Coordinatewise Convergence and Completeness of Rn

Coordinates and Euclidean Distance

A sequence of vectors has a limit when all of its coordinates settle down in a compatible way. The word “compatible” matters: Euclidean distance combines the coordinate errors by squaring, adding, and taking a square root. We will compare that distance with the individual coordinate errors, prove that these comparisons give an exact convergence test, and use the test to establish completeness.

Fix $m\in\mathbb{N}$. We give $\mathbb{R}^m$ its Euclidean metric, defined by $d(x,y)=\|x-y\|$, where for $x=(x_1,\dots,x_m)$ and $y=(y_1,\dots,y_m)$,

\|x-y\|=\sqrt{\sum_{i=1}^{m}(x_i-y_i)^2}.

A sequence $(x_k)$ in a metric space converges to $x$ if, for every $\varepsilon>0$, there is $N\in\mathbb{N}$ such that $d(x_k,x)<\varepsilon$ whenever $k\ge N$. It is Cauchy if, for every $\varepsilon>0$, there is $N\in\mathbb{N}$ such that $d(x_k,x_l)<\varepsilon$ whenever $k,l\ge N$. A metric space is complete if every Cauchy sequence in it converges to a point of that space.

Key inequalities: For $x,y\in\mathbb{R}^m$ and each $i\in\{1,\dots,m\}$, $$ |x_i-y_i|\le \|x-y\|\le \sqrt{m}\,\max_{1\le i\le m}|x_i-y_i|. $$

For the first inequality, the nonnegative term $(x_i-y_i)^2$ is one of the terms in the sum defining $\|x-y\|^2$. For the second, every squared coordinate difference is at most $(\max_i|x_i-y_i|)^2$, so their sum is at most $m(\max_i|x_i-y_i|)^2$. Taking square roots gives the stated bounds. These comparisons let us pass in both directions between a vector error and its coordinate errors.

Convergence Is Coordinatewise

Theorem (Coordinatewise convergence): Let $x_k=(x_{k,1},\dots,x_{k,m})\in\mathbb{R}^m$, and let $x=(x_1,\dots,x_m)\in\mathbb{R}^m$. Then $x_k\to x$ in the Euclidean metric if and only if $x_{k,i}\to x_i$ in $\mathbb{R}$ for every $i\in\{1,\dots,m\}$.

Proof. Suppose first that $x_k\to x$. Fix a coordinate $i$ and let $\varepsilon>0$. There is $N$ such that $\|x_k-x\|<\varepsilon$ for $k\ge N$. The first key inequality gives $|x_{k,i}-x_i|\le\|x_k-x\|<\varepsilon$. Thus the $i$-th coordinate sequence converges to $x_i$. Since $i$ was arbitrary, every coordinate sequence converges.

Conversely, suppose every coordinate sequence converges to its corresponding coordinate of $x$. Let $\varepsilon>0$. For each $i$, convergence gives an index $N_i$ such that $|x_{k,i}-x_i|<\varepsilon/\sqrt{m}$ whenever $k\ge N_i$. There are only finitely many coordinates, so $N=\max\{N_1,\dots,N_m\}$ is an index that works for all of them at once. For every $k\ge N$,

\|x_k-x\| =\sqrt{\sum_{i=1}^{m}|x_{k,i}-x_i|^2} <\sqrt{\sum_{i=1}^{m}\frac{\varepsilon^2}{m}} =\varepsilon.

Hence $x_k\to x$ in the Euclidean metric. Both implications hold, proving the equivalence. $\square$

Worked Example: Computing a Vector Limit

In $\mathbb{R}^3$, with the Euclidean metric, consider $x_k=(2+1/k,\,-1-2/k,\,5+3/k)$. Find its limit and verify convergence directly.

The three coordinate sequences converge respectively to $2$, $-1$, and $5$, so the coordinatewise theorem predicts the limit $x=(2,-1,5)$. Subtracting this proposed limit gives

x_k-x=\left(\frac{1}{k},-\frac{2}{k},\frac{3}{k}\right), \qquad \|x_k-x\| =\sqrt{\frac{1}{k^2}+\frac{4}{k^2}+\frac{9}{k^2}} =\frac{\sqrt{14}}{k}.

