Equivalence of Euclidean norm used in Ostrowski's Theorem

Question

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.

Answer 1

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.