Lemma 1

— Let $\mathfrak{F}$ be a Cauchy filter on $E$ (for $\mathscr{T}$); then there exists an integer $k$, such that for all $N \in \mathfrak{F}$ and every neighbourhood $V$ of $0$ in $E$, the subspace $E_k$ meets $N + V$.

We assume the contrary and obtain a contradiction. Suppose that for every $k$ there exists a convex neighbourhood $V_k$ of $0$ and a set $M_k \in \mathfrak{F}$ such that

Clearly we can suppose that $V_{k+1} \subset V_k$ for all $k$. Let $V$ be the convex envelope of $\bigcup_k (E_k \cap V_k)$, this is clearly a neighbourhood of $0$ for $\mathscr{T}$. For all $n$ we have $V \subset V_n + E_n$; in fact, every $x \in V$ can be written $\sum_i \lambda_i x_i$ where $\lambda_i \geq 0$, $\sum_i \lambda_i = 1$ and $x_i \in V_i \cap E_i$ for all $i$; now for $i < n$ we have $x_i \in E_n$, therefore $\sum_{i < n} \lambda_i x_i \in E_n$; and for $i \geq n$ we have $x_i \in V_n$, therefore $\sum_{i \geq n} \lambda_i x_i \in V_n$ since $V_n$ is convex, contains $0$ and $\sum_{i \geq n} \lambda_i \leq 1$. Hence $V + E_n \subset V_n + E_n$ for all $n$. This being so, let $M \in \mathfrak{F}$ be a set that is $V$-small. For some integer $m$, $E_m \cap M$ is not empty; and we conclude that as $\mathfrak{F}$ is a filter, the set $M_m$ meets $M$ and therefore $E_m + V_m$; we have a contradiction which establishes the lemma.