I am trying to prove that the space ${\mathbb{R}}^k$ with the $\infty$-metric is a complete metric space.

I know that I need to show that every Cauchy sequence in the metric space ${\mathbb{R}}^k$ with the $\infty$-metric converges to a point $x\in{\mathbb{R}}^k$. Part of my work so far involved proving that the space ${\mathbb{R}}^k$ with the old Pythagorean norm is a complete metric space, but I’m not sure if I should be using that at all in this proof. Here goes:


Suppose that $k\in{\mathbb{N}}$ and that $\left(x_{n}\right)$ is a Cauchy sequence in the space ${\mathbb{R}}^k$ with the $\infty$-metric. Suppose that $\epsilon>0$. Using the fact that $\left(x_{n}\right)$ is a Cauchy sequence, choose an integer $N$ such that the inequality $||x_{m}-x_{n}||_\infty<\frac{\epsilon}{2}$ holds for all integers $m$ and $n$ such that $m\ge{N}$ and $n\ge{N}$.

This is where I’m stuck. I feel like I want to use the triangle inequality from here…

Since the space ${\mathbb{R}}^k$ is a metric space, we know that the inequality $$||{x_n}-{x_m}||_{\infty}\le||{x_n}-{x}||_{\infty}+||{x_m}-{x}||_{\infty}\lt\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$ holds for all positive integers $n$ and $m$.

However, I know that this isn’t a good thing to state. I think that I’m assuming that there is some point $x$ in the space that the sequence converges to… this is exactly what I'm trying to prove! I’m totally stuck. Any advice would be GREATLY appreciated!

EDIT: I'm including the updated version of my proof based on advice from Brian M. Scott. Here it is:


Suppose that $k\in{\mathbb{N}}$ and that $\left(x_{n}\right)$ is a Cauchy sequence in the space ${\mathbb{R}}^k$ with the $\infty$-metric.

Suppose that $\epsilon>0$.

For each $n\in\mathbb{Z}^{+}$ we have $x_{n}= \left(x_{n,1},x_{n,2},\ldots,x_{n,k}\right)$.

Consider the real-valued sequences $\left(x_{n,j}\right)$ for each $j=1,2,\ldots,k$.

Using the fact that $\left(x_{n}\right)$ is a Cauchy sequence, choose $N\in\mathbb{Z}$ such that the inequality $||x_{m} – x_{n}||_{\infty}\lt \frac{\epsilon}{2}$ for all $n\ge N$ and $m\ge N$.

Since the inequality $||x_{m,j} – x_{n,j}||_{\infty}\le||x_{m} – x_{n}||_{\infty}$ holds for all $m$ and $n$, we have the inequality $||x_{m,j} – x_{n,j}||_{\infty}\lt\frac{\epsilon}{2}$ for all $m\ge N$ and $n\ge N$.

Thus for $j=1,2,\ldots,k$ and all $n\in\mathbb{Z}^{+}$ the sequence $\left( x_{n,j} \right)$ is a Cauchy sequence in $\mathbb{R}$.

Using the fact that $\mathbb{R}$ is complete, define $$x\left(j\right)=\lim_{n\to\infty}x_{n,j}$$ for $j=1,2,\ldots,k$.

The point $x=\left(x_{1},x_{2},\cdots,x_{k}\right)$ is the limit of $\left(x_{n}\right)$ under the Pythagorean metric.

From the inequality $$||x-x_{n}||_\infty\le||x-x_{m}||_{\infty}+||x_{m}-x_{n}||_{\infty}\lt\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$ that holds for all $n\ge N$ and $m\ge N$ we see that the sequence $\left(x_{n}\right)$ converges to the point $x\in\mathbb{R}^{k}$ and the space $\mathbb{R}^{k}$ is a complete metric space.


HINT: There are at least two reasonable approaches.

