The paperI^K Cauchy Functions by Sleziak, Toma, Das
Trying to prove Lemma $3.2$. Faced a little problem at $iv)\implies i).$
Lemma $3.2$. Let $(X, Φ)$ be a uniform space and let $I$ be an ideal on a set $S$. For a function $f : S → X$ following are equivalent.
$(i)$ $f\ $is $I$-Cauchy.
$(ii)$ For any $U ∈ Φ$ there is $m ∈ S$ such that $\{n ∈ S; (f(n), f(m)) ∈ U\} ∈ F(I)$.
$(iii)$ For every $U ∈ Φ$ there exists a set $A ∈ I$ such that $m, n \not∈ A$ implies $(f(m), f(n)) ∈ U.$
$(iv)$ The filter $f[F(I)]$ is a Cauchy filter.
Now $i)\implies ii)$ follows trivially from definitions.Need help proving $ii)\implies iii)\implies iv)\implies i).$
My Attempts:
$ii)\implies iii)$
By $ii),$ For any $U ∈ Φ$ there is $m ∈ S$ such that $$\{n ∈ S; (f(n), f(m)) ∈ U\} ∈ F(I)\\\implies \{n\in S: (f(n),f(m))\notin U\}\in I$$.
Let $$A=\{n\in S: (f(n),f(m))\notin U\}\in I$$. Easily, $m,n\notin A\implies (f(m),f(n))\in U.$
$iii)\implies iv)$
For first we need to show that $f\left[\mathcal F(I)\right]$ is a filter.
$1.$ Since $\phi$ is not in $F(I)$ we cannot map a nonempty set to $\phi$ so $\phi\notin f[f(I)].$
$2.$ $f(A)\in f[F(I)]$ s.t $f(A)\subset f(B)\in f(S).$To show $f(B)\in f[F(I)].$Now from set theory it is known that $f(A)\subset f(B)\implies A\subseteq B.$ Which implies $B\in F(I)\implies f(B)\in f[F(I)].$
$3.$ Let $f(A),f(B)\in f[F(I)]$To show $f(A)\bigcup f(B)\in f[F(I)].$ Now, $A,B\in F(I)\implies A\bigcup B\in F(I)\implies f(A\cup B)\in f[F(I)].$ Now, $f(A\cup B)\subseteq f(A)\cup f(B).$ From $2$ this implies $f(A)\cup f(B)\in f[F(I)].$
Thus $f(F(I))$ is a filter.To show it is Cauchy filter.Now whatever entourage $U$ we take there exists $A\in I$ s.t $m,n\notin A\implies (f(m),f(n))\in U\implies \exists A^c\in F(I) s.t. m,n\in A^c\implies (f(m),f(n))\in U.$Which implies it's a Cauchy Filter.
( Definition of Cauchy Filter used: Given a uniform space $X$, a filter $F$ on $X$ is called Cauchy filter if for every entourage $U$ there is an $A ∈ F$ with $(x,y) ∈ U$ for all $x,y ∈ A$.)
$iv)\implies i)$
Chose any $U\in \Phi,$ there is a $A\in F(I)$ s.t $m,n\in A\implies (f(m),f(n))\in U$ i.e $\{m,n\in S:(f(m),f(n))\in U\}\in F(I)\implies \{m,n\in S:(f(m),f(n))\notin U\}\in I$ Now what I need to find is a fixed $m_0\in S$ such that $\{m,n\in S:(f(m),f(n))\in U\}\in I$. Which is that $m_0\ ?$
Thanks.