[proofplan] We prove the equivalences cyclically: (3) $\Rightarrow$ (1) is immediate from the definition; (1) $\Rightarrow$ (2) follows from the topological characterisation of the open unit ball; and (2) $\Rightarrow$ (3) is the substantial direction, which proceeds by showing that the condition $|x| < 1 \Leftrightarrow |x|' < 1$ forces the ratio $\log|x|/\log|x|'$ to be a constant $s > 0$, using a density-of-rationals argument to derive a contradiction from any pair of elements where the ratio differs. [/proofplan] [step:Show (3) $\Rightarrow$ (1): a power relation implies equivalent topologies] Suppose there exists $s \in \mathbb{R}_{> 0}$ such that $|x|' = |x|^s$ for all $x \in K$. Then for any sequence $(x_n)$ in $K$ and any $x \in K$, \begin{align*} |x_n - x| \to 0 \quad \Longleftrightarrow \quad |x_n - x|^s \to 0 \quad \Longleftrightarrow \quad |x_n - x|' \to 0, \end{align*} since $t \mapsto t^s$ is a homeomorphism of $\mathbb{R}_{\geq 0}$ (as $s > 0$). Thus $|\cdot|$ and $|\cdot|'$ define the same topology on $K$, so they are equivalent. [/step] [step:Show (1) $\Rightarrow$ (2): equivalent topologies preserve the open unit ball condition] Suppose $|\cdot|$ and $|\cdot|'$ are equivalent, i.e., they induce the same topology on $K$. For any $x \in K$, \begin{align*} |x| < 1 \quad \Longleftrightarrow \quad x^n \to 0 \text{ in the } |\cdot|\text{-topology} \quad \Longleftrightarrow \quad x^n \to 0 \text{ in the } |\cdot|'\text{-topology} \quad \Longleftrightarrow \quad |x|' < 1. \end{align*} The first and last equivalences use the fact that $|x^n| = |x|^n \to 0$ if and only if $|x| < 1$ (and similarly for $|\cdot|'$). The middle equivalence uses the hypothesis that the two topologies coincide. [guided] The key insight is that the condition "$|x| < 1$" is a purely topological statement: it is equivalent to "$x^n \to 0$," which depends only on the topology, not on the specific absolute value. Suppose $|\cdot|$ and $|\cdot|'$ induce the same topology on $K$. Let $x \in K$ with $|x| < 1$. Then $|x^n| = |x|^n \to 0$ as $n \to \infty$, so $x^n \to 0$ in the $|\cdot|$-topology. Since the two topologies coincide, $x^n \to 0$ in the $|\cdot|'$-topology as well. This means $|x^n|' = |x|'^n \to 0$, which forces $|x|' < 1$. This argument is symmetric in $|\cdot|$ and $|\cdot|'$, so condition (2) holds. [/guided] [/step] [step:Show (2) $\Rightarrow$ (3): the condition on the unit ball forces a power relation] Assume that $|x| < 1$ implies $|x|' < 1$ for all $x \in K$. [step:Extend the condition to $|x| > 1$ and $|x| = 1$] Let $x \in K$ with $|x| > 1$. Then $|x^{-1}| = |x|^{-1} < 1$, so by hypothesis $|x^{-1}|' = |x|'^{-1} < 1$, giving $|x|' > 1$. By contrapositive, $|x|' \leq 1$ implies $|x| \leq 1$, and combining with the original condition: $|x| = 1$ if and only if $|x|' = 1$. [guided] We have assumed that $|x| < 1 \Rightarrow |x|' < 1$. We need to extend this to a complete characterisation of the relationship between the two absolute values. Suppose $|x| > 1$. Then $|x^{-1}| = |x|^{-1} < 1$, so by hypothesis $|x^{-1}|' < 1$, which gives $|x|' = |x^{-1}|'^{-1} > 1$. Now suppose $|x| = 1$. If $|x|' < 1$, then by the forward direction $|x| < 1$, a contradiction. If $|x|' > 1$, then $|x^{-1}|' < 1$, which forces $|x^{-1}| < 1$, i.e., $|x| > 1$, again a contradiction. So $|x|' = 1$. In summary: $|x| < 1 \Leftrightarrow |x|' < 1$, $|x| > 1 \Leftrightarrow |x|' > 1$, and $|x| = 1 \Leftrightarrow |x|' = 1$. [/guided] [/step] [step:Fix an element with $|y| > 1$ and define the exponent $s$] Since $|\cdot|$ is non-trivial, there exists $y \in K$ with $|y| \neq 0, 1$; replacing $y$ by $y^{-1}$ if necessary, we may assume $|y| > 1$ (and hence $|y|' > 1$ by the previous step). Define \begin{align*} s := \frac{\log |y|'}{\log |y|} \in \mathbb{R}_{> 0}. \end{align*} We will show $|x|' = |x|^s$ for all $x \in K$. [/step] [step:Show $\log|x|' / \log|x| = s$ for all $x$ with $|x| > 1$ by a density argument] Let $x \in K$ with $|x| > 1$. Define $\alpha := \log|x| / \log|y|$ and $\alpha' := \log|x|' / \log|y|'$. We claim $\alpha = \alpha'$, which gives $|x|' = |y|'^{\alpha'} = |y|'^{\alpha} = (|y|^s)^\alpha = |x|^s$. Suppose for contradiction that $\alpha \neq \alpha'$; say $\alpha < \alpha'$ (the case $\alpha > \alpha'$ is symmetric). By density of $\mathbb{Q}$ in $\mathbb{R}$, choose $m, n \in \mathbb{Z}$ with $n > 0$ and \begin{align*} \alpha < \frac{m}{n} < \alpha'. \end{align*} From $\alpha < m/n$, we get $n \log|x| < m \log|y|$, so $|x^n| < |y^m|$, hence $|x^n y^{-m}| < 1$. By condition (2), $|x^n y^{-m}|' < 1$. From $m/n < \alpha'$, we get $m \log|y|' < n \log|x|'$, so $|y^m|' < |x^n|'$, hence $|x^n y^{-m}|' > 1$. This contradicts $|x^n y^{-m}|' < 1$. Therefore $\alpha = \alpha'$, and $|x|' = |x|^s$. [guided] This is the heart of the proof. We want to show that the ratio $\log|x|' / \log|x|$ is the same for every $x \in K$ with $|x| > 1$. The strategy is to assume two elements give different ratios and use the density of $\mathbb{Q}$ in $\mathbb{R}$ to construct an element that simultaneously has absolute value less than $1$ under one norm and greater than $1$ under the other — contradicting condition (2). Let $x \in K$ with $|x| > 1$. Set $\alpha = \log|x| / \log|y|$ and $\alpha' = \log|x|' / \log|y|'$. Suppose $\alpha < \alpha'$. By density of $\mathbb{Q}$ in $\mathbb{R}$, there exist integers $m, n$ with $n > 0$ such that $\alpha < m/n < \alpha'$. The left inequality $\alpha < m/n$ means $\log|x| / \log|y| < m/n$, i.e., $n\log|x| < m\log|y|$ (both logarithms are positive since $|x|, |y| > 1$). Exponentiating: $|x|^n < |y|^m$, so $|x^n / y^m| = |x^n| / |y^m| < 1$. By condition (2), $|x^n / y^m|' < 1$. The right inequality $m/n < \alpha'$ means $m\log|y|' < n\log|x|'$, so $|y|'^m < |x|'^n$, giving $|x^n / y^m|' = |x|'^n / |y|'^m > 1$. These two conclusions contradict each other, so $\alpha = \alpha'$. Therefore \begin{align*} \frac{\log|x|'}{\log|y|'} = \frac{\log|x|}{\log|y|}, \end{align*} which rearranges to $\log|x|' = s \cdot \log|x|$, i.e., $|x|' = |x|^s$. [/guided] [/step] [step:Extend to all $x \in K$ to conclude $|x|' = |x|^s$] For $x \in K$ with $|x| = 1$, we have $|x|' = 1 = |x|^s$. For $x \in K$ with $|x| < 1$ (and $x \neq 0$), we have $|x^{-1}| > 1$, so $|x^{-1}|' = |x^{-1}|^s$, giving $|x|' = |x^{-1}|'^{-1} = |x^{-1}|^{-s} = |x|^s$. For $x = 0$, both sides are $0$. Thus $|x|' = |x|^s$ for all $x \in K$, establishing condition (3). [/step] [/step]