$\mathfrak S$-topology and the $\overline{\mathfrak S}$-topology are identical on $\mathcal L(E, F)$

Question

Recently I am reading the book Topological Vector Spaces by Schaefer and I came across this problem:

Let $E$ be a set and $F$ a topological vector space. Let $\mathfrak S$ be a directed(by inclusion) family of subsets of $E$ and $\mathcal B$ a neighborhood base of zero in $F$. For $S\in \mathfrak S$ and $V\in \mathcal B$ define $$M(S,V)=\{f\in F^E: f(S)\subset V\}$$

$M(S,V)$ is a $0$-neighborhood base in $F^E$ for a unique translation invariant topology called the $\mathfrak S$-topology.

It is clear that we could replace $\mathfrak S$ by any cofinal subfamily and $\mathfrak S$-topology does not depend on the particular choice of neighborhood base of $0$ in $F$.

Theorem 3.1 chapter 3: A vector space $G$ in $F^E$ is a topological vector space under an $\mathfrak S$-topology iff for all $f\in G$ and $S\in \mathfrak S$, $f(S)$ is bounded in $F$.

A family $\mathfrak S\neq \{\emptyset\}$ of bounded subsets of a locally convex space E is called saturated if

1. it contains arbitrary subsets of each of its members,
2. it contains all scalar multiples of each of its members, and
3. it contains the closed, convex, circled hull.of the union of each finite subfamily

Since the family of all bounded subsets of $E$ is saturated and since the intersection of any non-empty collection of saturated families is saturated, a given family $\mathfrak S$ of bounded sets in E determines a smallest saturated family $\overline{\mathfrak S}$ containing it; $\overline{\mathfrak S}$ is called the saturated hull of $\mathfrak S$.

My Question:

In the following the author claims:

$E$ and $F$ being locally convex, it is clear that for each family $\mathfrak S$ of bounded subsets of $E$, the $\mathfrak S$-topology and the $\overline{\mathfrak S}$-topology are identical on $\mathcal L(E, F)$(the space of all continuous linear operators from $E$ into $F$).

it is clear that $\overline{\mathfrak S}$-topology is finer than $\mathfrak S$-topology but the converse is not clear for me.

My Attempt:

I think it is enough to show that $\mathfrak S$ is cofinal in $\overline{\mathfrak S}$, i.e. for each $S\in \overline{\mathfrak S}$ there exists $S'\in \mathfrak S$ such that $S\subset S'$. However I am not sure if this is true.

Answer 1

Let $\mathcal B$ be a closed convex circled neighborhood base of $0$ since $\mathcal L(E,F)$ is the space of continuous linear operators then it is easy t show that $M(S, V)=M(S',V)$ if $S'$ is convex hull, circled hull or closure of $S$.

Let $S\in \overline{\mathfrak S}$ and $V\in \mathcal B$ then $S$ is a subset of a multiple of closed absolutely convex hulls of finite subfamilies $S_1,\cdots S_n$ of $\mathfrak S$ say $\alpha\overline{cco}(S_1\cup\cdots \cup S_n)$($\alpha\neq 0$).

Since $\mathfrak S$ is directed there is some $S'\in \mathfrak S$ such that $S_1\cup\cdots \cup S_n\subset S'$ and So $S\subset \alpha \overline{cco}(S')$ this yields $M(\alpha \overline{cco}(S'), V)\subset M(S, V)$.

since $M(\alpha S, V)=M(S, \frac 1\alpha V)$ and $M(\overline {cco}(S), V)=M(S, V)$we could take $V'\subset \frac 1\alpha V$ then $M(S', V')\subset M(S, V)$.