The strategy is to construct a topology $\tau_\mathcal{P}$ explicitly from translates of the members of $\mathcal{B}_0$ and verify that it has all the required properties. The proof proceeds in seven steps. Steps 1–2 establish the algebraic and [topological](/page/Topology) properties of $\mathcal{B}_0$ needed to form a base: the members are convex, balanced, absorbing, and satisfy an interior point property that ensures the translates cover intersections correctly. Step 3 verifies the base axioms for $\{y + V : y \in X, V \in \mathcal{B}_0\}$ and constructs $\tau_\mathcal{P}$. Steps 4–5 verify the TVS axioms ([continuity](/page/Continuity) of addition and scalar multiplication), using subadditivity and absolute homogeneity of seminorms. Step 6 establishes continuity of each seminorm via the reverse triangle inequality. Step 7 proves minimality by showing that any competing locally convex TVS topology must contain all basic neighbourhoods, using only the continuity of seminorms and the TVS translation structure. **Step 1: Algebraic properties of $\mathcal{B}_0$.** [claim:Algebraic Properties Of The Neighbourhood Filter Base] Every member of $\mathcal{B}_0$ contains $0$, is convex, balanced, and absorbing. The collection $\mathcal{B}_0$ is closed under finite intersections. [/claim] [proof] Let $V = \bigcap_{i=1}^{m} \{x \in X : p_{\alpha_i}(x) < \varepsilon_i\} \in \mathcal{B}_0$. *Contains $0$:* For each $i \in \{1, \ldots, m\}$, absolute homogeneity gives $p_{\alpha_i}(0) = p_{\alpha_i}(0 \cdot 0) = |0| \cdot p_{\alpha_i}(0) = 0 < \varepsilon_i$, so $0 \in V$. *Convex:* Let $x, y \in V$ and $t \in [0,1]$. For each $i$, subadditivity and absolute homogeneity give \begin{align*} p_{\alpha_i}(tx + (1-t)y) &\leq p_{\alpha_i}(tx) + p_{\alpha_i}((1-t)y) = t\, p_{\alpha_i}(x) + (1-t)\, p_{\alpha_i}(y). \end{align*} Since $x, y \in V$, we have $p_{\alpha_i}(x) < \varepsilon_i$ and $p_{\alpha_i}(y) < \varepsilon_i$, so the right-hand side satisfies $t\, p_{\alpha_i}(x) + (1-t)\, p_{\alpha_i}(y) < t\varepsilon_i + (1-t)\varepsilon_i = \varepsilon_i$. Therefore $tx + (1-t)y \in V$. *Balanced:* Let $x \in V$ and $\lambda \in \mathbb{R}$ (or $\mathbb{C}$) with $|\lambda| \leq 1$. For each $i$, absolute homogeneity gives $p_{\alpha_i}(\lambda x) = |\lambda|\, p_{\alpha_i}(x) \leq p_{\alpha_i}(x) < \varepsilon_i$, so $\lambda x \in V$. *Absorbing:* Let $x \in X$ be arbitrary. Define $t := 1 + \max_{1 \leq i \leq m} p_{\alpha_i}(x)/\varepsilon_i$. Since each $p_{\alpha_i}(x) \geq 0$ and each $\varepsilon_i > 0$, we have $t \geq 1 > 0$. For each $i$, absolute homogeneity gives $p_{\alpha_i}(x/t) = p_{\alpha_i}(x)/t$. Since $t \geq 1 + p_{\alpha_i}(x)/\varepsilon_i > p_{\alpha_i}(x)/\varepsilon_i$, we obtain $p_{\alpha_i}(x)/t < \varepsilon_i$, so $x/t \in V$, giving $x \in tV$. *Closed under finite intersections:* Let $V = \bigcap_{i=1}^{m} \{x : p_{\alpha_i}(x) < \varepsilon_i\}$ and $W = \bigcap_{j=1}^{l} \{x : p_{\beta_j}(x) < \delta_j\}$ be members of $\mathcal{B}_0$. Then \begin{align*} V \cap W &= \bigcap_{i=1}^{m} \{x : p_{\alpha_i}(x) < \varepsilon_i\} \cap \bigcap_{j=1}^{l} \{x : p_{\beta_j}(x) < \delta_j\}, \end{align*} which is a finite intersection of $m + l$ individual seminorm balls, hence a member of $\mathcal{B}_0$. [/proof] **Step 2: The interior point property.** The algebraic properties of Step 1 are not sufficient to guarantee that the translates of $\mathcal{B}_0$ form a topological base. The missing ingredient is that each member of $\mathcal{B}_0$ must be "open with respect to itself": every point inside $V$ must have room for a smaller basic neighbourhood around it that still fits inside $V$. This is the content of the following claim. [claim:Interior Point Property] For every $V \in \mathcal{B}_0$ and every $v \in V$, there exists $W \in \mathcal{B}_0$ such that $v + W \subseteq V$. [/claim] [proof] Write $V = \bigcap_{i=1}^{m} \{x \in X : p_{\alpha_i}(x) < \varepsilon_i\}$. Since $v \in V$, we have $p_{\alpha_i}(v) < \varepsilon_i$ for each $i \in \{1, \ldots, m\}$. Define the positive reals $\delta_i := \varepsilon_i - p_{\alpha_i}(v) > 0$ for each $i$, and set \begin{align*} W &:= \bigcap_{i=1}^{m} \{x \in X : p_{\alpha_i}(x) < \delta_i\} \in \mathcal{B}_0. \end{align*} For any $w \in W$, subadditivity gives \begin{align*} p_{\alpha_i}(v + w) &\leq p_{\alpha_i}(v) + p_{\alpha_i}(w) < p_{\alpha_i}(v) + \delta_i = p_{\alpha_i}(v) + (\varepsilon_i - p_{\alpha_i}(v)) = \varepsilon_i \end{align*} for every $i$, so $v + w \in V$. Therefore $v + W \subseteq V$. [/proof] **Step 3: The translates of $\mathcal{B}_0$ form a base for a topology.** Define the collection $\mathcal{B} := \{y + V : y \in X,\, V \in \mathcal{B}_0\}$. We now verify that $\mathcal{B}$ satisfies the two base axioms for a topology on $X$. [claim:Base Axioms] The collection $\mathcal{B}$ satisfies: (B1) Every point of $X$ belongs to some member of $\mathcal{B}$. (B2) If $B_1, B_2 \in \mathcal{B}$ and $z \in B_1 \cap B_2$, there exists $B_3 \in \mathcal{B}$ with $z \in B_3 \subseteq B_1 \cap B_2$. [/claim] [proof] *(B1):* Let $y \in X$. Pick any $V \in \mathcal{B}_0$ (the collection is non-empty since it contains, for example, $\{x : p_{\alpha_1}(x) < 1\}$ for any $\alpha_1 \in A$). By the [Algebraic Properties Of The Neighbourhood Filter Base] claim, $0 \in V$, so $y = y + 0 \in y + V \in \mathcal{B}$. *(B2):* Suppose $z \in (y_1 + V_1) \cap (y_2 + V_2)$ where $V_1, V_2 \in \mathcal{B}_0$. Then $z = y_1 + v_1 = y_2 + v_2$ for some $v_1 \in V_1$ and $v_2 \in V_2$. By the [Interior Point Property] claim, there exist $W_1, W_2 \in \mathcal{B}_0$ with $v_1 + W_1 \subseteq V_1$ and $v_2 + W_2 \subseteq V_2$. By the [Algebraic Properties Of The Neighbourhood Filter Base] claim, $W_1 \cap W_2 \in \mathcal{B}_0$. Set $W := W_1 \cap W_2$. Since $0 \in W$, we have $z = y_1 + v_1 \in y_1 + v_1 + W = z + W$, so $z \in z + W$. For any $w \in W$, we have $w \in W_1$, so $v_1 + w \in v_1 + W_1 \subseteq V_1$, giving $z + w = y_1 + (v_1 + w) \in y_1 + V_1$. Similarly $w \in W_2$, so $v_2 + w \in v_2 + W_2 \subseteq V_2$, giving $z + w = y_2 + (v_2 + w) \in y_2 + V_2$. Therefore $z + W \subseteq (y_1 + V_1) \cap (y_2 + V_2)$, and $z + W \in \mathcal{B}$. [/proof] Since $\mathcal{B}$ satisfies (B1) and (B2), there is a unique topology $\tau_\mathcal{P}$ on $X$ whose [open sets](/page/Open%20Set) are exactly the unions of members of $\mathcal{B}$. In this topology, $\{y + V : V \in \mathcal{B}_0\}$ is a neighbourhood base at each point $y \in X$: every open set containing $y$ is a union of members of $\mathcal{B}$, so it contains some $y + V$, and conversely each $y + V$ is a union of members of $\mathcal{B}$ (by the interior point property, for every $v \in V$ the set $y + v + W \subseteq y + V$ is a member of $\mathcal{B}$, so $y + V = \bigcup_{v \in V} (y + v + W_v)$ is open). In particular, $\tau_\mathcal{P}$ is locally convex: every neighbourhood of $y$ contains a basic neighbourhood $y + V$, and $y + V$ is convex because $V$ is convex (by the [Algebraic Properties Of The Neighbourhood Filter Base] claim) and translates of convex [sets](/page/Set) are convex. **Step 4: Continuity of addition.** [claim:Continuity Of Addition] The map $(x, y) \mapsto x + y$ from $(X \times X, \tau_\mathcal{P} \times \tau_\mathcal{P})$ to $(X, \tau_\mathcal{P})$ is continuous. [/claim] [proof] Fix $a, b \in X$ and let $V \in \mathcal{B}_0$. We must find neighbourhoods $U_1$ of $a$ and $U_2$ of $b$ in $\tau_\mathcal{P}$ such that $U_1 + U_2 \subseteq (a + b) + V$. Write $V = \bigcap_{i=1}^{m} \{x : p_{\alpha_i}(x) < \varepsilon_i\}$ and define \begin{align*} W &:= \bigcap_{i=1}^{m} \{x : p_{\alpha_i}(x) < \varepsilon_i / 2\} \in \mathcal{B}_0. \end{align*} Set $U_1 := a + W$ and $U_2 := b + W$, which are basic $\tau_\mathcal{P}$-neighbourhoods of $a$ and $b$ respectively. For any $u \in U_1$ and $v \in U_2$, write $u = a + w_1$ and $v = b + w_2$ with $w_1, w_2 \in W$. Then $u + v = (a + b) + (w_1 + w_2)$, and for each $i$, subadditivity gives \begin{align*} p_{\alpha_i}(w_1 + w_2) &\leq p_{\alpha_i}(w_1) + p_{\alpha_i}(w_2) < \frac{\varepsilon_i}{2} + \frac{\varepsilon_i}{2} = \varepsilon_i, \end{align*} so $w_1 + w_2 \in V$ and $u + v \in (a + b) + V$. Therefore $U_1 + U_2 \subseteq (a + b) + V$. [/proof] **Step 5: Continuity of scalar multiplication.