Equivalent filters converge to the same point

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 :-

If a filter $F$ is equivalent modulo $R$ to a filter which converges to $x$, then also $F$ converges to $x$.

An Attempt:-

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

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

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

update:-

let $\mathcal{F} \to t$ , so $\mathcal{F}- \mathcal{G} \to t-x $

but $F(0)⊂\mathcal{F}- \mathcal{G} \to t-x$, because $F ∼_R G$

$\implies$ $ t-x=0$

$\implies$ $x=t$

Answer 1

Let $U$ and $V$ be two neighborhoods of $0$ such that $V+V \subset U$. We want to show that $x+U \in \mathcal{F}$ to show the convergence. Since $\mathcal{F} \sim_R \mathcal{G}$ there exists $A \in \mathcal{F}, B \in \mathcal{G}$ such that $A - B \subseteq V$. We also know that $x+V \in \mathcal{G}$ as $\mathcal{F}(x) \subset \mathcal{G}$. Thus, $A-(B\cap(V+x))\subset A-B\subset V$. Therefore, $A \subset B\cap(V+x)+V \subset V+V+x\subset U+x$ which implies that $U+x \in \mathcal{F}$ since $\mathcal{F}$ is a filter and $A \in \mathcal{F}$.

Answer 2

Possible ideas: we want to show that $\mathcal{F} \to x$ as well. So take a neighbourhood $U$ of $0$ then $x + U$ is a typical neighbourhood of $x$, so you want to show that $x+U \in \mathcal{F}$ to show convergence.

Given $U$ we have $A \in \mathcal{F}, B \in \mathcal{G}$ such that $A - B \subseteq U$ by $\mathcal{F} \sim_R \mathcal{G}$. We also know that $x+U \in \mathcal{G}$ as $\mathcal{G} \to x$.

Another fact: $A \subseteq B + U$, so $B+U \in \mathcal{F}$, which is already close to what you need....

Answer 3

It is often good to start with what we want and see how the story unfolds naturally. Then at the end if we fall short of our goal, we can readjust. I will show this chain of reasoning instead of a polished proof which starts cleverly. We assume that $F\sim G$ and $G\to x$, and want to show $F\to x$. This requires that any neighborhood of $x$ is in $F$. Judging by your definitions, you are working in a normed space, so it is convenient to work with metric balls.

So, take any ball neighborhood $U=B(x,r)$ of $x$. Now $U-x$ is a neighborhood of the origin, so there exist $A\in F$ and $C\in G$ so that $A-C\subset U-x$. We want to argue that if $U$ is small, then $A$ and $C$ are almost the same.

To this end, want $A$ and $C$ to be small. Use the Cauchy property for any fixed $\epsilon>0$: There is $A'\in F$ and $C'\in G$ whose diameters are less than $\epsilon$. Now declare $A''=A\cap A'$ and $C''=C\cap C'$. By the properties of the filter $A''\in F$ and $B''\in G$. Also, we have $A''-C''\subset A-C\subset U-x$.

It follows that $A''\subset C''+U-x$ and therefore $C''+U-x\in F$. Now $G\to x$, so $x\in\overline{C''}$. We have proven that some set resembling $U$ is in $F$, not that $U$ itself is in $F$. We can make this more precise with balls: Since $x\in\overline{C''}$ and the diameter of $C''$ is less than $\epsilon$, we have $C''\subset B(x,\epsilon)$. Combining this with $U=B(x,r)$ gives $C''+U-x\subset B(x,r+\epsilon)$ and thus $B(x,r+\epsilon)\in F$.

We fall short of our goal by $\epsilon$, so we readjust. We want to show that for any $R>0$ the ball $B(x,R)$ is in the filter $F$. Now apply the above reasoning with $r=R/2$ and $\epsilon=R/2$. We get $B(x,R)\in F$ as desired.

If you want to polish the proof, start with a ball $B(x,R)$ with arbitrary radius $R>0$ and replace $r$ and $\epsilon$ in the proof with $r/2$. You could even start with an arbitrary neighborhood and then explain why it is enough to show the result for a ball. Also, it turned out that there was no need for $A''$ after all, so we could have just played with $A$. It was only necessary to shrink $C$ to $C''$.