  1. Start with a Cauchy sequence $\langle x^n:n\in\Bbb N\rangle$, where $x^n=\langle x_1^n,\dots,x_k^n\rangle$ for each $n\in\Bbb N$. Now look at the real-valued sequences $\langle x_i^n:n\in\Bbb N\rangle$ for $i=1,\dots,k$. Show that each is Cauchy in the usual metric on $\Bbb R$, let $x_i$ be the limit (since $\Bbb R$ is complete in the usual metric), and show that $\langle x^n:n\in\Bbb N\rangle$ converges to $\langle x_1,\dots,x_k\rangle$ in $\Bbb R^k$.

  2. Show that the $\infty$ metric is equivalent to the Euclidean metric. Then a sequence is Cauchy in one of these metrics if and only if it’s Cauchy in the other, it converges to a point $x$ in one if and only if it converges to $x$ in the other, and you already know that the Euclidean metric is complete.

  • $\begingroup$ In your second approach, what do you mean by "equivalent"? I thought they were different types of distance measure? I know that $||x||_{\infty}\le{||x||}$, but I'm not sure exactly what you mean. Thanks for the reply! $\endgroup$ May 20 '13 at 8:02
  • $\begingroup$ @Euclid'sCompass: You’re welcome. I mean this notion of strong equivalence: there are positive constants $\alpha$ and $\beta$ such that for all $x,y\in\Bbb R^k$, $\alpha\|x-y\|\le\|x-y\|_\infty\le\beta\|x-y\|$. (You may prefer the first approach: it’s much closer to what you had in mind initially.) $\endgroup$ May 20 '13 at 8:05
  • $\begingroup$ For the first approach, when you talk about the real-valued sequences $\langle x_i^n:n\in\Bbb N\rangle$ for $i=1,...,k$ are you basically saying focus on each axis individually? It seems like you're saying establish the fact that the coordinates on each axis will converge to a value in $\mathbb{R}$, and each limit on each axis will be the respective coordinate of the actual limit point. Am I interpreting that correctly? If so, how would I factor the $\infty$ metric into all of this? $\endgroup$ May 20 '13 at 8:29
  • $\begingroup$ @Euclid'sCompass: Yes, you’re interpreting it correctly. The $\infty$ metric is probably the easiest one with which to use this argument: for each coordinate $i$, $|x_i^m-x_i^n|\le\|x^m-x^n\|_\infty$ for all $m,n\in\Bbb N$. $\endgroup$ May 20 '13 at 16:46
  • $\begingroup$ $Brian M. Scott: I've revised my proof based on what you said, but I still feel like I'm not quite there. Will you let me know if I'm moving in the right direction? Please and thank you! By the way, the new proof is too long for this comment box, so I'm going to add it as an edit to the original post. Thanks again! $\endgroup$ May 21 '13 at 6:48

Since $$\max_{1\leq i\leq d}|x_i|^2\leq\sum_{i=1}^d |x_i|^2\leq d\ \max_{1\leq i\leq d}|x_i|^2$$ one has $$\|{\bf x}\|_\infty\leq\|{\bf x}\|_2\leq\sqrt{d}\>\|{\bf x}\|_\infty\qquad(x\in{\mathbb R}^d)\ .$$ Therefore the identity map $${\rm id}:\quad\bigl({\mathbb R}^d,\ \|\cdot\|_\infty\bigr)\ \to\ \bigl({\mathbb R}^d, \ \|\cdot\|_2\bigr)$$ is Lipschitz in both directions. This means that distances between points measured one way or the other differ by a factor at most $\sqrt{d}$. It follows that a sequence $({\bf x}_n)_{n\geq0}$ in ${\mathbb R}^d$ is Cauchy with respect to $\|\cdot\|_\infty$ iff it is Cauchy with respect to $\|\cdot\|_2$.

  • $\begingroup$ Thank you for the reply. This looks much simpler and quicker than the approach that I am trying to take, but I'm not sure what identity maps are, or what it means for one to be Lipschitz. I'm sure it'll make sense eventually though! Thanks again. $\endgroup$ May 22 '13 at 13:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.