Two filters which converge to the same point are equivalent modulo $R$

Question

Define the following relation on the collection of all Cauchy filters on X:

$F ∼_R G$ ⇔ $∀U$ nbhood of the origin in $X$, $∃A ∈ F$, $∃B ∈ G$ s.t. $A − B ⊂ U$.

show that :-

Two filters on $X$ which converge to the same point are equivalent modulo $R$.

An Attempt:-

Let $\mathcal F$ and $\mathcal G$ be two filter such that $\mathcal F \to x$ and $\mathcal G \to x$

$\implies$ $F(x)⊂ \mathcal F$ where $F(x)$ nhood filter of $x$ and

$\implies$ $F(x)⊂ \mathcal G$ where $F(x)$ nhood filter of $x$

$\implies$ $F(x)⊂\mathcal F \cap \mathcal G$ where $F(x)$ nhood filter of $x$

let $A_1 \in F(x)⊂ \mathcal F$ and $A_1 \in F(x)⊂ \mathcal G$

Now $∀U$ nbhood of the origin in $X$, $∃A_1 ∈ F$, $∃A_1 ∈ G$ s.t. $A_1 − A_1 ⊂ U$.

I think this is incorrect, and I appreciate any hints.

Answer 1

Suppose that $\mathcal{F} \to x, \mathcal{G} \to x$, and let $U$ be a neighbourhood of the origin. Find a neighbourhood $V$ of the origin such that $V - V \subseteq U$, which can be done in a TVS.

Then $x + V$ is a neighbourhood of $x$ (in any TVS) and so by convergence(s) $x+V \in \mathcal{F}, x+V \in \mathcal{G}$. And as $(x+V) - (x+V) = V - V \subseteq U$, we can take $A = x+V, B = x+V$ in the definition of $\sim_R$ for our $U$. As $U$ was arbitary, $\mathcal{F} \sim_R \mathcal{G}$.