Every topological space gives rise to a pseudotopological space. Conversely, if $X$ is a pseudotopological space then we can define a topology on $X$ such that every filter converging to $x$ in the original pseudotopology also converges in this topology.
Let $(X,T)$ be a topological space. Let $F$ a subspace of $(X,T)$ considered as a pseudotopological space. My question is whether $F$ necessarily is a subspace of the topological space $(X,T)$ or not.
We need to fix some notation. Let $(X, \to)$ be a pseudotopological space. Let $\mathrm{Fil}(X)$ be the set of all filters on $X$. So $\to$ is a subset of $\mathrm{Fil}(X)\times X$. Let $F\subseteq X$. Then we define a relation $\to|_F$ on $\mathrm{Fil}(F)\times F$ by
$\mathcal{H} \to|_F x :\Leftrightarrow (\exists \mathcal{G} \in \mathrm{Fil}(X)): \mathcal{G}\to x \textrm{ and } G\cap F \in\mathcal{H} \textrm{ for all }G\in\mathcal{G}.$
We can verify easily that $(F,\to|_F)$ satisfies the axioms of a pseudotopological space and call it a subspace of $(X,\to)$.
To any pseudotopological space $(X,\to)$ we associate a topology
$\tau_\to = \{U\subseteq X: \textrm{ if } x\in U \textrm{ and } \mathcal{F} \in \mathrm{Fil}(X) \textrm{ such that } \mathcal{F}\to x \textrm{ then } U\in\mathcal{F}\} \cup \{\emptyset\}$.
The last bit of notation we need is this. Let $(X,\tau)$ be a topological space and $F\subseteq X$. Then $\tau|_F = \{U\cap F: U \in \tau\}$.
The following statement is straightforward to prove and gives a positive answer to the question:
Observation: Let $(X,\tau)$ be a topological space and let $F\subseteq X$. Then $\tau|_F = \tau_{\to|_F}$.
