Let $F$ be a t.v.s., $T$ a set, and $\mathfrak S\subset 2^T$ a directed family. Then why is the $\mathfrak S$-topology on $F^T$ well-defined? Let $F$ be a topological vector space over some nontrivially-valued field $K$. Let $T$ be a set, and let $\mathfrak S$ be a family of subsets of $T$ which is directed under the inclusion relation $\subset$. Then Schaefer claims in the introduction to chapter 3 section 3 of his book Topological Vector Spaces that if we let $\mathfrak V$ be a neighborhood base of the origin in $F$, then the collection of sets $$M(S,V)=\{f\ :\ f(S)\subset V\}$$ defines a neighborhood base of a unique translation-invariant topology on $F^T$ as $S\in\mathfrak S$ and $V\in\mathfrak V$ vary.
However, while I see why we get an induced topology on $F^T$, I fail to see how this collection is a neighborhood base of the origin. My issue is that if we have $f_0\in M(S,V)$ for some $S,V$, then we should have that $$f_0+M(S',V')\subset M(S,V)$$ for some $S',V'$, which is equivalent to requiring that $$f(s)-f_0(s)\in V'\ \forall\ s\in S'\implies f(s)\in V\ \forall\ s\in S$$ for all $f:T\to F$.
However, I seem to have found a counterexample to this. Let $F=K=\Bbb R$, $T=\Bbb (-1,1)\subset\Bbb R$, and $\mathfrak S=\{T\}$. If we let $f_0:(-1,1)\to\Bbb R$ be the identity and $V=(-1,1)\subset\Bbb R$ so that $f_0\in M(T,V)$, then clearly $$f_0+M(T,V')\not\subset M(T,V)$$ for any neighborhood $V'$ of the origin in $\Bbb R$.
Am I misunderstanding something? Why does it make sense to talk about the induced translation-invariant topology in this case?
 A: The family $\mathfrak{M} = \{ M(S,V) : S \in \mathfrak{S}, V \in \mathfrak{V}\}$ need not consist of open sets. Some authors define a neighbourhood of a point $x$ as an open set containing $x$, but the other convention - more widespread according to my experience - is that a neighbourhood of $x$ is a set that contains an open set containing $x$, so every superset (open or not) of a neighbourhood is again a neighbourhood.
As your example shows, the sets $M(S,V)$ are in general not open, even if $V$ is open. And defining an open neighbourhood base in the $\mathfrak{S}$-topology is not entirely straightforward, one would have to demand that $V$ is a uniform neighbourhood of $f(S)$ rather than just $f(S) \subset V$.
What we need to get a (translation-invariant) topology from $\mathfrak{M}$ is the compatibility condition between neighbourhood filters. Generally, that condition is
$$\bigl(\forall U \in \mathscr{V}(x)\bigr)\bigl(\exists V \in \mathscr{V}(x)\bigr)\bigl(\forall y \in V\bigr)\bigl(U \in \mathscr{V}(y)\bigr).$$
Here, for a translation-invariant topology induced by $\mathfrak{M}$, that is equivalent to the condition that for all $M(S,V)$ there is an $M(S',V')$ such that for all $f \in M(S',V')$ there is an $M(S'',V'')$ such that $f + M(S'',V'') \subset M(S,V)$. Since $M(S,U) + M(S,W) \subset M(S, U+W)$, we can make a simple choice: If $V'$ is a neighbourhood of $0$ in $F$ with $V' + V' \subset V$, then
$$M(S,V') + M(S,V') \subset M(S, V' + V') \subset M(S,V).\tag{$\ast$}$$
Using $(\ast)$ and $-M(S,V) = M(S, -V)$, we can also see that the $\mathfrak{S}$-topology makes $F^T$ a topological group (Hausdorff iff $F$ is Hausdorff and $\bigcup \mathfrak{S} = T$). However, in general it doesn't make $F^T$ a topological vector space. Although for every scalar $\lambda$ the map $f \mapsto \lambda f$ is continuous, the map $\lambda \mapsto \lambda f$ is not continuous at $0$ if there is an $S\in \mathfrak{S}$ such that $f(S)$ is unbounded. The subspace
$$\mathscr{B}_{\mathfrak{S}} = \bigl\{ f \in F^T : \bigl(\forall S\in \mathfrak{S}\bigr)\bigl(f(S)\text{ is bounded}\bigr)\bigr\}$$
is a topological vector space in the subspace topology, and a subspace $E\subset F^T$ is a topological vector space in the $\mathfrak{S}$-topology if and only if $E \subset \mathscr{B}_{\mathfrak{S}}$.
