Problem Statement: Let $f : X \to Y$ be an open surjective mapping. Let $x \in X$ have a filter $\mathcal{F}$ such that $\mathcal{F} \to f(x)$ in $Y$. Show that there exists a filter $\mathcal{U}$ in $x$ where $f(\mathcal{U}) \vdash \mathcal{F}$ and $\mathcal{F} \vdash f(\mathcal{U})$. $ \newcommand{\nc}{\newcommand} \nc{\FF}{\mathcal{F}} \nc{\UU}{\mathcal{U}} \nc{\PP}{\mathcal{P}} \nc{\set}[1]{\left\{#1\right\}} \nc{\a}{\alpha} \nc{\b}{\beta} \nc{\g}{\gamma} \nc{\AA}{\mathcal{A}} \nc{\BB}{\mathcal{B}} \nc{\GG}{\mathcal{G}} $
Relevant Definitions:
Filter: A filter on a topological space $X$ is defined to be some subcollection $\FF := \set{A_\a}_{\a \in \AA}$ of $\PP(X)$ where each $A_\a \ne \varnothing$ and, for each $A_\a,A_\b$, there is an $A_\g \subseteq A_\a \cap A_\b$.
Filter Convergence: A filter $\FF$ is said to converge to a point $x$, denoted $\FF \to x$, if the following holds: for any open neighborhood of $x$, denoted $U_x$, $\exists \a \in \AA$ such that $A_\a \subseteq U_x$.
Filter Subordination: Let $\FF := \set{A_\a}_{\a \in \AA}$ and $\GG := \set{B_\b}_{\b \in \BB}$ be filters on the same space. We say that $\FF$ is subordinate to $\GG$, denoted $\FF \vdash \GG$, if $\forall A_\a$, $\exists B_\b$, such that $A_\a \subseteq B_\b$.
Neighborhood Filter: The neighborhood filter $\UU_x$ for a point $x \in X$ is the set of all open neighborhoods containing $x$.
Image of a Filter: For $\FF := \set{A_\a}_{\a \in \AA}$ our filter on $X$, and $f : X \to Y$, we define $f(\FF) := \set{f(A_\a)}_{\a \in \AA}$.
My Work So Far: So, my first feeling was to use the neighborhood filter to prove this claim, since there was no clear other constructible filter to use.
So, letting $x \in X$ and $\FF := \set{F_\a}_{\a \in \AA} \to f(x)$ be our filter, my goal is to prove $f(\UU_x) \vdash \FF$ and $\FF \vdash f(\UU_x)$, for $\UU_x$ the neighborhood filter of $x$.
The latter is a straightforward proof by contradiction, contradicting the convergence $\FF \to f(x)$ (since $f(U_x)$ is an open neighborhood of $f(x)$ by $f$'s openness).
However, I'm not sure what neighborhood of $x$ I should use for the claim $f(\UU_x) \vdash \FF$. My instict was to use $f^{-1}(\mathrm{int}(F_\a))$, since $f(f^{-1}(\mathrm{int}(F_\a))) = \mathrm{int}(F_\a) \subseteq F_\a$ by surjectivity. However, I'm not sure if $x \in f^{-1}(\mathrm{int}(F_\a))$ necessarily, nor if it is even necessarily open.
So I'm starting to feel that this is not the proper choice of filter. This begs the question then, what filter to use? Or can this approach be salvaged? Any insights are appreciated!
$\mathcal U_x$ does not always work.
Let $X$ be $\Bbb R$ in the indiscrete topology and $Y=\Bbb R$ in the discrete topology, and $f(x)=x$ for all $x$. This is clearly an open surjection. If we take $\mathcal F$ to be the fixed filter on $0=f(0)$, say (so all subsets that contain $0$) then $\mathcal F \to f(0)$ but $\mathcal{U}_0$ (on $X$) is just $\{\Bbb R\}$ and $ f(\mathcal U_0) \vdash \mathcal F$ is false. So look at such simple cases to maybe get some ideas..