** [claim:Continuity Of Scalar Multiplication] The map $(\lambda, x) \mapsto \lambda x$ from $(\mathbb{R} \times X, \tau_{\mathrm{std}} \times \tau_\mathcal{P})$ to $(X, \tau_\mathcal{P})$ is continuous (and similarly over $\mathbb{C}$). [/claim] [proof] Fix $\lambda_0 \in \mathbb{R}$, $x_0 \in X$, and $V = \bigcap_{i=1}^{m} \{x : p_{\alpha_i}(x) < \varepsilon_i\} \in \mathcal{B}_0$. We must find $\delta > 0$ and $W \in \mathcal{B}_0$ such that $|\lambda - \lambda_0| < \delta$ and $x - x_0 \in W$ together imply $\lambda x - \lambda_0 x_0 \in V$. The identity $\lambda x - \lambda_0 x_0 = \lambda(x - x_0) + (\lambda - \lambda_0)x_0$, combined with subadditivity and absolute homogeneity, gives \begin{align*} p_{\alpha_i}(\lambda x - \lambda_0 x_0) &\leq |\lambda| \cdot p_{\alpha_i}(x - x_0) + |\lambda - \lambda_0| \cdot p_{\alpha_i}(x_0) \end{align*} for each $i \in \{1, \ldots, m\}$. We choose the parameters to make each term less than $\varepsilon_i / 2$. *Controlling the second term:* For each $i$, if $p_{\alpha_i}(x_0) = 0$ then $|\lambda - \lambda_0| \cdot p_{\alpha_i}(x_0) = 0 < \varepsilon_i / 2$ for any $\delta$. If $p_{\alpha_i}(x_0) > 0$, the condition $|\lambda - \lambda_0| \cdot p_{\alpha_i}(x_0) < \varepsilon_i / 2$ requires $|\lambda - \lambda_0| < \varepsilon_i / (2 p_{\alpha_i}(x_0))$. Since there are finitely many indices $i$, we may choose \begin{align*} \delta &:= \min\left(1,\; \min_{i:\, p_{\alpha_i}(x_0) > 0} \frac{\varepsilon_i}{2\, p_{\alpha_i}(x_0)}\right) > 0. \end{align*} With $|\lambda - \lambda_0| < \delta$, the second term satisfies $|\lambda - \lambda_0| \cdot p_{\alpha_i}(x_0) < \varepsilon_i / 2$ for every $i$. The choice $\delta \leq 1$ also gives $|\lambda| \leq |\lambda_0| + 1$. *Controlling the first term:* Define $\eta_i := \varepsilon_i / (2(|\lambda_0| + 1)) > 0$ for each $i$, and set \begin{align*} W &:= \bigcap_{i=1}^{m} \{h \in X : p_{\alpha_i}(h) < \eta_i\} \in \mathcal{B}_0. \end{align*} If $x - x_0 \in W$, then $p_{\alpha_i}(x - x_0) < \eta_i$ for each $i$, so $|\lambda| \cdot p_{\alpha_i}(x - x_0) \leq (|\lambda_0| + 1) \eta_i = \varepsilon_i / 2$. *Combining:* If $|\lambda - \lambda_0| < \delta$ and $x - x_0 \in W$, then for every $i$, \begin{align*} p_{\alpha_i}(\lambda x - \lambda_0 x_0) &\leq |\lambda| \cdot p_{\alpha_i}(x - x_0) + |\lambda - \lambda_0| \cdot p_{\alpha_i}(x_0) < \frac{\varepsilon_i}{2} + \frac{\varepsilon_i}{2} = \varepsilon_i, \end{align*} so $\lambda x - \lambda_0 x_0 \in V$, and therefore $\lambda x \in \lambda_0 x_0 + V$. [/proof] Steps 3–5 together establish Part 1 of the theorem: $\tau_\mathcal{P}$ is a locally convex TVS topology on $X$ with neighbourhood base $\{y + V : V \in \mathcal{B}_0\}$ at each point $y$. **Step 6: Each seminorm is $\tau_\mathcal{P}$-continuous (Part 2).