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

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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

1.1L1

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.

1.2L1

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

1.3L3L7L8

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.

1.4L4

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

1.5L5

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.

2.1step 1.1step 1.2L2

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 $y : = u - v$ .

2.2step 1.4step 1.5L5L6

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.

3.1step 2.1L3

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 $x_{k}^{(j)} \to x_{k}$ .

3.2step 2.1step 1.3L1L7

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 $x^{(j)} \to x$ .

3.3step 2.1L3L7L8

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 $j, l \geq K$ .

4.1step 3.1step 3.2

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

5.1step 4.1step 3.3step 2.2∎

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

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

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

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

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

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