Lemma 1

Under the hypothesis of prop. 7, the topology of $E$ is the finest topology for which all the mappings $v_n : E_n \to E$ are continuous.

We need to prove that, if $U$ is a subset of $E$ such that $U \cap E_n$ is open in $E_n$ for every $n$, then $U$ is open in $E$; in other words, we must prove that, for every $x \in U$, there exists a convex balanced set $V$ such that $x + V \subset U$ and that $V \cap E_n$ is a neighbourhood of 0 in $E_n$ for every large enough $n$ (II, p. 27, prop. 5). For every $n$, let $W_n$ be a convex balanced neighbourhood of 0 in $E_n$ such that the closure $H_n$ of $W_n$ in $E_{n+1}$ is compact. Let $x \in U$ and let $n_0$ be an integer such that $x \in E_{n_0}$. We shall construct, by induction, a sequence $(\varepsilon_n)_{n \geq 0}$ of scalars $> 0$ such that $x + \sum_{n_0 \leq i \leq n} \varepsilon_i H_i$ is contained in $U$ for $n \geq n_0$. Suppose that the $\varepsilon_i$ for $i < n$ have been constructed. If $n = n_0$, set $V_{n-1} = \{0\}$; if not, set

Then $V_{n-1}$ is compact in $E_n$, and a fortiori in $E_{n+1}$. Since $U \cap E_{n+1}$ is open in $E_{n+1}$, there exists a scalar $\varepsilon_n > 0$ such that $x + V_n = x + V_{n-1} + \varepsilon_n H_n$ is contained in $U$ (GT, II, § 4, No. 3, cor.). Let $V = \bigcup_{n \geq n_0} V_n$. Then $V$ is convex and balanced; for $n \geq n_0$, we have $V \cap E_n \supset \varepsilon_n H_n \cap E_n \supset \varepsilon_n W_n$, hence $V \cap E_n$ is a neighbourhood of 0 in $E_n$. This completes the proof of the lemma.

The set $U = E - \{0\}$ is such that the set $U \cap E_n = E_n - \{0\}$ is open in $E_n$ for every $n$, hence $U$ is open in $E$, which proves that $E$ is Hausdorff (GT, III, § 1, No. 3, prop. 2). It is clear that property (ii) implies (iii) and that (iii) implies (i). Finally we show that (i) implies (ii). For this it is enough to show that if a subset $A$ of $E$ is not absorbed by any of the sets $\sum_{0 \leq i \leq n} H_i$, then $A$ is not bounded. But then there exists a sequence $(x_n)_{n \geq 1}$ of points of $A$ such that, for every $n$, we have $x_n \notin n^2 \sum_{0 \leq i \leq n} H_i$.

Then the set of the $x_n / n$ is closed by virtue of lemma 1, since its intersection with $E_m$ is discrete for every integer $m$. The complement of the set of the $x_n / n$ is then an open neighbourhood of 0 which does not absorb the sequence $(x_n)$, hence $A$ is not bounded.