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:
Proof:
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:
Proof:
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.
Best Answer
HINT: There are at least two reasonable approaches.
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 $\Bbb R^k$.
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.