Lemma 2

Let $V$ be a convex neighbourhood of 0 in $M$, a vector subspace of a locally convex space $F$. Then there exists a convex neighbourhood $W$ of 0 in $F$ such that $W \cap M = V$. Further, if $M$ is closed in $F$, then, for every point $x_0 \in \complement M$, there exists a convex neighbourhood $W_0$ of 0 in $F$ such that $W_0 \cap M = V$ and $x_0 \notin W_0$.

In fact, by hypothesis there exists a convex neighbourhood $U$ of 0 in $F$ such that $U \cap M \subset V$. Clearly, the convex envelope $W$ of $U \cup V$ in $F$ is a neighbourhood of 0 in $F$; we show that $W \cap M = V$. For, every point $z \in W$ is of the form $\lambda x + (1-\lambda) y$ with $x \in V, y \in U$, and $0 \leq \lambda \leq 1$ (II, p. 9, prop. 8); if $z \in M$, and $\lambda \neq 1$ then necessarily $y \in M$, therefore $y \in U \cap M \subset V$ and hence $z \in V$; the result is obviously true if $\lambda = 1$. If $M$ is closed in $F$, the space $F/M$ is Hausdorff, thus there exists a convex neighbourhood $U_0 \subset U$ of 0 in $F$ such that $U_0$ does not meet $x_0 + M$; then the convex envelope $W_0$ of $U_0 \cup V$ fulfils the required conditions.

Returning to the theorem, to prove the second part of (i) note that if $x \in E$ then $x \in E_n$ for some $n$; if $x \neq 0$ and $\mathscr{T}_n$ is Hausdorff then there is a neighbourhood $V_n$ of 0 for $\mathscr{T}_n$, which does not contain $x$. We see that there is a neighbourhood $V$ of 0 for $\mathscr{T}$ such that $V \cap E_n = V_n$, hence $x \notin V$, and it follows that $\mathscr{T}$ is Hausdorff.

(ii) Let $x \in E - E_n$; there exists $m > n$ such that $x \in E_m$, thus, as $E_n$ is closed in $E_m$ for $\mathscr{T}_m$ (because of the hypothesis that $\mathscr{T}_{n+1}$ induces $\mathscr{T}_n$ on $E_n$ for every $n$) there exists in the topology $\mathscr{T}_m$ a convex neighbourhood $V_m$ of 0 in $E_m$ such that $(x + V_m) \cap E_n$ is empty. Now we saw in (i) that there exists a convex neighbourhood $V$ of 0 for $\mathscr{T}$ such that $V \cap E_m = V_m$; and thus $(x + V) \cap E_m = x + V_m$, therefore $(x + V) \cap E_n = \varnothing$, which proves (ii).

(iii) From lemma 1, if $\mathfrak{F}$ is a minimal Cauchy filter for $\mathscr{T}$ (GT, II, § 3.2) then there exists a $k$ such that the trace of $\mathfrak{F}$ on $E_k$ is a filter $\mathfrak{F}_k$; from (i) this last is a Cauchy filter for $\mathscr{T}_k$ and thus $\mathfrak{F}_k$ converges in $E_k$ by hypothesis; but as the filter on $E$ generated by $\mathfrak{F}_k$ is finer than $\mathfrak{F}$, we see that $\mathfrak{F}$ has a cluster point for $\mathscr{T}$ and thus converges for $\mathscr{T}$.

When for all $n$ the topology induced on $E_n$ by $\mathscr{T}_{n+1}$ is just $\mathscr{T}_n$ we say that $\mathscr{T}$ is the strict inductive limit of the sequence $(\mathscr{T}_n)$ and that the space $E$ with the topology $\mathscr{T}$ is the strict inductive limit of the sequence of locally convex spaces $E_n$.