Proposition 6

*Let $E$ be a locally convex space which is the strict inductive limit of an increasing sequence $(E_n)$ of closed vector subspaces of $E$ (II, p. 33). A subset $B$ of $E$ is bounded if and only if it is contained in one of the subspaces $E_n$, and is bounded in this subspace.

The condition is sufficient, since the topology induced on $E_n$ by that of $E$ is precisely the given topology of $E_n$ (II, p. 32, prop. 9). To see that the condition is necessary, it is enough (III, p. 4, prop. 3) to prove that if a sequence $(x_m)$ of points of $E$ is not contained in any of the subspaces $E_n$, then it cannot tend to 0. By extracting a subsequence of the sequence $(x_m)$, we can assume that there exists a strictly increasing sequence $(n_k)$ of integers such that, for every index $k$, we have $x_k \notin E_{n_k}$ and $x_k \in E_{n_{k+1}}$. Then there exists (II, p. 33, lemma 2) an increasing sequence $(V_k)$ of convex sets such that $V_k$ is a neighbourhood of 0 in $E_{n_k}$, $V_{k+1} \cap E_{n_k} = V_k$ and such that $x_k \notin V_{k+1}$ for every index $k$. The union $V$ of the $V_k$ is then a neighbourhood of 0 in $E$, and we have $x_k \notin V$ for all $k$. This proves that the sequence $(x_k)$ does not tend to 0.*

The conclusion of prop. 6 is not necessarily true for a space E which is the inductive limit of a non-denumerable directed set of closed subspaces of E (cf. III, p. 38, exerc. 7).