Given $\varepsilon>0$, choose $N\in\mathbb{N}$ with $N>\sqrt{14}/\varepsilon$. Then for $k\ge N$, $\|x_k-x\|=\sqrt{14}/k\le\sqrt{14}/N<\varepsilon$, proving convergence by the metric definition.

Check: At $k=2$, $x_2=(5/2,-2,13/2)$, so $x_2-x=(1/2,-1,3/2)$ and $\|x_2-x\|=\sqrt{1/4+1+9/4}=\sqrt{14}/2$. This agrees with the formula $\sqrt{14}/k$ at $k=2$.

Cauchy Sequences Can Also Be Tested by Coordinates

Proposition: A sequence $(x_k)$ in Euclidean $\mathbb{R}^m$ is Cauchy if and only if each coordinate sequence $(x_{k,i})$ is Cauchy in $\mathbb{R}$.

Proof. Suppose $(x_k)$ is Cauchy. Given $\varepsilon>0$, choose $N$ such that $\|x_k-x_l\|<\varepsilon$ whenever $k,l\ge N$. The first key inequality gives $|x_{k,i}-x_{l,i}|\le\|x_k-x_l\|<\varepsilon$ for every coordinate $i$, so each coordinate sequence is Cauchy.

Conversely, suppose every coordinate sequence is Cauchy, and fix $\varepsilon>0$. For each $i$, choose $N_i$ such that $|x_{k,i}-x_{l,i}|<\varepsilon/\sqrt{m}$ whenever $k,l\ge N_i$. Set $N=\max\{N_1,\dots,N_m\}$. For $k,l\ge N$,

\|x_k-x_l\| =\sqrt{\sum_{i=1}^{m}|x_{k,i}-x_{l,i}|^2} <\sqrt{\sum_{i=1}^{m}\frac{\varepsilon^2}{m}} =\varepsilon.

Thus $(x_k)$ is Cauchy in the Euclidean metric. $\square$

Worked Example: A Vector Cauchy Sequence

In Euclidean $\mathbb{R}^2$, let $x_k=(1-2^{-k},\,3+3\cdot2^{-k})$, where $k\in\mathbb{N}$. Show that the sequence is Cauchy and find its limit.

For $k,l\in\mathbb{N}$, subtracting coordinates gives $x_k-x_l=(2^{-l}-2^{-k},\,3(2^{-k}-2^{-l}))$. Therefore

\|x_k-x_l\| =\sqrt{10}\,|2^{-k}-2^{-l}| \le \sqrt{10}(2^{-k}+2^{-l}).

If $k,l\ge N$, this is at most $2\sqrt{10}\,2^{-N}$. For any $\varepsilon>0$, choose $N\in\mathbb{N}$ with $2\sqrt{10}\,2^{-N}<\varepsilon$. The defining Cauchy inequality then holds for all $k,l\ge N$.

The coordinates converge to $1$ and $3$, so the coordinatewise theorem gives $x_k\to(1,3)$. Directly, the error is $x_k-(1,3)=(-2^{-k},\,3\cdot2^{-k})$, and hence $\|x_k-(1,3)\|=\sqrt{10}\,2^{-k}\to0$.

Check: At $k=2$, $2^{-k}=1/4$, so $x_2=(3/4,15/4)$. Its difference from $(1,3)$ is $(-1/4,3/4)$, whose norm is $\sqrt{1/16+9/16}=\sqrt{10}/4$. This matches $\sqrt{10}\,2^{-k}$ at $k=2$.

Completeness of Euclidean Space

Completeness of $\mathbb{R}$ means that every Cauchy sequence of real numbers has a real limit. Using that standard result from the earlier material on completeness of the real line, completeness in each coordinate will give completeness of the whole finite-dimensional space.

Theorem (Completeness of $\mathbb{R}^m$): For every $m\in\mathbb{N}$, $\mathbb{R}^m$ with the Euclidean metric is complete.

