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In a proof of Ostrowski's theorem (The only nontrivial norms on $\mathbb{Q}$ are $\|\cdot\|_{p}$ and $\|\cdot\|_{\infty}$), we come to the point where we have shown a certain norm, $\|\cdot\|$, has $\|n\|=n^{\alpha}$ for all natural numbers $n$ and a posititve constant $\alpha$. How is it that we can immediately conclude that $\|\cdot\|$ is equivalent to $\|\cdot\|_{\infty}$? (Two norms are equivalent if they induce equivalent metrics. Two metrics are equivalent if they have the same Cauchy sequences.) Thank you for any help or insight you can give.

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$||\cdot||_{\infty}$ is the Euclidean norm on $\mathbb{Q}$ –  Bendito Jan 7 '12 at 17:14
...isn't it usually the max norm? –  gnometorule Jan 7 '12 at 17:17 The definition in the link says that two absolute values on a field are equivalent if one is a power of the other. Maybe the definition used in your proof is the same. Then there is nothing to prove. –  Beni Bogosel Jan 7 '12 at 17:18
Thanks Beni. Now that I see we have proved $||n||=||n||_{\infty}^{\alpha}$ for all natural numbers, we have it for all rationals, and I am just left with changing my defintion of equivalence –  Bendito Jan 7 '12 at 17:26

1 Answer 1

up vote 1 down vote accepted

Given $\| n \| = n^{\alpha}$ and the property $\| x y \| = \| x \| \cdot \|y \|$, we can derive $\| x \| = \| x \|_{\infty}^{\alpha}$ for all $x \in \mathbb Q$. To prove this, take $x = \frac{m}{n}$, or $m = x \cdot n$; then $\| m \| = \| x \| \cdot \| n \|$.

It remains to show that if two norms $\| \cdot \|$ and $\| \cdot \|_{\infty}$ on a field are (positive) powers of each other, then they are topologically equivalent.

Suppose the sequence $(x_n)$ is Cauchy w.r.t. $\| \cdot \|$. Therefore, given $\varepsilon > 0$, there exists $N$ such that for all $n, m \geqslant N$, we have $\| x_n - x_m \| \leqslant \varepsilon$. Therefore, $\| x_n - x_m \|_{\infty}^{\alpha} \leqslant \varepsilon$, or $\| x_n - x_m \|_{\infty} \leqslant \varepsilon^{1/\alpha}$ for all $n, m \geqslant N$. Since the final statement is true for all $\varepsilon > 0$, it follows that $(x_n)$ is Cauchy w.r.t. $\| \cdot \|_{\infty}$.

Reversing the above argument, we conclude that if $(x_n)$ is Cauchy w.r.t. $\| \cdot \|_{\infty}$, then it is Cauchy w.r.t. $\| \cdot \|$ as well. Thus the norms $\| \cdot \|$ and $\| \cdot \|_{\infty}$ have the same set of Cauchy sequences; i.e., they are topologically equivalent.

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Very nice. Thanks for the help! –  Bendito Jan 7 '12 at 17:28

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