**Proof plan.** We first verify that $d$ is a well-defined translation-invariant metric by checking the metric axioms using properties of the [function](/page/Function) $t \mapsto t/(1+t)$. We then show that the identity map $(X, \tau_{\mathcal{P}}) \to (X, \tau_d)$ is a homeomorphism by verifying that each topology's basic [open sets](/page/Open%20Set) are open in the other. The convergence and Cauchy characterisations follow directly from the equivalence of [topologies](/page/Topology). **Step 1 (The function $\varphi$ is subadditive and monotone).** Define $\varphi: [0, \infty) \to [0, \infty)$ by $\varphi(t) = t/(1+t)$. Then $\varphi$ is strictly increasing, $\varphi(0) = 0$, and $0 \le \varphi(t) < 1$ for all $t \ge 0$. [claim:Subadditivity Of Phi] For all $s, t \ge 0$, $\varphi(s + t) \le \varphi(s) + \varphi(t)$. [/claim] [proof] We compute directly: \begin{align*} \varphi(s) + \varphi(t) - \varphi(s+t) &= \frac{s}{1+s} + \frac{t}{1+t} - \frac{s+t}{1+s+t}. \end{align*} Bringing the three fractions to a common denominator $(1+s)(1+t)(1+s+t)$, the numerator becomes \begin{align*} &s(1+t)(1+s+t) + t(1+s)(1+s+t) - (s+t)(1+s)(1+t) \\ &= s(1+t+s+st+t+t^2) + t(1+s+s+s^2+t+st) \\ &\quad - (s+t)(1+s+t+st) \\ &= st(s+t) + st \\ &\ge 0. \end{align*} A cleaner approach: since $\varphi$ is concave on $[0,\infty)$ (as $\varphi''(t) = -2/(1+t)^3 < 0$), it is subadditive on $[0,\infty)$ with $\varphi(0) = 0$. [/proof] **Step 2 ($d$ is a translation-invariant metric).** Each term $2^{-n} \varphi(p_n(x-y))$ is non-negative and bounded by $2^{-n}$, so the [series](/page/Series) converges and $d(x,y) \le 1$. Translation invariance is immediate: $p_n((x+z)-(y+z)) = p_n(x-y)$. We verify the metric axioms: - *Positivity and identity of indiscernibles:* $d(x,y) \ge 0$ is clear. If $d(x,y) = 0$, then each term $\varphi(p_n(x-y)) = 0$, so $p_n(x-y) = 0$ for every $n$ (since $\varphi$ is strictly increasing with $\varphi(0)=0$). Because $\{p_n\}$ is separating, $x - y = 0$. - *Symmetry:* $p_n(x - y) = p_n(y - x)$ by the seminorm axiom $p_n(\lambda z) = |\lambda| p_n(z)$ with $\lambda = -1$. - *Triangle inequality:* For any $x, y, z \in X$, \begin{align*} d(x,z) &= \sum_{n=1}^\infty 2^{-n} \varphi(p_n(x - z)) \\ &\le \sum_{n=1}^\infty 2^{-n} \varphi(p_n(x-y) + p_n(y-z)) \\ &\le \sum_{n=1}^\infty 2^{-n} \bigl[\varphi(p_n(x-y)) + \varphi(p_n(y-z))\bigr] \\ &= d(x,y) + d(y,z), \end{align*} where the first inequality uses the triangle inequality for $p_n$ and the monotonicity of $\varphi$, and the second uses the subadditivity of $\varphi$ (Claim 1). **Step 3 (The topologies coincide).** We show each topology is contained in the other. [claim:Seminorm Topology Is Finer Than Metric Topology] Every $d$-ball $B_d(x, r)$ is open in $\tau_{\mathcal{P}}$. [/claim] [proof] Fix $x \in X$ and $r > 0$. Choose $N$ large enough that $\sum_{n > N} 2^{-n} < r/2$. For each $n \le N$, the map $y \mapsto p_n(x - y)$ is $\tau_{\mathcal{P}}$-[continuous](/page/Continuity). The [set](/page/Set) $\{y : \sum_{n=1}^N 2^{-n} \varphi(p_n(x - y)) < r/2\}$ is $\tau_{\mathcal{P}}$-open (as a finite intersection of preimages of open sets under continuous maps), and it is contained in $B_d(x, r)$ because \begin{align*} d(x,y) &= \sum_{n=1}^N 2^{-n}\varphi(p_n(x-y)) + \sum_{n > N} 2^{-n}\varphi(p_n(x-y)) < \frac{r}{2} + \frac{r}{2} = r. \end{align*} [/proof] [claim:Metric Topology Is Finer Than Seminorm Topology] For every basic $\tau_{\mathcal{P}}$-neighbourhood $U = \{y : p_{n_1}(x - y) < \varepsilon, \ldots, p_{n_m}(x - y) < \varepsilon\}$ of $x$, there exists $\delta > 0$ such that $B_d(x, \delta) \subseteq U$. [/claim] [proof] Let $N = \max(n_1, \ldots, n_m)$ and let $\varepsilon > 0$ be given. If $d(x, y) < \delta$, then in particular $2^{-n_k} \varphi(p_{n_k}(x - y)) < \delta$ for each $k$, so $\varphi(p_{n_k}(x-y)) < 2^{n_k} \delta$. Since $\varphi^{-1}(s) = s/(1-s)$ is increasing, choosing $\delta$ small enough that $2^{N}\delta/(1 - 2^{N}\delta) < \varepsilon$ ensures $p_{n_k}(x - y) < \varepsilon$ for all $k$. [/proof] **Step 4 (Convergence and Cauchy characterisations).** Since $\tau_d = \tau_{\mathcal{P}}$, a [sequence](/page/Sequence) $x_k \to x$ in $(X, d)$ if and only if $x_k \to x$ in $\tau_{\mathcal{P}}$. By definition of the seminorm topology, $x_k \to x$ in $\tau_{\mathcal{P}}$ if and only if $p_n(x_k - x) \to 0$ for every $n$. The Cauchy characterisation follows identically: $\{x_k\}$ is $d$-Cauchy if and only if it is Cauchy in every seminorm, because the metric balls and seminorm balls generate the same topology. $\blacksquare$