Proof. Let $(x_k)$ be a Cauchy sequence in Euclidean $\mathbb{R}^m$, and write $x_k=(x_{k,1},\dots,x_{k,m})$. By the Cauchy coordinate proposition, each real sequence $(x_{k,i})$ is Cauchy. Completeness of $\mathbb{R}$ gives a real number $x_i$ such that $x_{k,i}\to x_i$, for each $i\in\{1,\dots,m\}$. Thus $x=(x_1,\dots,x_m)$ is a point of $\mathbb{R}^m$. By the coordinatewise convergence theorem, $x_k\to x$ in the Euclidean metric. Every Cauchy sequence in $\mathbb{R}^m$ therefore converges to a point of $\mathbb{R}^m$, which is precisely completeness. $\square$

Worked Example: A Cauchy Sequence with a Coordinate Limit

Consider $y_k=(4+1/k,\,-2+1/k)$ in Euclidean $\mathbb{R}^2$. Verify the Cauchy property and identify the limit.

For any $k,l\in\mathbb{N}$, the coordinate differences are $1/k-1/l$ in both coordinates. Thus

\|y_k-y_l\| =\sqrt{2}\left|\frac{1}{k}-\frac{1}{l}\right| \le \sqrt{2}\left(\frac{1}{k}+\frac{1}{l}\right).

For $k,l\ge N$, this is at most $2\sqrt{2}/N$. Given $\varepsilon>0$, choose $N\in\mathbb{N}$ with $N>2\sqrt{2}/\varepsilon$. Then $\|y_k-y_l\|\le2\sqrt{2}/N<\varepsilon$, so the sequence is Cauchy. Its coordinates converge to $4$ and $-2$; consequently its limit is $y=(4,-2)$. Direct substitution gives $\|y_k-y\|=\sqrt{2}/k\to0$.

Check: At $k=2$, $y_2=(9/2,-3/2)$. Subtracting $y$ gives $(1/2,1/2)$, with norm $\sqrt{1/4+1/4}=\sqrt{2}/2$, as the error formula $\sqrt{2}/k$ gives.

A Common Pitfall: Forgetting the Dimension Factor

The Euclidean norm is not generally bounded above by the largest coordinate error alone. For example, in $\mathbb{R}^2$, let $v=(2/5,2/5)$. The largest coordinate error from the origin is $2/5$, but the Euclidean norm is $\|v\|=\sqrt{(2/5)^2+(2/5)^2}=2\sqrt{2}/5$, which is greater than $1/2$ because $8/25>1/4$. So a proposed argument that coordinate errors below $1/2$ automatically force Euclidean error below $1/2$ is false.

The correction is to control each coordinate error by $\varepsilon/\sqrt{m}$, not merely by $\varepsilon$. If there are $m$ coordinates and each error is strictly less than $\varepsilon/\sqrt{m}$, then the sum of their squares is strictly less than $m(\varepsilon/\sqrt{m})^2=\varepsilon^2$. This is the step that makes the reverse direction of the convergence and Cauchy criteria work.

Both results depend on having finitely many coordinates: the common index $N$ is chosen as the maximum of finitely many indices. The proof therefore makes no claim about sequences with infinitely many coordinates. In finite-dimensional Euclidean space, however, coordinatewise convergence and the Cauchy property exactly match their metric versions, and completeness follows directly from completeness of the real line.

Check Your Understanding

Use the Euclidean norm and the coordinate criteria proved above to answer each question.

For $x_k=(3-2/k,\,1+4/k)$ in $\mathbb{R}^2$, find the limit and compute $\|x_k-x\|$.
Explain why the estimate $\|x-y\|\le\max_i|x_i-y_i|$ fails in $\mathbb{R}^2$, giving a specific vector.
Show that if $(x_k)$ is Cauchy in Euclidean $\mathbb{R}^3$, then its second-coordinate sequence is Cauchy in $\mathbb{R}$.
Identify the exact use of completeness of $\mathbb{R}$ in the proof that $\mathbb{R}^m$ is complete.
For $z_k=(1/k,\,1/k,\,1/k)$ in $\mathbb{R}^3$, find the limit and calculate the Euclidean error.