Proposition 9

Suppose that the topology induced on $E_n$ by $\mathscr{T}_{n+1}$ is identical with $\mathscr{T}_n$ for every integer $n$. Then (i) The topology induced by $\mathscr{T}$ on $E_n$ is identical with $\mathscr{T}_n$ for each $n$; if the $\mathscr{T}_n$ are Hausdorff then $\mathscr{T}$ is Hausdorff. (ii) If, for every $n$, $E_n$ is closed in $E_{n+1}$ (for $\mathscr{T}_{n+1}$), then $E_n$ is closed in $E$ (using $\mathscr{T}$) for every $n$. (iii) If each $E_n$ is complete (using $\mathscr{T}_n$) then $E$ is complete using $\mathscr{T}$.

(i) To prove the first assertion, it is sufficient to prove that the topology $\mathscr{T}'_n$ induced by $\mathscr{T}$ on $E_n$ is finer than $\mathscr{T}_n$. For this, let $V_n$ be a convex neighbourhood of $0$ in $E_n$ for the topology $\mathscr{T}_n$; we are going to construct an increasing sequence of convex neighbourhoods of $0$ in $E_{n+p}$ for $\mathscr{T}_{n+p}$, say $(V_{n+p})_{p \geq 1}$, such that $V_{n+p} \cap E_n = V_n$ for every index $p \geq 1$. Then the union $V$ of the increasing sequence $(V_{n+p})$ will be a convex set such that $V \cap E_k$ is a neighbourhood of $0$ in $E_k$ (using $\mathscr{T}_k$), for every index $k$; therefore $V$ will be a neighbourhood of $0$ in $E$ for $\mathscr{T}$ and as $V \cap E_n = V_n$, we have proved that $\mathscr{T}'_n$ is finer than $\mathscr{T}_n$.

To define the $V_{n+p}$ we proceed by induction on $p$ using the following lemma: