Prove that $\hat{\mathscr{F}} \rightarrow \hat{x}$, where $\hat{x}=\{\mathscr{F} \in \mathscr{C}_E: \mathscr{F} \sim_R i^{-1}(\mathscr{F}')\}$.

Question

Let $\mathscr{C}_E$ be the collection of all Cauchy filters on the TVS $E$ and define the following relation in $\mathscr{C}_E$:

$\mathscr{F} \sim_R \mathscr{G} \Leftrightarrow$ for all neighborhood $U$ of the origin in $E$, there exists $A \in \mathscr{F}$, $B \in \mathscr{G}$ such that $A − B ⊂ U$.

Consider $\mathscr{F}(0)$ be the filter of neighborhoods of $0$ in $E$ and $\hat{E}=\mathscr{C}_E/\sim_R$.

Let $i:E \rightarrow \hat{E}$ given by $i(x)=\{\mathscr{F} \in \mathscr{C}_E:\mathscr{F}\rightarrow x\}=\{\mathscr{F} \in \mathscr{C}_E: \mathscr{F} \sim_R \mathscr{F}(x)\}$.

Take any Cauchy Cauchy filter $\hat{\mathscr{F}} \in \hat{E}$ and consider $$\hat{\mathscr{F}}'=\{\hat{G} \subset \hat{E}: \hat{G} \supset \hat{M}+\hat{U} ~ (\exists \hat{M} \in \hat{\mathscr{F}}) ~ (\exists \hat{U} \in \hat{\mathscr{F}}(0))\}$$ where $$\hat{\mathscr{F}}(0)=\{\hat{W} \subset \hat{E}: \hat{W} \supset \hat{U}\}$$ and $$\hat{U}=\{\hat{x} \in \hat{E}: U \hbox{ belongs to some representative of } \hat{x}\},$$ with $U \in \mathscr{F}(0)$. It's easy to see that $\hat{\mathscr{F}}' \subset \hat{\mathscr{F}}$ and $\hat{\mathscr{F}}'$ is a Cauchy filter on $\hat{E}$.

Consider the family $\mathscr{F}'=\{\hat{W} \subset i(E): \hat{W} \supset \hat{A}\cap i(E) ~ (\exists \hat{A} \in \hat{\mathscr{F}}')\}$ which is also a Cauchy filter. Assume that $i$ is a topological isomorphism between $E$ and $i(E)$. Then, $i^{-1}(\mathscr{F}')$ is a Cauchy filter on $X$.

My question: How to prove that $\hat{\mathscr{F}} \rightarrow \hat{x}$, where $\hat{x}=\{\mathscr{F} \in \mathscr{C}_E: \mathscr{F} \sim_R i^{-1}(\mathscr{F}')\}$.

We aim to prove that $\hat{\mathscr{F}}(\hat{x}) \subset \hat{\mathscr{F}}$. Thus, it is sufficient to prove that $\hat{\mathscr{F}}(\hat{x}) \subset \hat{\mathscr{F}}'$. I think we should have an inclusion of the type $\hat{W}+\hat{M} \subset \hat{U} +\hat{x}$, where $\hat{W}, \hat{U} \in \hat{\mathscr{F}}(0)$, $\hat{M} \in \hat{\mathscr{F}}$ and $U \in \mathscr{F}(0)$, to prove the desired.

Let $\mathscr{F} \in \hat{x}$, since $\mathscr{F} \sim_R i^{-1}(\mathscr{F}')$, for all open neighborhood $U$ of $0$ there exists $A \in \mathscr{F}$ and $B \in i^{-1}(\mathscr{F}')$ such that $B-A \subset U$. Thus, $B-A \subset U$ which implies that $$i(B)-i(A) \subset \hat{U}$$ since $i(U^\circ) \subset \hat{U} \cap i(E) \subset i(\overline{U})$. Let $x \in A$, then $i(B) \subset i(x)+\hat{U}\cap i(E)$. From the definition of $i^{-1}(\mathscr{F}')$ we obtain $B \supset i^{-1}((\hat{M}+\hat{V})\cap i(E))$ for some $\hat{M} \in \hat{\mathscr{F}}$ and $\hat{V} \in \hat{\mathscr{F}}(0).$ But I don't know how to proceed.

Answer 1

It seems the situation is overcomplicated, in particular, by a notation, and a possible proof should be like a proof of the following

Lemma. Let $E$ be a dense linear subspace of a topological vector space $\hat E$ such that each Cauchy filter on $E$ converges in $\hat E$. Then $\hat E$ is complete.

Proof. Let $\hat{\mathscr{F}}$ be a Cauchy filter on $\hat E$ and $\hat{\mathscr{F}}(0)$ be the filter of neighborhoods of the zero of $\hat E$. It is easy to show that a filter $\hat{\mathscr{F}}’=\hat{\mathscr{F}}+\hat{\mathscr{F}}(0)=\{F+U:F\in \hat{\mathscr{F}}, U\in \hat{\mathscr{F}}(0) \}$ is a Cauchy filter contained in $\hat{\mathscr{F}}$. Since $\hat{\mathscr{F}}’$ has an open base, its restriction $\mathscr{F}$ to $E$ is a (Cauchy) filter on $E$. Therefore $\mathscr{F}$ converges to some point $\hat x\in\hat E$. Since $\hat{\mathscr{F}}’$ is a Cauchy filter with a base $\mathscr{F}$, the filter $\hat{\mathscr{F}}’$ converges to $\hat x$ and so does its supfilter $\hat{\mathscr{F}}$. $\square$

Answer 2

Let $\mathscr{F}'$ be a Cauchy filter in $i(E)$. Since $i(E)$ is dense in $\hat{E}$, it is sufficient to prove that $\mathscr{F}'$ converges in $\hat{E}$. It's is easy to see that $i^{-1}(\mathscr{F})$ is a Cauchy filter in $E$. Consider $\hat{x}=\{\mathscr{F} \in \mathscr{C}_E:\mathscr{F} \sim_R i^{-1}(\mathscr{F}')\}$. Let's show that $\mathscr{F}' \rightarrow \hat{x}$. Let $U \in \mathscr{F}(0)$ and consider the neighborhood $\hat{U}+\hat{x}$ of $\hat{x}$. Since $i^{-1}(\mathscr{F}')$ is a Cauchy filter, there exists $A \in i^{-1}(\mathscr{F}')$ such that $A-A\subset U$. Let $A_0=i(A)$, we claim that $A_0 \subset \hat{x} +\hat{U}$. Indeed, let $\tilde{x} \in A_0$ and $x \in E$ such that $i(x)=\tilde{x}$. Then, $\{x\}-A \subset U$. Since $\{x\} \in \mathscr{G}=\{V \subset E: V \ni x\}$ and $\mathscr{F}(x) \subset \mathscr{G}$, we obtain that $\mathscr{G} \in i(x)$. Therefore, $\{x\}-A \in \mathscr{G}-i^{-1}(\mathscr{F}')$, i.e., $\{x\}-A$ belongs to a representative of $i(x)-\hat{x}$. Thus, $U$ belongs to a representative of $i(x)-\hat{x}$, or yet $\tilde{x}-\hat{x} =i(x)-\hat{x} \in \hat{U}$. Therefore, $\tilde{x} \in \hat{x}+\hat{U}$ which implies that $A_0 \subset \hat{x}+\hat{U}$. Since $A_0 \in \mathscr{F}'$ and $A_0 \subset \hat{U}+\hat{x}$ for all $u \in \mathscr{F}(0)$ we obtain that $\mathscr{F}' \rightarrow \hat{x}$ in $\hat{E}$. From the above Lemma the result follows.