Proposition 8

Let E be a Montel space and $\mathfrak{T}$ a filter on E, which converges to a point $x_0$ in E for the weakened topology. If $\mathfrak{T}$ is a countable base, or contains a bounded set, then $\mathfrak{T}$, converges to $x_0$ for the initial topology also.

Assume first that there exists a bounded set $\overline{B}$ in $\mathfrak{T}$. The closure $\overline{B}$ of B for the initial topology of E is bounded; in addition, $\overline{B}$ is compact because E is a Montel space. The topology on $\overline{B}$ induced by $\sigma(E, E')$ is Hausdorff and coarser than the topology induced by the initial topology; they therefore coincide (GT, I, § 9, No. 4). This prove the proposition for this case.

Next assume that $\mathfrak{T}$ has a countable base. It is enough (GT, I, § 6, No. 8, prop. 11) to consider the case of a sequence $(x_n)_{n \geq 1}$ tending to $x_0$ for $\sigma(E, E')$. Let B be the set of all $x_n$ for $n \geq 0$. This set is bounded for $\sigma(E, E')$, hence also for the initial topology (III, p. 27, cor. 3). Thus we have reduced to the first case of the proof.

Every Montel space is reflexive: this follows from def. 4 and from th. 2 of IV, p. 16. Further: