Let $R$ be a ring and $M$ be a right $R$-module. $M$ is called $\textbf{isoartinian}$ if, for every descending chain $M \geq M_1 \geq M_2 \cdots$ of submodules of $M$, there can be found an index $n$ $\geq1$ such that $M_n \cong M_i$ for every $i$ $\geq$ $n$.
$M$ is an isoartinian module if and only if for each non-empty set $F$ of submodules of $M$, there is a submodule $N$ of $M$ in $F$ such that, for each $N'$ $\leq$ $N$, if $N'$ is in $F$, then $N$ $\cong$ $N'$. Identically, $M$ is an isoartinian module if and only if for each non-empty chain $C$ of submodules of $M$, there is a submodule $N$ of $M$ in $C$ such that, for each $N'$ $\leq$ $N$, if $N'$ is in $C$, then $N$ $\cong$ $N'$.
I've tried to prove this statement, but I'm not sure it's true or not. I just need some check about my proof.
($\Rightarrow$) Let $F$ be a non-empty set of submodules of $M$. Assume by the contrary that there does not exist a minimal element. Let $N$ be an element of $F$. Since $N$ is not minimal, there exists $N'$ such that $N$ > $N'$. Since $N'$ is not minimal, it can be found $N''$ such that $N$ > $N'$ > $N''$. Continuing in this way, we get a sequence in $F$ which does not stop up to isomorphism.
($\Leftarrow$) Conversely, suppose for each non-empty set $F$ of submodules of $M$, it can be found an element $N$ in $F$ such that, for every $N'$ $\leq$ $N$ (N $\leq$ $N'$, respectively), if $N'$ is an element of $F$, then $N \cong N'$. Let $F$ be a family of submodules of $M$. Assume that $N$ > $N'$ > $N''$ > $\cdots$ is a decreasing chain in $F$. Then either that chain will stop at $N$ or a submodule containing $N$. Therefore, $M$ is an isoartinian module.