TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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For $n \ge 1$ a sequence in $\mathbb{R}^n$ converges iff each coordinate sequence converges, is Cauchy iff each coordinate sequence is Cauchy, and $\mathbb{R}^n$ is complete in every norm

Statement

Let $n \in \mathbb{N}$ with $n \ge 1$ , let $\mathbb{R}^{n}$ carry the Euclidean metric $d_2$ of $\mathbb{R}^n$ as the set of functions $n \to \mathbb{R}$ , and $d_1$ , $d_2$ , $d_\infty$ are metrics on it, and let $\bigl(x^{(j)}\bigr)_{j\in\mathbb{N}}$ be a sequence in $\mathbb{R}^{n}$ (Convergence of a sequence in a metric space: $x_k \to x$ iff $d(x_k, x) \to 0$ in $\mathbb{R}$ ). For $k < n$ write $\bigl(x^{(j)}_k\bigr)_{j\in\mathbb{N}}$ for the $k$ -th coordinate sequence, a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences). Then:

Convergence is componentwise. For $x \in \mathbb{R}^{n}$ , $x^{(j)} \to x$ in $(\mathbb{R}^{n}, d_2)$ if and only if $x^{(j)}_k \to x_k$ in $\mathbb{R}$ for every $k<n$ (Limits and Cauchy sequences of reals).
Cauchyness is componentwise. $\bigl(x^{(j)}\bigr)$ is Cauchy in $(\mathbb{R}^{n},d_2)$ (Cauchy sequence in a metric space) if and only if every coordinate sequence is Cauchy in $\mathbb{R}$ .
Completeness in every norm. For every norm $N$ on $\mathbb{R}^{n}$ (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms) the metric space $(\mathbb{R}^{n}, d_N)$ is complete (Complete metric space: every Cauchy sequence converges in the space).

Clause 3 is obtained by citation and is not reproved here. $\mathbb{R}$ and $\mathbb{R}^n$ for $n \ge 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $\mathbb{R}$ clause 2 states that $(\mathbb{R}^{n},d_2)$ is complete, for $n \ge 1$ 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 $n \ge 1$ all norms on $\mathbb{R}^n$ are equivalent and the dictionary of Equivalent norms, and the dictionary with equivalent metrics.

Facts & Assumptions

Given: A natural $n \ge 1$ ; the space $\mathbb{R}^{n}$ with the norms of The $p$ -norms $\lVert x\rVert_p$ for rational $p \ge 1$ , and $\lVert x\rVert_\infty$ and the metric $d_2$ ; a sequence $\bigl(x^{(j)}\bigr)$ in $\mathbb{R}^{n}$ ; a point $x \in \mathbb{R}^{n}$ ; a norm $N$ on $\mathbb{R}^{n}$ ; and a rational $\varepsilon > 0$ .

[L1]

The comparison chain for $n \ge 1$ (The finite and reverse triangle inequalities for a norm; and for $n \ge 1$ every norm $N$ on $\mathbb{R}^n$ satisfies $N(x) \le C\lVert x\rVert_1$ and is Lipschitz, hence continuous, for $d_2$ clause 3): $\lVert y\rVert_\infty \le \lVert y\rVert_2 \le \lVert y\rVert_1 \le \iota(n)\lVert y\rVert_\infty$ for every $y \in \mathbb{R}^{n}$ , where $\lVert y\rVert_\infty = \max\{|y_k| : k<n\}$ (The $p$ -norms $\lVert x\rVert_p$ for rational $p \ge 1$ , and $\lVert x\rVert_\infty$ , Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L2]

The dictionary $d_2(u,v) = \lVert u-v\rVert_2$ , and $(u-v)_k = u_k - v_k$ (Each $\lVert\cdot\rVert_p$ is a norm on $\mathbb{R}^n$ , 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 $j \ne i$ is an ordered basis of $F^{n}$ ; hence $\dim_F F^{n} = n$ , and $F^{0}$ is the zero space with basis $\varnothing$ and dimension $0$ ).

[L3]

Convergence and Cauchyness in a metric space, and their agreement on $\mathbb{R}$ with the real notions (Convergence of a sequence in a metric space: $x_k \to x$ iff $d(x_k, x) \to 0$ in $\mathbb{R}$ , Cauchy sequence in a metric space, Limits and Cauchy sequences of reals, The absolute value makes $\mathbb{R}$ 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 $\varepsilon$ may be used interchangeably in both.

[L4]

$(\mathbb{R}^{n}, d_2)$ is complete for $n \ge 1$ ( $\mathbb{R}$ and $\mathbb{R}^n$ for $n \ge 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $\mathbb{R}$ clause 2, Complete metric space: every Cauchy sequence converges in the space).

