∣ x_{k}^{(j)} - x_{k}^{(l)} ∣ < ε / ι (n)$ for $j, l \geq K_{k}$ and taking $K : = \max {K_{0}, \dots, K_{n - 1}}$ gives $d_{2} (x^{(j)}, x^{(l)}) < ε$ for $j, l \geq K$ .

step 2.1L3L7L8

4.1

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

step 3.1step 3.2

5.1

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

step 4.1step 3.3step 2.2∎

Remarks

No choice principle is used. The only place a family of indices is produced is steps 1.3 and 3.3, where finitely many indices are obtained, each as the least element of a nonempty set of naturals (The well-ordering principle). A least element is determined by the set, not selected from it.
What happens at $n = 0$ , stated separately because the theorem does not cover it. $R^{0}$ has exactly one element, the empty function, and is the zero vector space (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 $0$ clause 5); by A norm on a real vector space, the induced metric, and the dictionary with the metric axioms it carries exactly one norm, the zero function, whose induced metric is constantly $0$ . Every sequence in a one-point metric space is Cauchy and converges to that point, so $(R^{0}, d_{N})$ is complete. That statement is proved here from scratch in this remark and is not obtained from $R$ and $R^{n}$ for $n \geq 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $R$ , which is stated for $n \geq 1$ only because $d_{\infty}$ is a maximum over an empty index set at $n = 0$ . Clauses 1 and 2 are vacuous at $n = 0$ , there being no index $k < 0$ .
Clause 1 is the reason the rest of this page can work coordinatewise. Every later item that reduces a statement about $R^{n}$ or $R^{m}$ to $n$ or $m$ statements about $R$ passes through it, and each such item therefore carries the hypothesis $n \geq 1$ or $m \geq 1$ in its own statement.

Depends on

Used by

Dependency tree · two levels

103 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

Complete metric space (Wikipedia) (standard reference, not scraped)
Euclidean space (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 a sequence in RnRn converges iff each coordinate sequence converges, is Cauchy iff each coordinate sequence is Cauchy, and RnRn is complete in every norm