Equivalence for artinian and noetherian vector spaces I'm trying to prove the next proposition:

For a vector space $V$ over a filed $F$, the next are equivalent:
a) $V$ has a finite dimension
b) $V$ is a finitely generated module
c) $V$ is a Noetherian module
d) $V$ is finitely cogenerated module
e) $V$ is a Artinian module

But I'm stuck when it's about to deal with a finitely cogenerated, whose definition is:

A $R$-Module $M$ is finitely cogenerated iff for every  family of submodules of $M$, $\hspace{2mm} \{N_{i}\}_{i\in I}$ such that $\cap_{i\in I}N_{i}=\{0\}$ exist $ I_{0}\subseteq I$ finite such thath $\cap_{i\in I_{0}}N_{i}=\cap_{i\in I}N_{i}$

I can prove that a) $\implies$ b) $\implies$ c), and c) $\implies$ b) $\implies$ a) $\implies$ e) but I don't see how to conect c) and d).
Any help is appreciated, thank you.
 A: I would think that proving $e)\Rightarrow d)$ would be much easier than connecting $c)$ and $d)$.
Suppose $V$ is not finitely cogenerated. Then there exists a collections $\{N_i\}_{i\in I}$ of submodules of $M$ such that $\cap_i N_i=\{0\}$ but no finite intersection yields $\{0\}$. Suppose $L_0,\ldots,L_{n-1}$ have been defined. Then $\cap_{j=0}^{n-1}L_i\neq\{0\}$, so there exists $L_n\in\{N_i\}_{i\in I}$ such that $\cap_{j=0}^{n-1}L_i\nsubseteq L_n$ and hence $\cap_{j=0}^n L_i\subsetneq\cap_{j=0}^{n-1}L_i$. Then this process defines an infinite descending sequence of submodules $L_0\supsetneq L_0\cap L_1\supsetneq L_0\cap L_1\cap L_2\supsetneq\cdots$ so that $V$ is not Artinian. Hence $e)\Rightarrow d)$ (to be explicit, we used the axiom of dependent choice to guarantee the existence of this sequence).
On the other hand, suppose $V$ is not finite dimensional. Then there is a basis $B$ for $V$ which is not finite. Then $\{\text{span}(B\setminus\{b\})\}_{b\in B}$ is a collection of submodules of $V$ (all with codimension $1$) such that any finite intersection is nontrivial, but whose total intersection is the trivial submodule. Thus, $V$ is not finitely cogenerated, and hence $d)\Rightarrow a)$.
Now, using the directions you have already proven, this shows that all five conditions are equivalent.