** [claim:Seminorm Continuity] For every $\alpha \in A$, the seminorm $p_\alpha: (X, \tau_\mathcal{P}) \to [0, \infty)$ is continuous. [/claim] [proof] We establish the reverse triangle inequality for seminorms: for all $x, y \in X$, \begin{align*} |p_\alpha(x) - p_\alpha(y)| &\leq p_\alpha(x - y). \end{align*} Indeed, subadditivity gives $p_\alpha(x) = p_\alpha((x - y) + y) \leq p_\alpha(x - y) + p_\alpha(y)$, so $p_\alpha(x) - p_\alpha(y) \leq p_\alpha(x - y)$. Exchanging $x$ and $y$ gives $p_\alpha(y) - p_\alpha(x) \leq p_\alpha(y - x) = p_\alpha(x - y)$, where the last equality uses absolute homogeneity with $\lambda = -1$. Combining yields $|p_\alpha(x) - p_\alpha(y)| \leq p_\alpha(x - y)$. Now fix $y \in X$ and $\varepsilon > 0$. The set $U := y + \{h \in X : p_\alpha(h) < \varepsilon\}$ is a basic $\tau_\mathcal{P}$-neighbourhood of $y$ (the case $m = 1$, $\alpha_1 = \alpha$, $\varepsilon_1 = \varepsilon$). For every $x \in U$, we have $p_\alpha(x - y) < \varepsilon$, and the reverse triangle inequality gives $|p_\alpha(x) - p_\alpha(y)| \leq p_\alpha(x - y) < \varepsilon$. Since $y$ and $\varepsilon$ were arbitrary, $p_\alpha$ is continuous at every point. [/proof] **Step 7: Minimality — $\tau_\mathcal{P}$ is the coarsest such topology (Part 3).** [claim:Minimality] If $\tau'$ is any locally convex TVS topology on $X$ in which every $p_\alpha$ is continuous, then $\tau_\mathcal{P} \subseteq \tau'$. [/claim] [proof] We show that every member of $\mathcal{B}_0$ is a $\tau'$-neighbourhood of $0$, and then use the TVS structure of $\tau'$ to extend this to all translates. Fix $\alpha \in A$ and $\varepsilon > 0$. Since $p_\alpha: (X, \tau') \to [0, \infty)$ is $\tau'$-continuous by hypothesis and $[0, \varepsilon)$ is open in $[0, \infty)$, the preimage $p_\alpha^{-1}([0, \varepsilon)) = \{x \in X : p_\alpha(x) < \varepsilon\}$ is $\tau'$-open. This set contains $0$ (since $p_\alpha(0) = 0$), so it is a $\tau'$-neighbourhood of $0$. Since any finite intersection of $\tau'$-neighbourhoods of $0$ is again a $\tau'$-neighbourhood of $0$ (a standard property of topological neighbourhoods), every $V \in \mathcal{B}_0$ is a $\tau'$-neighbourhood of $0$. Now let $y \in X$ and $V \in \mathcal{B}_0$. Since $\tau'$ is a TVS topology, the addition map is $\tau'$-continuous, and in particular the translation map $x \mapsto y + x$ is a $\tau'$-homeomorphism (its inverse is translation by $-y$, which is also continuous). Therefore $y + V$ is a $\tau'$-neighbourhood of $y$. Finally, let $G \in \tau_\mathcal{P}$. For every $y \in G$, the definition of $\tau_\mathcal{P}$ provides $V_y \in \mathcal{B}_0$ with $y + V_y \subseteq G$. Since $y + V_y$ is a $\tau'$-neighbourhood of $y$, it contains a $\tau'$-open set $O_y$ with $y \in O_y \subseteq y + V_y \subseteq G$. Then $G = \bigcup_{y \in G} O_y$ is a union of $\tau'$-open sets, hence $\tau'$-open. Therefore $\tau_\mathcal{P} \subseteq \tau'$. [/proof] Uniqueness follows from minimality: if $\tau$ and $\tau'$ are both coarsest locally convex TVS topologies making all seminorms continuous, then $\tau \subseteq \tau'$ (by minimality of $\tau$) and $\tau' \subseteq \tau$ (by minimality of $\tau'$), so $\tau = \tau'$.