Proposition 5

*Let $(E_i)_{i \in I}$ be a family of Hausdorff locally convex spaces, and let $E$ be the topological direct sum of this family (II, p. 29). In order that a subset $B$ of $E$ be bounded, it is necessary and sufficient that there exists a finite subset $J$ of $I$ such that $\operatorname{pr}_i(B)$ is bounded in $E_i$ for $i \in J$ and $\operatorname{pr}_i(B) \subset \{0\}$ for all $i \notin J$.

Let $J$ be a finite subset of $I$. Since the topology of $E$ induces the product topology on $\prod_{j \in J} E_j$ (II, p. 30, prop. 7 and p. 31, prop. 8), it follows from III, p. 4, cor. 2 that the condition is sufficient.

Conversely, let $B$ be a bounded subset of $E$. Then $\operatorname{pr}_i(B)$ is bounded for all $i$ (III, p. 4, cor. 1). Therefore it is enough to prove that there exists a finite subset $J$ of $I$ such that $\operatorname{pr}_i(B) \subset \{0\}$ for all $i \notin J$. If not, then there exists an infinite sequence $(i_n)$ of distinct elements of $I$ and an infinite sequence $(x_n)$ of elements of $B$ such that $\operatorname{pr}_{i_n}(x_n) \neq 0$. Since $E_{i_n}$ is Hausdorff, there exists a continuous semi-norm $p_n$ on $E_{i_n}$ such that $p_n(\operatorname{pr}_{i_n}(x_n)) \geq n$. Hence $p = \sum_{n \geq 1} p_n \circ \operatorname{pr}_{i_n}$ is a continuous semi-norm on $E$ and $p$ is not bounded on $B$, which is a contradiction.*