**Proof plan.** The proof uses the [Baire category theorem](/theorems/630), applied to the complete [metric space](/page/Metric%20Space) $X$. The pointwise boundedness condition defines, for each generating seminorm $q_m$ on $Y$ and each $k \in \mathbb{N}$, a closed set $F_k^m$ on which all operators are simultaneously bounded in $q_m$. Baire's theorem forces one of these [sets](/page/Set) to have nonempty interior, from which equicontinuity follows by translation. **Step 1 (Constructing the Baire sets).** Fix a generating seminorm $q_m$ on $Y$ and define, for each $k \in \mathbb{N}$, \begin{align*} F_k^m &:= \{x \in X : q_m(T_\alpha x) \le k \text{ for all } \alpha \in A\}. \end{align*} [claim:The Sets Are Closed] Each $F_k^m$ is closed in $X$. [/claim] [proof] For each $\alpha \in A$, the map $x \mapsto q_m(T_\alpha x)$ is [continuous](/page/Continuity) (as the composition of the continuous map $T_\alpha$ with the continuous seminorm $q_m$). Therefore $\{x : q_m(T_\alpha x) \le k\}$ is closed, and $F_k^m = \bigcap_{\alpha \in A} \{x : q_m(T_\alpha x) \le k\}$ is an intersection of [closed sets](/page/Closed%20Set), hence closed. [/proof] **Step 2 (Applying Baire).** The pointwise boundedness condition states that for every $x \in X$, $\sup_{\alpha} q_m(T_\alpha x) < \infty$. This means $x \in F_k^m$ for some $k$, so $X = \bigcup_{k=1}^\infty F_k^m$. By the [Baire category theorem](/theorems/630), since $X$ is a complete metric space, at least one $F_{k_0}^m$ has nonempty interior: there exist $x_0 \in X$ and a basic neighbourhood $V = \{x : p_{n_1}(x - x_0) < \varepsilon, \ldots, p_{n_N}(x - x_0) < \varepsilon\}$ with $V \subseteq F_{k_0}^m$. **Step 3 (Extracting the seminorm bound).** [claim:Equicontinuity At The Origin] For all $x \in X$ with $p_{n_i}(x) < \varepsilon$ for $i = 1, \ldots, N$, and all $\alpha \in A$, \begin{align*} q_m(T_\alpha x) &\le 2k_0. \end{align*} [/claim] [proof] If $p_{n_i}(x) < \varepsilon$ for each $i$, then $x_0 + x \in V \subseteq F_{k_0}^m$, so $q_m(T_\alpha(x_0 + x)) \le k_0$. Also $x_0 \in F_{k_0}^m$, so $q_m(T_\alpha x_0) \le k_0$. By the triangle inequality, \begin{align*} q_m(T_\alpha x) &= q_m(T_\alpha(x_0 + x) - T_\alpha x_0) \le q_m(T_\alpha(x_0 + x)) + q_m(T_\alpha x_0) \le 2k_0. \end{align*} [/proof] **Step 4 (Rescaling to the equicontinuity estimate).** Let $C = 2k_0/\varepsilon$. For any $x \in X$ with $\max_{i} p_{n_i}(x) > 0$, set $\lambda = \varepsilon/(2\max_i p_{n_i}(x))$. Then $p_{n_i}(\lambda x) < \varepsilon$ for each $i$, so by Claim 2, $q_m(T_\alpha(\lambda x)) \le 2k_0$. By homogeneity, $q_m(T_\alpha x) \le (2k_0/\lambda) = (2k_0/\varepsilon) \cdot 2\max_i p_{n_i}(x)$. For $x$ with $\max_i p_{n_i}(x) = 0$, the same bound holds for $\lambda x$ for all $\lambda > 0$, giving $q_m(T_\alpha x) = 0$. In either case, \begin{align*} q_m(T_\alpha x) &\le C' \max_{1 \le i \le N} p_{n_i}(x) \quad \text{for all } x \in X, \text{ all } \alpha \in A, \end{align*} where $C' = 4k_0/\varepsilon$. This is exactly the equicontinuity condition: every $T_\alpha$ satisfies the same seminorm bound with the same constant and the same controlling seminorms. $\blacksquare$