Equivalence of Euclidean norm used in Ostrowski's Theorem 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.
 A: 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. 