[L5]

All norms on $\mathbb{R}^{n}$ are equivalent for $n \ge 1$ (For $n \ge 1$ all norms on $\mathbb{R}^n$ 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]

$\iota(n) > 0$ for $n \ge 1$ , and $u > 0$ gives $u^{-1} > 0$ (The canonical natural $\iota(n) = n \cdot 1_F$ 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 $y \in \mathbb{R}^{n}$ and every $k<n$ : $|y_k| \le \lVert y\rVert_\infty \le \lVert y\rVert_2$ , the first inequality because $\lVert y\rVert_\infty$ bounds the set it is the maximum of.

1.2

For every $y \in \mathbb{R}^{n}$ : $\lVert y\rVert_2 \le \iota(n)\lVert y\rVert_\infty$ , and $\lVert y\rVert_\infty = |y_{k_0}|$ for some $k_0 < n$ .

1.3

Conversely suppose $x^{(j)}_k \to x_k$ for every $k<n$ . Given a rational $\varepsilon > 0$ , the real $\varepsilon/\iota(n)$ is positive, so for each $k<n$ the set of indices $K$ such that $|x^{(j)}_k - x_k| < \varepsilon/\iota(n)$ for all $j \ge K$ 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

$(\mathbb{R}^{n},d_2)$ is complete, by citation and for $n \ge 1$ only.

1.5

Let $N$ be any norm on $\mathbb{R}^{n}$ . By [L5], $N$ and $\lVert\cdot\rVert_2$ 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 $u,v \in \mathbb{R}^{n}$ and $k<n$ : $|u_k - v_k| \le d_2(u,v) \le \iota(n)\max\{|u_k-v_k| : k<n\}$ , by steps 1.1 and 1.2 applied to $y := u - v$ .

step 1.1step 1.2L2

2.2

Hence a Cauchy sequence in $(\mathbb{R}^{n},d_N)$ is Cauchy in $(\mathbb{R}^{n},d_2)$ , converges there by step 1.4, and therefore converges in $(\mathbb{R}^{n},d_N)$ to the same point; so $(\mathbb{R}^{n},d_N)$ is complete, which is clause 3.

step 1.4step 1.5L5L6

3.1

Suppose $x^{(j)} \to x$ in $(\mathbb{R}^{n},d_2)$ and fix $k<n$ . Given a rational $\varepsilon > 0$ , take $K$ with $d_2(x^{(j)},x) < \varepsilon$ for $j \ge K$ ; then $|x^{(j)}_k - x_k| \le d_2(x^{(j)},x) < \varepsilon$ for $j \ge K$ , so $x^{(j)}_k \to x_k$ .

step 2.1L3

3.2

For $j \ge K$ and every $k<n$ we have $|x^{(j)}_k - x_k| < \varepsilon/\iota(n)$ ; the maximum of these $n$ numbers is one of them, so $\max\{|x^{(j)}_k - x_k| : k<n\} < \varepsilon/\iota(n)$ and hence $d_2(x^{(j)},x) < \iota(n)\cdot\varepsilon/\iota(n) = \varepsilon$ 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 $d_2(x^{(j)},x^{(l)}) < \varepsilon$ for $j,l \ge K$ then $|x^{(j)}_k - x^{(l)}_k| < \varepsilon$ for $j,l \ge K$ and every $k<n$ ; and conversely, choosing for each $k<n$ the least $K_k$ beyond which $|x^{(j)}_k - x^{(l)}_k| < \varepsilon/\iota(n)$ for $j,l \ge K_k$ and taking $K := \max\{K_0,\dots,K_{n-1}\}$ gives $d_2(x^{(j)},x^{(l)}) < \varepsilon$ for $j,l \ge K$ .

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. $\mathbb{R}^{0}$ 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 $j \ne i$ is an ordered basis of $F^{n}$ ; hence $\dim_F F^{n} = n$ , and $F^{0}$ is the zero space with basis $\varnothing$ 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 $(\mathbb{R}^{0}, d_N)$ is complete. That statement is proved here from scratch in this remark and is not obtained from $\mathbb{R}$ and $\mathbb{R}^n$ for $n \ge 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $\mathbb{R}$ , which is stated for $n \ge 1$ 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 $\mathbb{R}^{n}$ or $\mathbb{R}^{m}$ to $n$ or $m$ statements about $\mathbb{R}$ passes through it, and each such item therefore carries the hypothesis $n \ge 1$ or $m \ge 1$ in its own statement.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 192 results over 36 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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 \ge 1n≥1 a sequence in Rn\mathbb{R}^nRn converges iff each coordinate sequence converges, is Cauchy iff each coordinate sequence is Cauchy, and Rn\mathbb{R}^nRn is complete in every norm